Exponential and Logarithmic Functions

In This Chapter

1.6 Exponential Functions

2.6 Logarithmic Functions

3.6 Exponential and Logarithmic Equations

4.6 Exponential and Logarithmic Models

5.6 Hyperbolic Functions

Chapter 6 Review Exercises

The curve that describes the shape of a telephone wire hanging between two poles is the graph of a hyperbolic function.

A Bit of History In this chapter we will consider two types of functions that appear often in applications: exponential and logarithmic functions.

The wealthy English political and religious polemicist John Napier (1550-1617) is remembered today principally for one of his mathematical diversions, the logarithm. We will see in Section 6.2 that logarithms are essentially exponents and that there are two important types. The invention of the natural logarithm is often attributed to Napier. His friend and collaborator, the English mathematician Henry Briggs (1561-1631), devised the base ten or common logarithm. The word logarithm comes from two Greek words: logos, which means reasoning or reckoning, and arithmos, which means number. Thus a logarithm is a “reckoning number.” For several hundred years logarithms were used primarily as an aid in performing complex and tedious arithmetic computations. A calculator prior to 1967 was an analog instrument called the slide rule. The operations performed on a slide rule were based entirely on the properties of logarithms. With the invention of the electronic handheld calculator in 1966 and the evolution of the personal computer, the slide rule and the use of logarithms as a pencil-and-paper computing method have gone the way of the dinosaurs.

6.1 Exponential Functions

Introduction In the preceding chapters we considered functions such as f (x) = x², that is, a function with a variable base x and constant power or exponent 2. We now examine functions having a constant base b and a variable exponent x.

The domain of an exponential function f defined in (1) is the set of all real numbers (—∞, ∞).

In (1) the base b is restricted to positive numbers in order to guarantee that b^x is always a real number. For example, with this restriction we avoid complex numbers such as (—4)^1/2. Also, the base b = l is of little interest to us since it can be shown that f is the constant function f (x) = 1^x = 1. Moreover, for b > 0, we have f (0) = b⁰ = 1.

Exponents As just mentioned, the domain of an exponential function (1) is the set of all real numbers. This means that the exponent x can be either a rational or an irrational number. For example, if the base b = 3 and the exponent x is a rational number, say, x = 1/5 and x = 1.4, then

For an exponent x that is an irrational number, b^x is defined, but its precise definition is beyond the scope of this text. We can, however, suggest a procedure for defining a number such as . From the decimal representation = 1.414213562…we see that the rational numbers

are successively better approximations to . By using these rational numbers as exponents, we would expect that the numbers

are then successively better approximations to . In fact, this can be shown to be true with a precise definition of b^x for an irrational value of x. But on a practical level, we can use the key on a calculator to obtain the approximation 4.72804388 to .

The precise definition of b^x for x an irrational number requires the concept of a limit. The notion of a limit of a function is the backbone of differential and integral calculus.

Laws of Exponents As we saw in Chapter 002, the laws of exponents are stated first for integer exponents and then for rational exponents. Since b^x can be defined for all real numbers x when b > 0, it can be proved that these same laws of exponents hold for all real number exponents. If a > 0, b > 0 and x, x₁, and x₂ denote real numbers, then

EXAMPLE 1 Rewriting a Function

At times, we will use the laws of exponents to rewrite a function in a different form. For example, neither f(x) = 2^3x nor g(x) = 4^—2x has the precise form of the exponential function defined in (1). However, by the laws of exponents given in Theorem 6.1.1, f can be rewritten as f(x) = 8^x (b = 8 in (1)), and g can be recast as . The details are shown below:

Graphs We distinguish two types of graphs for (1) depending on whether the base b satisfies b > 1or 0 < b < 1. The next two examples illustrate, in turn, the graphs of f(x) = 3^x and f(x) = ()^x. Before graphing, we can make some intuitive observations about both functions. Since the bases b = 3 and b = are positive, the values of 3^x and ()^x are positive for every real number x. As a consequence, there are no real numbers x₁ and x₂ for which 3^x₁ and ()^x₂ are zero. Graphically, this means that the graphs of f(x) = 3^x and f(x) = ()^x have no x-intercepts. Also, 3⁰ = 1 and ()⁰ = 1, and so f(0) = 1 in each case. This means that the graphs of f(x) = 3^x and f(x) = ()^x have the same y-intercept (0, 1).

EXAMPLE 2 Graph for b > 1

Graph the function f(x) = 3^x

Solution We first construct a table of some function values corresponding to preselected values of x. As shown in FIGURE 6.1.1, we plot the corresponding points obtained from the table and connect them with a continuous curve. The graph shows that f is an increasing function on the interval (—∞,∞).

FIGURE 6.1.1 Graph of function in Example 2

FIGURE 6.1.2 Graph of function in Example 3

EXAMPLE 3 Graph for 0 < b < 1

Graph the function f (x) = ()^x.

Solution Proceeding as in Example 2, we construct a table of some function values corresponding to preselected values of x. Note, for example, by the laws of exponents

As shown in FIGURE 6.1.2, we plot the corresponding points obtained from the table and connect them with a continuous curve. In this case the graph shows that f is a decreasing function on the interval (-∞,∞).

Reflections Exponential functions with bases satisfying 0 < b < 1, such as b = , are frequently written in an alternative manner. We note that y = ()^x is the same as y = 3^-x. From this last result we see that the graph of y = 3^-x is simply the graph of y = 3^x reflected in the y-axis.

f increasing for b > 1; f decreasing for 0 < b < 1

Review Theorem 4.2.3 in Section 4.2 for reflections in the x- and y-axes.

Horizontal Asymptote FIGURE 6.1.3 illustrates the two general shapes that the graph of an exponential function f(x) = b^x can have; but there is one more important aspect of all such graphs. Observe in FIGURE 6.1.3 that for b > 1,

In other words, the line y = 0 (the x-axis) is a horizontal asymptote for both types of exponential graphs.

Properties The following list summarizes some of the important properties of the exponential function f(x) = b^x. Reexamine the graphs in Figures 6.1.1-6.1.3 as you read this list.

Although the graphs y = b^x in the case, say, when b > 1, all share the same basic shape and all pass through the same point (0, 1), there are subtle differences. The larger the base b the more steeply the graph rises as x increases. In FIGURE 6.1.4 we compare the graphs of y = 5^x, y = 3^x, y = 2^x, and y = (1.2)^x in green, blue, gold, and red, respectively, on the same coordinate axes. We see from its graph that the values of y = (1.2)^x increase slowly as x increases. For example, for y = (1.2)^x, f(3) = (1.2)³ = 1.728, whereas, for y = 5^x, f(3) = 5³ = 125.

The fact that (1) is a one-to-one function, follows from the horizontal line test discussed in Section 4.6. Note in Figures 6.1.1-6.1.4 that a horizontal line can cross or intersect an exponential graph in at most one point.

Of course, we can obtain other kinds of graphs by rigid and nonrigid transformations, or when an exponential function is combined with other functions by either an arithmetic operation or by function composition. In the next several examples we examine variations of the exponential graph.

FIGURE 6.1.4 Graphs of y = b^x for b = 1.2, 2, 3, 5

EXAMPLE 4 Horizontally Shifted Graph

Graph the function f(x) = 3^{x + 2}.

Solution From the discussion in Section 4.2 you should recognize that the graph of f (x) = 3^{x + 2} is the graph of y = 3^x shifted 2 units to the left. Recall, since the shift is a rigid transformation to the left, the points on the graph of f(x) = 3^{x + 2} are the points on the graph of y = 3^x moved horizontally 2 units to the left. This means that the y-coordinates of points (x, y) on the graph of y = 3^x remain unchanged but 2 is subtracted from all the x-coordinates of the points. Thus we see from FIGURE 6.1.5 that the points (0, 1) and (2, 9) on the graph of y = 3^x are moved, in turn, to the points (-2, 1) and (0, 9) on the graph of f(x) = 3^{x + 2}.

FIGURE 6.1.5 Shifted graph in Example 4

The function f(x) = 3^{x + 2} in Example 4 can be rewritten, if desired, as f(x) = 9 · 3^x. By (i) of the laws of exponents, 3^{x + 2} = 3²3^x = 9 · 3^x. In this manner we can reinterpret the graph of f(x) = 3^{x + 2} as a vertical stretch of the graph of y = 3^x by a factor of 9. For example, (1, 3) is on the graph of y = 3^x, whereas (1, 9 · 3) = (1, 27) is on the graph of f(x) = 3^{x + 2}.

The Number e Most every student of mathematics has heard of, and has likely worked with, the famous irrational number = 3.141592654…. Recall, an irrational number is a nonrepeating and nonterminating decimal. In calculus and applied mathematics the irrational number

arguably plays a role more important than the number . The usual definition of the number e is the number that the function f(x) = (1 + 1/x)^x approaches as we let x become large without bound in the positive direction, that is,

See Problems 39 and 40 in Exercises 6.1.

The Natural Exponential Function When the base in (1) is chosen to be b = e, the function

is called the natural exponential function. Since b = e > 1 and b = 1/e < 1, the graphs of y = e^x and y = e^-x (or y = (1/e)^x = 1/e^x) are given in FIGURE 6.1.6.

FIGURE 6.1.6 Graphs of the natural exponential function (in (a)) and its reciprocal (in (b))

On the face of it, the natural exponential function (2) possesses no noticeable graphical characteristic that distinguishes it from, say, the function f(x) = 3^x, and has no special properties other than the ones given in (i)-(vii) above. Questions as to why (2) is a “natural” and frankly, the most important exponential function, can only be answered fully in courses in calculus and beyond. We will explore some of the importance of the number e in Sections 6.3 and 6.4.

EXAMPLE 5 Reflection and Vertical Shift

Graph the function f(x) = 2 - e^-x. State the range.

Solution We first draw the graph of y = e^-x as shown in Figure 6.1.7(a). Then we reflect the first graph in the x-axis to obtain the graph of y = -e^-x in Figure 6.1.7(b). Finally, the graph in Figure 6.1.7(c) is obtained by shifting the graph in part (b) upward 2 units.

The y-intercept (0, -1) of y = -e^-x when shifted upward 2 units returns us to the original y-intercept in Figure 6.1.7(a). Finally, because of the vertical shift the horizontal asymptote, which was y = 0 in parts (a) and (b) of the figure, becomes y = 2 in Figure 6.1.7(c). From the last graph we can conclude that the range of the function f(x) = 2 - e^-x is the set of real numbers defined by y < 2, that is, the interval (—∞, 2) on the y-axis.

Graph of function in Example 5

In the next example we graph the function composition of the natural exponential function y = e^x with the simple quadratic polynomial function y = —x².

EXAMPLE 6 A Function Composition

Graph the function f(x) = e^-x2.

Solution Because f(0) = e^—02 = e⁰ = 1, the y-intercept of the graph is (0, 1). Also, f(x) ≠ 0 since e^—x2 ≠ 0 for every real number x. This means that the graph of f has no x-intercepts. Then from

we conclude that f is an even function and so its graph is symmetric with respect to the y-axis. Lastly, observe that

By symmetry we can also conclude that f (x) → 0 as x → — ∞. This shows that y = 0 is a horizontal asymptote for the graph of f. The graph of f is given in FIGURE 6.1.8.

Graph of function in Example 6

Bell-shaped graphs such as that given in FIGURE 6.1.8 are very important in the study of probability and statistics.

In Problems 1-12, sketch the graph of the given function f. Find the y-intercept and the horizontal asymptote of the graph. State whether the function is increasing or decreasing.

f(x) = -2^x

f(x) = -2^-x

f(x) = 2^x+1

f(x) = 2^2-x

f(x) = -5 + 3^x

f(x) = 2 + 3^-x

f(x) = 9 - e^x

f(x) = -1 + e^x-3

f(x) = -3 - e^x+5

In Problem 13-16, find an exponential function f(x) = b^x such that the graph of f passes through the given point.

(3, 216)

(-1, 5)

(-1, e²)

(2, e)

In Problems 17 and 18, determine the range of the given function.

f(x) = 5 + e^-x

f(x) = 4 - 2^-x

In Problems 19 and 20, find the x- and y-intercepts of the graph of the given function. Do not graph.

f(x) = xe^x + 10e^x

f (x) = x²2^x - 2^x

In Problems 21-24, use a graph to solve the given inequality.

2^x > 16

e^x ≤ 1

e^x-2 < 1

In Problems 25 and 26, use the graph in FIGURE 6.1.8 to sketch the graph of the given function f.

Graph for Problem 37

Graph for Problem 38

f (x) = e^-(x-3)2

f (x) = 3 - e^-(x+1)2

In Problems 27 and 28, use f(—x) = f(x) to demonstrate that the given function is even. Sketch the graph of f.

f (x) = e^x2

f (x) = e^-|x|

In Problems 29-32 use the graphs obtained in Problems 27 and 28 as an aid in sketching the graph of the given function f.

f (x) = 1 - e^x2

f (x) = 2 + 3e^|x|

f (x) = -e^|x-3|

f (x) = e^(x+2)2

Show that f (x) = 2^x + 2^-x is an even function. Sketch the graph of f.

Show that f (x) = 2^x - 2^-x is an odd function. Sketch the graph of f.

In Problems 35 and 36, sketch the graph of the given piecewise-defined function f.

Find an equation of the red line in FIGURE 6.1.9.

Find the total area of the shaded region in FIGURE 6.1.10.

Calculator Problems

Use a calculator to fill in the given table.

Use a graphing utility to graph the functions f(x) = (1 + 1/x)^x and g(x) = e on the same set of coordinate axes. Use the intervals (0, 10], (0,100], and (0, 1000]. Describe the behavior of f for large values of x. In graphical terms, what is the constant function g(x) = e?

Graph the function f in part (a) on the interval [—10, 0). Superimpose that graph with the graph of f on (0, 10] obtained in part (a). Is f a continuous function?

In Problems 41 and 42, use a graphing utility as an aid in determining the x-coordinates of the points of intersection of the graphs of the functions f and g.

f(x) = x², g(x) = 2^x

f(x) = x³, g(x) = 3^x

For Discussion

Suppose 2^t = a and 6^t = b. Using the laws of exponents given in this section, answer the following questions in terms of the symbols a and b.

What does 12^t equal?

What does 3^t equal?

What does 6^-t equal?

What does 6^3t equal?

What does 2^-3t2^7t equal?

What does 18^t equal?

Discuss: What does the graph of y = e^ex look like? Do not use a calculator.

6.2 Logarithmic Functions

Introduction Since an exponential function y = b^x is one-to-one, we know that it has an inverse function. To find this inverse, we interchange the variables x and y to obtain x = b^y. This last formula defines y as a function of x:

y is that exponent of the base b that produces x.

By replacing the word exponent with the word logarithm, we can rephrase the preceding line as

y is that logarithm of the base b that produces x.

This last line is abbreviated by the notation y = log_bx and is called the logarithmic function.

For b > 0 there is no real number y for which b ^y can be either 0 or negative. It then follows from x = b^y that x > 0. In other words, the domain of a logarithmic function y = log_bx is the set of positive real numbers (0, ∞).

For emphasis, all that is being said in the preceding sentences is:

The logarithmic expression y = log_bx and the exponential expression x = b^y are equivalent.

That is, both symbols mean the same thing. As a consequence, within a specific context such as solving a problem, we can use whichever form happens to be more convenient. The following table lists several examples of equivalent logarithmic and exponential statements.

Graphs Recall from Section 4.6 that the graph of an inverse function can be obtained by reflecting the graph of the original function in the line y = x. This technique was used to obtain the red graphs from the blue graphs in FIGURE 6.2.1. As you inspect the two graphs in Figure 6.2.1(a) and in Figure 6.2.1(b), remember that the domain (—∞, ∞) and range (0, ∞) of y = b^x become, in turn, the range (—∞, ∞) and domain (0, ∞) of y = log_bx. Also note that the y-intercept (0, 1) for the exponential function (blue graphs) becomes the x-intercept (1, 0) for the logarithmic function (red graphs).

Graphs of logarithmic functions

Vertical Asymptote When the exponential function is reflected in the line y = x, the horizontal asymptote y = 0 for the graph of y = b^x becomes a vertical asymptote for the graph of y = log_b x. In FIGURE 6.2.1 we see that for b > 1,

From (7) of Section 4.6 we conclude that x = 0, which is the equation of the y-axis, is a vertical asymptote for the graph of y = log_bx.

Properties The following list summarizes some of the important properties of the logarithmic function f(x) = log_bx.

We would like to call attention to the third entry in the foregoing list for special emphasis:

also

Thus, in addition to (1, 0) the graph of any logarithmic function (1) with base b also contains the point (b, 1). The equivalence of y = log_bx and x = b^y also yields two sometimes-useful identities. By substituting y = log_b x into x = b^y, and then x = b^y into y = log_bx gives

For example, from (4), 8^log₈10 = 10 and log₁₀10⁵ = 5.

EXAMPLE 1 Logarithmic Graph for b > 1

Graph f(x) = log₁₀(x + 10).

Solution This is the graph of y = log₁₀ x, which has the shape shown in Figure 6.2.1(a), shifted 10 units to the left. To reinforce the fact that the domain of a logarithmic function y = log₁₀x is the set of positive real numbers, that is, x > 0, we can obtain the domain of f(x) = log₁₀(x + 10) by replacing x by x + 10 and requiring that x + 10 > 0 or x > -10. In interval notation, the domain of f is (—10, ∞). In the short accompanying table, we have chosen convenient values of x in order to plot a few points.

Notice,

The vertical asymptote x = 0 for the graph of y = log₁₀x becomes x = —10 for the shifted graph. This asymptote is the red dashed vertical line in FIGURE 6.2.2.

Graph of function in Example 1

Graph of the natural logarithm is shown in red

Natural Logarithm Logarithms with base b = 10 are called common logarithms and logarithms with base b = e are called natural logarithms. Furthermore, it is customary to write the natural logarithm

The symbol “ln x” is usually read phonetically as “ell-en of x.” Since b = e > 1, the graph of y = ln x has the characteristic logarithmic shape shown in Figure 6.2.1(a). See FIGURE 6.2.3. For base b = e, (1) becomes

The analogs of (2) and (3) for the natural logarithm are

The identities in (4) become

For example, from (8), e^{ln 13} = 13.

Common and natural logarithms can be found on all calculators.

Laws of Logarithms The laws of exponents given in Theorem 6.1.1 can be restated in an equivalent manner as the laws of logarithms. To see this, suppose we write M = b^x₁ and N = b^x₂. Then by (1), x₁ = log _b M and x₂ = log _b N.

Product: By (i) of Theorem 6.1.1, MN = b^x₁+x₂. Expressed as a logarithm this is x₁ + x₂ = log _bMN. Substituting for x₁ and x₂ gives

Quotient: By(ii)of Theorem 6.1.1, M/N = b^{x₁ - x₂}. Expressed as a logarithm this is x₁ - x₂ = log _b(M/N). Substituting for x₁ and x₂ gives

Power: By (iv) of Theorem 6.1.1, M^c = b^cx₁. Expressed as a logarithm this is cx₁ = log _b M^c. Substituting for x₁ gives

For convenience and future reference, we summarize these product, quotient, and power laws of logarithms next.

EXAMPLE 2 Using the Laws of Logarithms

Simplify and write as a single logarithm

Solution There are several ways to approach this problem. Note, for example, that the second and third terms can be combined arithmetically as

Alternatively, we can use law (iii) followed by law (ii) to combine these terms:

Hence,

EXAMPLE 3 Rewriting Logarithmic Expressions

Use the laws of logarithms to rewrite each expression and evaluate.

(a) ln

(b) ln 5e

(c)

Solution

(a) Since = e^1/2 we have from (iii) of the laws of logarithms:

(b) From (i) of the laws of logarithms and a calculator:

(c) From (ii) of the laws of logarithms:

Note that (iii) of the laws of logarithms can also be used here:

EXAMPLE 4 Value of a Logarithm

If log _b 2 = 0.4307 and log _b 3 = 0.6826, then find log _b

Solution We begin by rewriting as (18)^1/3. Then by the laws of logarithms

NOTES FROM THE CLASSROOM

i Students often struggle with the concept of a logarithm. It may help if you repeat to yourself a few dozen times, “A logarithm is an exponent.” It may also help if you begin reading a statement such as 3 = log₁₀1000 as “3 is the exponent of 10 that….”

ii Be very careful applying the laws of logarithms. The logarithm does not distribute over addition,

In other words, the exponent of a sum is not the sum of the exponents.

Also,

In general, there is no way that we can rewrite either

iii In calculus, the first step in a procedure known as logarithmic differentiation requires the student to take the natural logarithm of both sides of a complicated The idea is to use the laws of logarithms to transform powers into constant multiples, products into sums, and quotients into differences. See Problems 61-64 in Exercises 6.2.

iv You may see different notations for the natural exponential function and for the natural logarithm. For example, on some calculators you may see y = exp x instead of y = e^x. In the computer algebra system Mathematica the natural exponential function is written Exp[x] and the natural logarithm is written Log[x].

6.2 Exercises Answers to selected odd-numbered problems begin on page ANS-15.

In Problems 1-6, rewrite the given exponential expression as an equivalent logarithmic expression.

4^-1/2 =

9⁰ = 1

10⁴ = 10,000

10^0.3010 = 2

t^-s = v

(a + b)² = a² + 2ab + b²

In Problems 7-12, rewrite the given logarithmic expression as an equivalent exponential expression.

log₂128 = 7

log _bu = v

log _bb² = 2

In Problems 13-18, find the exact value of the given logarithm.

log₁₀(0.0000001)

log₄64

log₂(2² + 2²)

ln e^e

n(e⁴e⁹)

In Problems 19-22, find the exact value of the given expression.

10^{log₁₀ 62}

25^{log₅ 8}

e^{-ln 7}

In Problems 23 and 24, find a logarithmic function f(x) = log _bx such that the graph of f passes through the given point.

(49, 2)

In Problems 25-32, find the domain of the given function f. Find the x-intercept and the vertical asymptote of the graph. Use transformations to graph the given function f.

f(x) = -log₂x

f(x) = -log₂(x + 1)

f(x) = log₂(-x)

f(x) = log₂(3 - x)

f(x) = 3 - log₂(x + 3)

f(x) = 1 - 2log₄(x - 4)

f(x) = -1 + ln x

f(x) = 1 + ln(x - 2)

In Problems 33 and 34, use a graph to solve the given inequality.

ln(x + 1) < 0

log₁₀(x + 3) > 1

Show that f(x) = ln |x| is an even function. Rewrite f as a piecewise-defined function and sketch its graph. Find the x-intercepts and the vertical asymptote of the graph.

Use the graph obtained in Problem 35 to sketch the graph of y = ln |x - 2|.

Find the x-intercept and the vertical asymptote of the graph.

In Problems 37 and 38, sketch the graph of the given function f.

f(x) = |ln x|

f(x) = |ln(x + 1)|

In Problems 39-42, find the domain of the given function f.

f(x) = ln(2x - 3)

f(x) = ln(3 - x)

f(x) = ln(9 - x²)

f(x) = ln(x² - 2x)

In Problems 43-48, use the laws of logarithms to rewrite the given expression as one logarithm.

log₁₀2 + 2log₁₀5

ln(x⁴ - 4) 2 ln(x² + 2)

ln 5 + ln 5² + ln 5³ - ln 5⁶

5 ln 2 + 2 ln 3 - 3 ln 4

In Problems 49-60, use log _b4 = 0.6021 and log _b 5 = 0.6990 to evaluate the given logarithm. Round your answer to four decimal places.

log _b2

log _b20

log _b64

log _b625

log _b80

log _b0.8

log _b3.2

log ₄b

log _b5b

In Problems 61-64, use the laws of logarithms so that ln y contains no products, quotients, or powers.

For Discussion

FIGURE 6.2.4 Graph for Problem 65

In science it is sometimes useful to display data using logarithmic coordinates. Which of the following equations determines the graph shown in FIGURE 6.2.4?

y = 2x + 1

y = e + x²

y = ex²

x²y = e

Use a graphing utility to obtain the graph of the function

Show that f is an odd function, that is, f (-x) = -f (x).

If a > 0and b > 0, a ≠ b, then log _a x is a constant multiple of log _b x. That is, log _a x = k log _b x. Find k.

Show that (log₁₀e) (log _e 10) = 1.Can you generalize this result?

Discuss: How can the graphs of the given function be obtained from the graph of f(x) = ln x by means of a rigid transformation (a shift or a reflection)?

y = ln 5x

y = ln x^-1

y = ln (-x)

Find the vertical asymptotes for the graph of Sketch the graph of f. Do not use a calculator.

Using correct mathematical notation, rewrite the statement:

c is the exponent of = that gives the number N,

in two different, but equivalent, ways.

Find the zeros of the function f(x) = 5 - log₂|-x + 4|. Check your answers.

6.3 Exponential and Logarithmic Equations

Introduction Since exponential and logarithmic functions appear in the context of many different applications, we are often called upon to solve equations that involve these functions. While we postpone applications until Section 6.4, we examine in the present section some of the ways that can be used to solve a variety of exponential and logarithmic equations.

Solving Equations Here is a brief list of equation-solving strategies.

Of course, this list is not comprehensive and does not reflect the fact that in solving equations involving exponential and logarithmic functions we may also have to employ standard algebraic procedures such as factoring and using the quadratic formula.

In the first two examples we use the equivalence

to toggle between logarithmic and exponential expressions.

EXAMPLE 1 Rewriting an Exponential Expression

Solve e^10k = 7 for k.

Solution We use (1), with b = e, to rewrite the given exponential expression as a logarithmic expression:

Therefore, with the aid of a calculator

EXAMPLE 2 Rewriting a Logarithmic Expression

Solve log₂x = 5 for x.

Solution We use (1), with b = 2, to rewrite the logarithmic expression in its equivalent exponential form:

One-to-One Properties Recall from (1) of Section 4.6 that a one-to-one function f possesses the property that if f(x₁) = f(x₂), then necessarily x₁ = x₂. We have seen in Sections 6.1 and 6.2 that both the exponential function y = b^x, b > 0, b ≠ 1, and the logarithmic function y = log _bx are one-to-one. As a consequence we have:

EXAMPLE 3 Using the One-to-One Property (1)

Solve 2^{x - 3} = 8^x+1 for x.

Solution Observe on the right-hand side of the given equality that 8 can be written as a power of 2, that is, 8 = 2³. Furthermore, by (iv) of the laws of exponents given in Theorem 6.1.1,

Thus, the equation is the same as

From the one-to-one property (2) it follows that the exponents are equal, that is, x - 3 = 3x + 3. Solving for x then gives 2x = -6 or x = -3. You are encouraged to check this answer by substituting - 3 for x in the original equation.

EXAMPLE 4 Using the One-to-One Property (2)

Solve 7^{2(x + 1)} = 343 for x.

Solution By noting that 343 = 7³, we have the same base on both sides of the equality:

Thus by (2) we can equate exponents and solve for x:

EXAMPLE 5 Using the One-to-One Property (3)

Solve ln 2 + ln(4x - 1) = ln(2x + 5) for x.

Solution By (i) of the laws of logarithms in Theorem 6.2.1, the left-hand side of the equation can be written

The original equation is then

Since two logarithms with the same base are equal, it follows immediately from the one-to-one property (3) that

For logarithmic equations, especially of the kind in Example 5, you should get accustomed to checking your answer by substituting it back into the original equation. It is possible for a logarithmic equation to have an extraneous solution.

EXAMPLE 6 An Extraneous Solution

Solve log₂ x + log₂(x - 2) = 3.

Solution We start using again that the sum of logarithms on the left-hand side of the equation is the logarithm of a product:

With b = 2 we use (1) to rewrite the last equation in the equivalent exponential form

By ordinary algebra we then have

From the last equation we conclude that either x = 4 or x = -2. However, we must rule out x = -2 as a solution. In other words, the number x = -2 is an extraneous solution because, when substituted into the original equation, the very first term, log₂(-2), is not defined. Thus the only solution of the given equation is x = 4.

Check:

When we use the phrase “take the logarithm of both sides of an equality” we are actually using the property that if M and N are two positive numbers such that M = N, then log _bM = log _bN.

EXAMPLE 7 Taking the Natural Logarithm of Both Sides

Solve e^2x = 3^x-4.

Solution Since the bases of the exponential expression on each side of the equality are different, one way to proceed is to take the natural logarithm (the common logarithm could also be used) of both sides. From the equality

and (iii) of the laws of logarithms in Theorem 6.2.1, we get

Now using ln e = 1 and the distributive law, the last equation becomes

Gathering the terms involving the symbol x to one side of the equality then gives

You are encouraged to verify the calculation that x ≈ -4.8752.

EXAMPLE 8 Using the Quadratic Formula

Solve 5^x - 5^2x = 2.

Solution Because 5^-x = 1/5^x, the equation is

Multiplying both sides of the foregoing equation by 5^x then gives

If we let X — 5^X, then the last equation can be interpreted as a quadratic equation X² - 2X - 1 = 0. Using the quadratic formula to solve for X yields

Because 1 - is a negative number there are no real solutions of 5^x = 1 - and so

Now by taking the natural logarithm of both sides of the equality we obtain

Using the key of a calculator, the division yields x ≈ 0.548.

Change of Base In (4) of Example 8 it follows from (1) that a perfectly valid solution of the equation 5^x - 5^-x = 2is x — log₅(1 + ).But from a computational viewpoint (that is, expressing x as a number), the last answer is not desirable since no calculator has a logarithmic function with base 5. But by equating x = log₅(1 + ) with the result in (5) we have discovered that logarithm with base 5 can be expressed in terms of the natural logarithm:

The result given in (6) is just a special case of a more general result known as the change-of-base formula.

PROOF: If we let y = log _aM, then from (1) a^y = M. Then

In order to obtain the numerical value of a logarithm using a calculator, we usually choose b = 10 or b = e in (7):

EXAMPLE 9 Changing the Base

Find the numerical value of log₂ 50.

Solution We can use either formula in (8). If we choose the first formula in (8) with M = 50 and a = 2, we have

Using the key to calculate the two common logarithms and then dividing yields the approximation

Alternatively, the second formula in (8) gives the same result:

We can check the answer in Example 9 on a calculator by using the key. You are urged to verify that 2^5.6439 ≈ 50.

EXAMPLE 10 Changing the Base

Find the x in the domain of f(x) = 6^x for which f (x) = 73.

Solution We must find a solution of the equation 6^x = 73. One way of proceeding is to rewrite the exponential expression as an equivalent logarithmic expression:

x = log₆ 73.

Then with the identification a = 6 it follows from the second equation in (8) and the aid of a calculator that

You should verify that f(2.3946) = 6²³⁹⁴⁶ ≈ 73.

6.3 Exercises Answers to selected odd-numbered problems begin on page ANS-15.

In Problems 1-20, solve the given exponential equation.

5^x-2 = 1

3^x = 27^x2

e^5x-2 = 30

2^x. 3^x = 36

2^x2 = 8^2x–3

5 – 10^2x = 0

7^-x = 9

3^2(x–1) = 7²

4^x = 5^2x+1

3^x+4 = 2^{x –16}

In Problems 21–40, solve the given logarithmic equation.

log₃ 5x = log₃ 160

ln(10 + x) 5 ln(3 + 4x)

ln x = ln 5 + ln 9

3 log₈ x = log₈ 36 + log₈ 12 - log₈2

log₂(log₃x) = 2

log₃ 81^x - log₃ 3^2x = 3

log₂(x - 3) - log₂(2x+1)=-log₂4

log₂x + log₂(10 - x) = 4

log₈ x + log₈x² = 1

log₆2x - log₆(x + 1) = 0

log₁₀ x² + log₁₀ x³ + log₁₀ x⁴ + log₁₀ x⁵ = log₁₀ 16

ln 3 + ln(2x - 1) = ln 4 + ln(x + 1)

ln(x + 3) + ln(x - 4) - ln x = ln 3

In Problems 41-50, either use factoring or the quadratic formula to solve the given equation.

(5^x)² – 26(5^x) + 25 = 0

64^x - 10(8^x) + 16 = 0

log ₄x² = (log₄ X)²

(log₁₀ x)² + log₁₀x = 2

(5^x)² – 2(5^x) – 1 = 0

2^2x - 12(2^x) + 35 = 0

(ln x)² + ln x = 2

(log₁₀ 2x)² = log₁₀(2x)²

2^x + 2^–x = 2

10^2x - 103(10^x) + 300 = 0

In Problems 51-56, find the x-intercepts of the graph of the given function.

51. f(x) = e^x+4 – e

f(x) = 4^x–1 – 3

f (x) = – 3^2x + 5

f (x) = x³8^x + 5x²8^x + 6x8^x

In Problems 57-62, graph the given functions. Determine the approximate x-coordinates of the points of intersection of their graphs.

f(x) = 4e^x, g(x) = 3^-x

f (x) = 2^x, g(x) = 3 - 2^x

f (x) = 3^x2, g(x) = 2(3^x)

In Problems 63-66, solve the given equation.

x^lnx = e⁹

log_x81 = 2

log₅ 125^x = -2

In Problems 67 and 68, use the natural logarithm to find x in the domain of the given function for which f takes on the indicated value.

f(x) = 6^x; f(x) = 51

f(x) = (1)^x; f (x) = 7

For Discussion

Discuss: How would you find the x-intercepts of the graph of the function f(x) = e^2x - 5e^x + 4?

A Curiosity The logarithm developed by John Napier was actually

Express this logarithm in terms of the natural logarithm.

6.4 Exponential and Logarithmic Models

Introduction In this section we consider some mathematical models utilizing exponential or logarithmic functions. Roughly speaking, a mathematical model is a mathematical description of something that we will call a system. To construct a mathematical model we start with a set of reasonable assumptions about the system that we are trying to describe. These assumptions include any empirical laws that are applicable to the system. The end result could be a description as simple as a single function.

Exponential Models In the physical sciences, the exponential expression Ce^kt, where C and k are constants, frequently appears in mathematical models of systems that change with time t. As a consequence, mathematical models are often used to predict a future state of a system. For example, extremely complicated mathematical models are used to predict the weather over various regions of the country for, say, the next week.

Population Growth In one model of a growing population, it is assumed that the rate of growth of the population is proportional to the number present at time t. If P(t) denotes the population or number present at time t, then with the aid of calculus it can be shown that this assumption gives rise to

where t is time, and P₀ and k are constants. The function (1) is used to describe the growth of populations of bacteria, small animals, and, in some rare circumstances, humans. Setting t = 0 gives P(0) = P₀, and so P₀ is called the initial population. The constant k > 0 is called the growth constant or growth rate. Since e^kt, k > 0, is an increasing function on the interval [0, ∞), the model in (1) describes uninhibited growth.

EXAMPLE 1 Bacterial Growth

It is known that the doubling time* E. Coli bacteria, which reside in the large intestine (colon) of healthy people, is just 20 minutes. Use the exponential growth model (1) to find the number of E. Coli bacteria in a culture after 6 hours.

E. Coli bacteria

Solution Let us use hours as our unit of time, so that 20 min = h. Because the initial number of E. Coli in the culture is not specified, we will simply denote the initial size of the culture as P₀. Now using (1), a function interpretation of the first sentence in this example is P () = 2P₀. This means P₀e^k/3 = 2P₀or e^k/3 = 2. Solving this last equation for k gives the growth constant

A model for the size of the culture after t hours is then P(t) = P₀e^2.0794t. Setting t = 6 gives P(6) = P₀e^2.0794(6) ≈ 262,144P₀. Put another way, if the culture consists of only one bacterium at t = 0, then (with P₀ = 1) the model predicts that there will be 262,144 cells 6 hours later.

In the early nineteenth century the English clergyman and economist Thomas R. Malthus used the growth model (1) to predict the world population. For specific values of P₀ and k, the function values P(t) were actually reasonable approximations to the world population for a period of time during the nineteenth century. Since P(t) is an increasing function, Malthus predicted that the future population growth would surpass the world's ability to produce food. As a consequence he also predicted wars and worldwide famine. More a doomsayer than a seer, Malthus failed to foresee that the food supply would keep pace with the increased population through simultaneous advances in science and technology.

Thomas R. Malthus (1776-1834)

In 1840, a more realistic model for predicting human populations in small countries was advanced by the Belgian mathematician/biologist P. F. Verhulst (1804-1849). The so-called logistic function

where K, c, and r are constants, has over the years proved to be an accurate growth model for populations of protozoa, bacteria, fruit flies, water fleas, and animals confined to limited spaces. In contrast to uninhibited growth of the Malthusian model (1), (2) exhibits bounded growth. More specifically, the population predicted by (2) will not increase beyond the number K, called the carrying capacity of the system. For r < 0, e^rt → 0 and P(t) → K as t → ∞. You are asked to graph a special case of (2) in Problem 7 in Exercises 6.4.

Pierre and Marie Curie

Radioactive Decay Element 88, better known as radium, was discovered by Pierre and Marie Curie in 1898. Radium is a radioactive element, which means that a radium atom spontaneously decays, or disintegrates, by emitting radiation in the form of alpha particles, beta particles, and gamma rays. When an atom disintegrates in this manner, its nucleus is transmuted into a nucleus of another element. For example, the nucleus of an atom of the most stable isotope of radium, Ra-226, is transmuted into the nucleus of a radon atom Rn-222. Radon is a heavy, odorless, colorless, and highly dangerous radioactive gas that usually originates in the ground. Because it can penetrate a sealed concrete floor, radon frequently accumulates in the basements of some new and highly insulated homes. Some medical organizations have claimed that after cigarette smoking, exposure to radon gas is the second leading cause of lung cancer.

If it is assumed that the rate of decay of a radioactive substance is proportional to the amount remaining or present at time t, then we arrive at basically the same model as in (1). The important difference is that k < 0. If A(t) represents the amount of the decaying substance that remains at time t, then

where A₀ is the initial amount of the substance present, that is, A(0) = A₀. The constant k < 0 in (3) is called the decay constant or decay rate.

EXAMPLE 2 Decay of Radium

Suppose there are 20 grams of radium on hand initially. After t years the amount remaining is modeled by the function A(t) = 20e^-0.000418t. Find the amount of radium remaining after 100 years. What percent of the original 20 grams has decayed after 100 years?

Solution Using a calculator, we find that after 100 years there remains

Thus, only

of the initial 20 grams has decayed.

Half-Life The half-life of a radioactive substance is the time T it takes for one-half of a given amount of that element to disintegrate and change into a new element. See FIGURE 6.4.1. Half-life is a measure of the stability of an element; that is, the shorter the half-life, the more unstable the element. For example, the half-life of the highly radioactive strontium 90, Sr-90, produced in nuclear explosions, is 29 days, whereas the half- life of the uranium isotope U-238 is 4,560,000 years. The half-life of californium, Cf-244, first discovered in 1950, is only 45 minutes. Polonium, Po-213, has a half-life of 0.000001 second.

FIGURE 6.4.1 Time T is the half-life

EXAMPLE 3 Half-Life of Radium

Use the exponential model in Example 2 to determine the half-life of radium.

Solution If A(t) = 20e^-0.000418t, then we must find the time T for which

From 20e^-0.000418T = 10 we get e^-0.000418T = By rewriting the last expression in the logarithmic form -0.000418 T we can solve for T:

A careful reading of Example 3 reveals that the initial amount present plays no part in the actual calculation of the half-life. Since the solution of A(T) = A₀e^-0.000418T = A₀ leads to e^-0.000418T = , we see that T is independent of A₀. In other words, the half-life of 1 gram, 20 grams, or 10,000 grams of radium is the same. It takes about 1660 years for one-half of any given quantity of radium to transmute into radon.

Ibuprofen is an NSAID

Medications also have half-lives. In this case, the half-life of a drug is the time T that it takes for the body to eliminate, by metabolism or excretion, one-half of the amount of the drug taken. For example, the most popular NSAIDs (non-steroidal anti-inflammatory drugs such as aspirin and ibuprofen) taken for the relief of continuing pain, have relatively short half-lives of a few hours and as a consequence must be taken several times a day. The NSAID naproxen has a longer half-life and is usually taken once every 12 hours. See Problem 31 in Exercises 6.4.

Willard Libby (1908-1980)

Carbon Dating The approximate age of fossils of once-living matter can sometimes be determined by a method known as carbon dating. The radioactive isotope of carbon, carbon-14 or C-14, is formed presumably at a constant rate in the atmosphere by the interaction of cosmic rays on nitrogen-14. The carbon-dating method, invented by the chemist Willard Libby around 1950, is based on the fact that a plant or an animal absorbs C-14 through the process of breathing and eating, and ceases to absorb C-14 when it dies. As the next example shows, the carbon-dating procedure is based on the knowledge that the half-life of C-14 is about 5730 years. Carbon-14 decays back to the original nitrogen-14.

The Psalms scroll

Libby won the 1960 Nobel Prize in chemistry for his work. Libby's method has been used to date wooden furniture found in Egyptian tombs, the Dead Sea scrolls written on papyrus and animal skin, the famous linen Shroud of Turin, and a recently discovered copy of the Gnostic Gospel of Judas written on papyrus.

EXAMPLE 4 Carbon Dating a Fossil

A fossilized bone is found to contain of the initial amount of C-14 that the organism contained while it was alive. Determine the approximate age of the fossil.

Solution If A₀ denotes an initial amount A₀, measured in grams, of C-14 in the organism, then t years after its death there are A(t) = A₀e^kt grams remaining. When t = 5730, A(5730)= A₀, and so A₀ = A₀e^5730k Solving this last equation for the decay constant k gives

Hence a model for the amount of C-14 remaining is A(t)= A₀e^-0.00012097t. Using this model, we now solve A(t) = A₀ for t:

The age determined in the last example is actually beyond the border of accuracy for the carbon-14-dating method. After 9 half-lives of the isotope, or about 52,000 years, about 99.7% of carbon-14 has decayed making its measurement in a fossil nearly impossible.

Newton's Law of Cooling/Warming Suppose an object or body is placed in a medium (air, water, etc.) that is held at constant temperature T_m, called the ambient temperature. If the initial temperature T₀ of the body or object at the moment it is placed into the medium is greater than the ambient temperature T_m, then the body will cool. On the other hand, if T₀ is less than T_m, then it will warm up. For example, in an office kept at, say, 70°F, a steaming cup of coffee will cool off, whereas a glass of ice water will warm up. The usual cooling/warming assumption is that the rate at which an object cools/warms is proportional to the difference T(t) – T_m, where T(t) represents the temperature of the object at time t. In either case, cooling or warming, this assumption leads to T(t) - T_m = (T₀ – T_m)e^kt, where k is a negative constant. Observe that since e^kt → 0 for k < 0, the last expression is consistent with one's intuitive expectation that T(t) - T_m → 0, or equivalently T(t) → T_m, as t → ∞ (the coffee cools to room temperature; the ice water warms to room temperature). Solving for T(t) we obtain a function for the temperature of the object,

The mathematical model in (4), named after its discoverer, is called Newton's law of cooling/warming. Note that T(0) = T₀.

EXAMPLE 5 Cooling of a Cake

A cake is removed from an oven where the temperature was 350°F into a kitchen where the temperature is 75°F. One minute later the temperature of the cake is measured to be 300°F. Assume that the temperature of the cake in the kitchen is given by (4).

(a) What is the temperature of the cake after 6 minutes?

(b) At what time is the temperature of the cake 80°F?

(c) Graph T(t).

Cake will cool off to room temperature

Solution (a) When the cake is removed from the oven its temperature is also 350°F, that is, T₀ = 350. The ambient temperature is the temperature of the kitchen T_m = 75. Thus (4) becomes T(t) = 75 + 275e^kt. The measurement that T(1) = 300 is the condition that determines the value of k. From T(1) = 75 + 275 e^k = 300 we find

The mathematical model T(t) = 75 + 275e ^-0.2007t then predicts that the temperature of the cake in 6 minutes after it is removed from the oven will be

(b) To determine when the temperature of the cake will be 80°F, we solve the equation T (t) = 80 for t. Rewriting T(t) = 75 + 275e^-02007t = 80 as

(c) With the aid of a graphing utility we obtain the graph of T(t) shown in blue in FIGURE 6.4.2. Since T(t) = 75 + 275e^-02007t → 75 as t → ∞, T = 75, shown in red in FIGURE 6.4.2, is a horizontal asymptote for the graph of T(t) = 75 + 275e^-02007t.

FIGURE 6.4.2 Graph of T(t) in Example 5

Compound Interest Investments such as savings accounts pay an annual rate of interest that can be compounded annually, quarterly, monthly, weekly, daily, and so on. In general, if a principal of P dollars is invested at an annual rate r of interest that is compounded n times a year, then the amount S accrued at the end of t years is given by

S is called the future value of the principal P. If the number n is increased without bound, then interest is said to be compounded continuously. To find the future value of P in this case, we let m = n/r. Then n = mr and

Since n → ∞ implies that m → ∞ we see from page 283 of Section 6.1 that (1 + 1/m)^m → e. The right-hand side of (6) becomes

Thus, if an annual rate r of interest is compounded continuously, the future value S of a principal P in t years is

EXAMPLE 6 Comparison of Future Values

Suppose that $1000 is deposited in a savings account whose annual rate of interest is 3%. Compare the future value of this principal in 10 years (a) if interest is compounded monthly and (b) if interest is compounded continuously.

Solution (a) Since there are 12 months in a year, we identify n = 12. Furthermore, with P = 1000, r = 0.03, and t = 10, (6) becomes

(b) From (7)

Thus over 10 years we have gained only $0.51 by compounding continuously rather than monthly.

Logarithmic Models Probably the most famous application of the base 10 logarithm, or common logarithm, is the Richter scale. In 1935, the American seismologist Charles F. Richter devised a logarithmic scale for comparing the energies of different earthquakes. The magnitude M of an earthquake is defined by

where A is the amplitude of the largest seismic wave of the earthquake and A₀ is a reference amplitude that corresponds to the magnitude M = 0. The number M is calculated to one decimal place. Earthquakes of magnitude 6 or greater are considered potentially destructive.

Charles F. Richter (1900-1985)

EXAMPLE 7 Comparing Intensities

The earthquake on December 26, 2004, off the west coast of Northern Sumatra, which spawned a tsunami causing over 200,000 deaths, was initially classified as a 9.3 on the Richter scale. On March 28, 2005, an aftershock in the same area was classified as an 8.7 on the Richter scale. How many times more intense was the 2004 earthquake?

Solution From (8) we have

This means, in turn, that

Now, since 9.3 = 0.6 + 8.7, it follows from the laws of exponents that

Thus the original earthquake in 2004 was approximately 4 times as intense as the aftershock in 2005.

You can see from Example 7 that if, say, one earthquake is a 6.0 and another is a 4.0 on the Richter scale, then the 6.0 earthquake is 10² = 100 times more intense than the 4.0 earthquake.

pH of a Solution In chemistry, the hydrogen potential, or pH, of a solution is defined as

where the symbol [H ⁺] denotes the concentration of hydrogen ions in a solution measured in moles per liter. The pH scale was invented in 1909 by the Danish biochemist Søren Sørensen. Solutions are classified according to their pH value as acidic, base, or neutral. A solution with a pH in the range 0 < pH < 7 is said to be acid; when pH < 7, the solution is base (or alkaline). In the case when pH = 7, the solution is neutral. Water, if uncontaminated by other solutions or by acid rain, is an example of a neutral solution, whereas undiluted lemon juice is highly acid and has a pH in the range pH ≤ 3.A solution with pH = 6 is ten times more acidic than a neutral solution. See Problems 47-50 in Exercises 6.4.

Søren Sørensen (1868-1939)

As the next example illustrates, pH values are usually calculated to one decimal place.

EXAMPLE 8 pH of Human Blood

The concentration of hydrogen ions in the blood of a healthy person is found to be [H⁺] = 3.98 X 10^-8 moles/liter. Find the pH of blood.

Solution From (9) and the laws of logarithms (Theorem 6.2.1),

With the help of the base 10 log key on a calculator, we find that

pH ≈ — [0.5999 — 8] ≈ 7.4.

Human blood in usually a base solution. The pH values of blood usually fall within the rather narrow range 7.2 < pH < 7.6.A person with a blood pH outside these limits can suffer illness and even death.

6.4 Exercises Answers to selected odd-numbered problems begin on page ANS-15.

Population Growth

After 2 hours the number of bacteria in a culture is observed to have doubled.

(a) Find an exponential model (1) for the number of bacteria in the culture at time t.

(b) Find the number of bacteria present in the culture after 5 hours.

(c) Find the time that it takes the culture to grow to 20 times its initial size.

A model for the number of bacteria in a culture after t hours is given by (1).

(a) Find the growth constant k if it is known that after 1 hour the colony has expanded to 1.5 times its initial population.

(b) Find the time that it takes for the culture to quadruple in size.

A model for the population in a small community is given by P(t) = 1500e^kt. If the initial population increases by 25% in 10 years, what will the population be in 20 years?

A model for the population in a small community after t years is given by (1).

(a) If the initial population has doubled in 5 years, how long will it take to triple? To quadruple?

(b) If the population of the community in part (a) is 10,000 after 3 years, what was the initial population?

A model for the number of bacteria in a culture after t hours is given by P(t) = P₀e^kt. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?

In genetic research a small colony of drosophila (small two-winged fruit flies) is grown in a laboratory environment. After 2 days it is observed that the population of flies in the colony has increased to 200. After 5 days the colony has 400 flies.

(a) Find a model P(t) = P₀e^kt for the population of the fruit-fly colony after t days.

(b) What will be the population of the colony in 10 days?

(c) When will the population of the colony be 5000 fruit flies?

A student sick with a flu virus returns to an isolated college campus of 2000 students. The number of students infected with the flu t days after the student's return is predicted by the logistic function

(a) According to this model, how many students will be infected with the flu after 5 days?

(b) How long will it take for one-half of the student population to become infected?

(c) How many students does the model predict will become infected after a very long period of time?

(d) Sketch a graph of P(t).

In 1920, Pearl and Reed proposed a logistic model for the population of the United States based on the years 1790, 1850, and 1910. The logistic function they proposed was

where P is measured in thousands and t represents the number of years past 1780.

(a) The model agrees quite well with the census figures between 1790 and 1910. Determine the population figures for 1790, 1850, and 1910.

(b) What does this model predict for the population of the United States after a very long time? How does this prediction compare with the 2000 census population of 281 million?

Radioactive Decay and Half-Life

Initially 200 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. Construct an exponential model A(t) = A₀e^kt for the amount remaining of the decaying substance after t hours. Find the amount remaining after 24 hours.

Determine the half-life of the substance in Problem 9.

Do this problem without using the exponential model (3). Initially there are 400 grams of a radioactive substance on hand. If the half-life of the substance is 8 hours, give an educated guess of how much remains (approximately) after 17 hours. After 23 hours. After 33 hours.

Construct an exponential model A(t) = A₀e^kt for the amount remaining of the decaying substance in Problem 11. Compare the predicted values A(17), A(23), and A(33) with your guesses.

Iodine 131, used in nuclear medicine procedures, is radioactive and has a half-life of 8 days. Find the decay constant k for iodine 131. If the amount remaining of an initial sample after t days is given by the exponential model A(t) = A₀e^kt, how long will it take for 95% of the sample to decay?

The amount remaining of a radioactive substance after t hours is given by A(t) = 100e^kt. After 12 hours, the initial amount has decreased by 7%. How much remains after 48 hours? What is the half-life of the substance?

The half-life of polonium 210, Po-210, is 140 days. If A(t) = A₀e^kt represents the amount of Po-210 remaining after t days, what is the amount remaining after 80 days? After 300 days?

Strontium 90 is a dangerous radioactive substance found in acid rain. As such it can make its way into the food chain by polluting the grass in a pasture on which milk cows graze. The half-life of strontium 90 is 29 years.

(a) Find an exponential model (3) for the amount remaining after t years.

(b) Suppose a pasture is found to contain Str-90 that is 3 times a safe level A₀. How long will it be before the pasture can be used again for grazing cows?

Charcoal drawing in Problem 17

Carbon Dating

Charcoal drawings were discovered on walls and ceilings in a cave in Lascaux, France. Determine the approximate age of the drawings, if it was found that 86% of C-14 in a piece of charcoal found in the cave had decayed through radioactivity.

Analysis on an animal bone fossil at an archeological site reveals that the bone has lost between 90% and 95% of C-14. Give an interval for the possible ages of the bone.

Shroud image in Problem 19

The shroud of Turin shows the negative image of the body of a man who appears to have been crucified. It is believed by many to be the burial shroud of Jesus of Nazareth. In 1988 the Vatican granted permission to have the shroud carbon dated. Several independent scientific laboratories analyzed the cloth and the consensus opinion was that the shroud is approximately 660 years old, an age consistent with its historical appearance. This age has been disputed by many scholars. Using this age, determine what percentage of the original amount of C-14 remained in the cloth as of 1988.

In 1991 hikers found a preserved body of a man partially frozen in a glacier in the Austrian Alps. Through carbon-dating techniques it was found that the body of ötzi—the iceman, as he came to be called—contained 53% as much C-14 as found in a living person. What is the approximate date of his death?

The iceman in Problem 20

Newton's Law of Cooling/Warming

Suppose a pizza is removed from an oven at 400°F into a kitchen whose temperature is a constant 80°F. Three minutes later the temperature of the pizza is found to be 275°F.

(a) What is the temperature T(t) of the pizza after = minutes?

(b) Determine the time when the temperature of the pizza is 150°F.

(c) After a very long period of time, what is the approximate temperature of the pizza?

A glass of cold water is removed from a refrigerator whose interior temperature is 39°F into a room maintained at 72°F. One minute later the temperature of the water is 43°F. What is the temperature of the water after 10 minutes? After 25 minutes?

A thermometer is brought from the outside, where the air temperature is -20°F, into a room where the air temperature is a constant 70°F. After 1 minute inside the room the thermometer reads 0°F. How long will it take for the thermometer to read 60°F?

A thermometer is taken from inside a house to the outside, where the air temperature is 5°F. After 1 minute outside the thermometer reads 59°F, and after 5 minutes it reads 32°F. What is the temperature inside the house?

A dead body was found within a closed room of a house where the temperature was a constant 70°F. At the time of discovery, the core temperature of the body was determined to be 85°F. One hour later a second measurement showed that the core temperature of the body was 80°F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6°F. Determine how many hours elapsed before the body was found.

Repeat Problem 25 if evidence indicated that the dead person was running a fever of 102°F at the time of death.

Thermometer in Problem 24

Compound Interest

Suppose that 1¢ is deposited in a savings account paying 1% annual interest compounded continuously. How much money will have accrued in the account after 2000 years? What is the future value of 1¢ in 2000 years if the account pays 2% annual interest compounded continuously?

Suppose that $100,000 is invested at an annual interest rate of 5%. Use (6) and (7) to compare the future values of that amount in 1 year by completing the following table.

Suppose that $5000 is deposited in a savings account paying 6% annual interest compounded continuously. How much interest will be earned in 8 years?

Present Value If (7) is solved for P, that is, P = Se^-rt, we obtain the amount that should be invested now at an annual rate r of interest in order to be worth S dollars after t years. We say that P is the present value of the amount S. What is the present value of $100,000 at an annual rate of 3% compounded continuously for 30 years?

Miscellaneous Exponential Models

Effective Half-life Radioactive substances are removed from living organisms by two processes: natural physical decay and biological metabolism. Each process contributes to an effective half-life E that is defined by

1/E = 1/P 1 1/B,

where P is the physical half-life of the radioactive substance and B is the biological half-life.

(a) Radioactive iodine, I-131, is used to treat hyperthyroidism (overactive thyroid). It is known that for human thyroids, P = 8 days and B = 24 days. Find the effective half-life of I-131 in the thyroid.

(b) Suppose the amount of I-131 in the human thyroid after t days is modeled by A(t) = A₀e^kt, k < 0. Use the effective half-life found in part (a) to determine the percentage of radioactive iodine remaining in the human thyroid gland 2 weeks after its ingestion.

Newton's Law of Cooling Revisited The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modification of (4) is

T(t) = T_m + (T₀ - T_m)e^kSt, k < 0.

Suppose two cups A and B are filled with coffee at the same time. Initially the temperature of the coffee is 150°F. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min, the temperature of the coffee in cup A is 100°F. If T_m = 70°F, what is the temperature of the coffee in cup B after 30 min?

Series Circuit In a simple series circuit consisting of a constant voltage E, an inductance of L henries, and a resistance of R ohms, it can be shown that the current I(t) is given by

Solve for t in terms of the other symbols.

Drug Concentration Under some conditions the concentration of a drug at time t after injection is given by

Here a and b are positive constants and C₀ is the concentration of the drug at t = 0. Determine the steady-state concentration of a drug, that is, the limiting value of C(t) as t → ∞. Determine the time t at which C(t) is one-half the steady-state concentration.

Richter Scale

Two of the most devastating earthquakes in the San Francisco Bay area occurred in 1906 along the San Andreas fault and in 1989 in the Santa Cruz Mountains near Loma Prieta peak. The 1906 and 1989 earthquakes measured 8.5 and 7.1 on the Richter scale, respectively. How much greater was the intensity of the 1906 earthquake compared to the 1989 earthquake?

Marina district in San Francisco, 1989

How much greater was the intensity of the 2004 Northern Sumatra earthquake (Example 7) compared to the 1964 Alaskan earthquake of magnitude 8.9?

If an earthquake has a magnitude 4.2 on the Richter scale, what is the magnitude on the Richter scale of an earthquake that has an intensity 20 times greater? [Hint: First solve the equation 10^x = 20.]

Show that the Richter scale defined in (8) of this section can be written

pH of a Solution

In Problems 39-42, determine the pH of a solution with the given hydrogen-ion concentration [H⁺].

10^-6

4 × 10^-7

2.8 × 10^-8

5.1 × 10^-5

In Problems 43-46, determine the hydrogen-ion concentration [H+] of a solution with the given pH.

3.3

7.3

8.1

In Problems 47-50, determine how many more times acidic the first substance is compared to the second substance.

47. lemon juice: pH = 2.3; vinegar, pH = 3.3

48. battery acid: pH = 1; lye, pH = 13

49. clean rain: pH = 5.6; acidic rain, pH = 3.8

50. NaOH: [H ⁺] = 10^-14; HCl, [H ⁺] = 1

Miscellaneous Logarithmic Models

51. Richter Scale and Energy Charles Richter working with Beno Gutenberg developed the model

that relates the Richter magnitude M of an earthquake and its seismic energy E (measured in ergs). Calculate the seismic energy E of the 2004 Northern Sumatra earthquake where M = 9.3.

52. Intensity Level The intensity level b of a sound measured in decibels (dB) is defined by

where I is the intensity of the sound measured in watts/cm² and I₀ = 10^-16 watts/cm² is the intensity of the faintest sound that can be heard (0 dB). Use (10) and complete the following table.

53. Threshold of Pain The threshold of pain is generally taken to be around 140 dB. Find the intensity of sound I corresponding to 140 dB.

54. Intensity Levels The intensity of sound I is inversely proportional to the square of the distance d from its source, that is,

where k is the constant of proportionality. Suppose d₁ and d₂ are distances from a source of sound, and that the corresponding intensity levels of the sounds are b₁ and b₂. Use (11) in (10) to show that b₁ and b₂ are related by

55. Intensity Level When a plane P₁, flying at an altitude of 1500 ft, passed over a point on the ground its intensity level b₁ was measured as 70 dB. Use (12) to find the intensity level b₂ of a second plane P₂, flying at an altitude of 2600 ft, when it passed over the same point.

56. Talking Politics At a distance of 4 ft, the intensity level of an animated political conversation is 60 dB. Use (12) to find the intensity level 14 ft from the conversation.

57. Pupil of the Eye An empirical model devised by DeGroot and Gebhard relates the diameter d of the pupil of the eye (measured in millimeters, mm) to the luminance B of light source (measured in millilambert's, mL):

See FIGURE 6.4.3

FIGURE 6.4.3 Pupil diameter in Problem 57

(a) The average luminance of clear sky is approximately B = 255 mL. Find the corresponding pupil diameter.

(b) The luminance of the Sun varies from approximately B = 190,000 mL at sunrise to mL at noon. Find the corresponding pupil diameters.

(c) Find the luminance B corresponding to a pupil diameter of 7 mm.

Body Surface Area Medical researchers use the empirical mathematical model

to estimate body surface area A (measured in square meters), given a person's mass m (in kilograms) and height h (in centimeters).

(a) Estimate the body surface area of a person whose mass is m = 70 kg and who is h = 175 cm tall.

(b) Determine your mass and height and estimate your own body surface area.

6.5 Hyperbolic Functions

Introduction We have already seen in Section 6.4 the usefulness of the exponential function e^x in various mathematical models. As a further application, consider a long rope or a flexible wire, such as a telephone wire hanging only under its own weight between two fixed supports or telephone poles. It can be shown that under certain conditions the hanging wire assumes the shape of the graph of the function

The symbol c stands for a positive constant that depends on the physical characteristics of the wire. Functions such as (1), consisting of certain combinations of e^x and e^-x, appear in so many applications that mathematicians have given them names.

Hyperbolic Functions In particular, when c = 1 in (1), the resulting function is called the hyperbolic cosine

Telephone wires

Analogous to the trigonometric functions tan x, cot x, sec x, and csc x that are defined in terms of sin x and cos x, there are four additional hyperbolic functions tanh x, coth x, sech x, and csch x that are defined in terms of sinh x and cosh x:

Gateway arch in St. Louis, MO

Graphs The graph of the hyperbolic cosine, shown in FIGURE 6.5.1, is called a catenary. The word catenary derives from the Latin word for a chain, catena. The shape of the famous 630-ft tall Gateway arch in St. Louis, Missouri, is an inverted catenary. Compare the shape in FIGURE 6.5.1 with that in the accompanying photo. The graph of y = sinh x is given in FIGURE 6.5.2.

FIGURE 6.5.1 Graph of y = cosh x

FIGURE 6.5.2 Graph of y = sinh x

The graphs of the hyperbolic tangent, cotangent, secant, and cosecant are given in FIGURE 6.5.3. Observe that y = 1 and y = -1 are horizontal asymptotes for the graphs of y = tanh x and y = coth x and that x = 0 is a vertical asymptote for the graphs of y = coth x and y = csch x.

FIGURE 6.5.3 Graphs of the hyperbolic tangent (a), cotangent (b), secant (c), and cosecant (d)

Identities Although the hyperbolic functions are not periodic, they possess identities that are very similar to trigonometric identities. Analogous to the basic Pythagorean identity of trigonometry cos² x + sin² x = 1, for the hyperbolic sine and cosine we have

See Problems 1-6 in Exercises 6.5.

6.5 Exercises Exercises: Answers to selected odd-numbered problems begin on page ANS-16.

In Problems 1-6, use the definitions of cosh x and sinh x in (2) and (3) to verify the given identity.

1. cosh²x - sinh²x = 1

2. 1 - tanh²x = sech²x

3. cosh(-x) = cosh x

4. sinh(-x) = -sinh x

5. sinh 2x = 2sinh x cosh x

6. cosh 2x = cosh²x + sinh²x

7. (a) If sinh x = use the identity given in Problem 1 to find the value of cosh x.

(b) Use the result of part (a) to find the numerical values of tanh x, coth x, sech x, and csch x.

8. (a) If tanh x = use the identity given in Problem 2 to find the value of sech x.

(b) Use the result of part (a) to find the numerical values of tanh x, coth x, sech x, and csch x.

9. As can be seen in FIGURE 6.5.2, the hyperbolic sine function y = sinh x is one-to-one. Use (2) in Definition 6.5.1 in the form e^x - 2y - e^-x = 0 to show that the inverse hyperbolic sine sinh^-1x is given by

10. The function y = cosh x on the restricted domain [0, ∞) is one-to-one. Proceed as in Problem 9 to find the inverse of y = cosh x, x ≥ 0.

A. True/False_______________________________________

In Problems 1-14, answer true or false.

1. y = ln x and y = e^x are inverse functions._______

2. The point (b, 1) is on the graph of f(x) = log _bx._______

3. y = 10^-x and y = (0.1)^x are the same function._______

4. If f(x) = e^x2 - 1, then f(x) = 1 when x = ±ln ._______

5. 4^x/2 = 2^x_________

6.

7. 2^x + 2^-x = (2 + 2^-1)^x_________

8. 2^3+3x = 8^1+x_______

9. -ln 2 = ln ()______

10.

11. ln (ln e) = 1________

12.

13. ln (e + e) = 1 + ln 2_________

14. log₆(36)^-1 = -2________

B. Fill in the Blanks_________________________________

In Problems 1-22, fill in the blanks.

1. The graph of y = 6 - e^-x has the y-intercept ____________and horizontal asymptote y =__________.

2. The x-intercept of the graph of y = -10 + 10^5x is_____________.

3. The graph of y = ln (x + 4) has the x-intercept __________ and vertical asymptote x = __________.

4. The y-intercept of the graph of y = log₈(x + 2) is__________.

5. log₅ 2 - log₅10 =________

6.

7. e^{3 ln 10} = ________

8. 10^{log₁₀4.89} = _______

9. log₄(4 · 4² · 4³) = ________

10. = _____

11. If log₃N = -2, then N = _________.

12. If log _b 6 = , then b = _______.

13. If ln e³ = y, then y =_________.

14. If ln 3 + ln(x - 1) = ln 2 + ln x, then x = _______.

15. If -1 + ln(x - 3) = 0, then x = ________.

16. If ln (ln x) = 1, then x = _______.

17. If 100 - 20e^-015t = 35, then to four rounded decimals t = ________.

18. If 3^x = 5, then 3^-2x = __________.

19. f(x) = 4^3x = (______)^x

20. f(x) = (e²)^x/6 = (______)^x

21. If the graph of y = e^x-2 + C passes through (2, 9), then C = ______.

22. By rigid transformations, the point (0, 1) on the graph of y = e^x is moved to the point ________ on the graph of y = 4 + e^{x - 3}.

C. Exercises____________________________________________________________

In Problems 1 and 2, rewrite the given exponential expression as an equivalent logarithmic expression.

1. 5^-1 = 0.2

2.

In Problems 3 and 4, rewrite the given logarithmic expression as an equivalent exponential expression.

3. log₉27 = 1.5

4. log₆(36)^-2 = -4

In Problems 5-12, solve for x.

5. 2^1-x = 8

6. 3^2x = 81

7. e ^1-2x = e²

8. e^x2 - e⁵e^x-1 = 0

9. 2^1-x = 7

10. 3^x = 7^x-1

11. e^x+2 = 6

12. 3e^x = 4e^-3x

In Problems 13 and 14, solve for the indicated variable

13. P = Se^-rm; for m

14. for t

In Problems 15 and 16, graph the given functions on the same coordinate axes.

15. y = 4^x, y = log₄x

16. y = ()^x, y = log_1/2x

17. Match the letter of the graph in Figure 6.R.1 with the appropriate function.

Figure 6.R.1 Graphs for Problem 17

(i) f(x) = b^x, b > 2

(ii) f(x) = b^x, 1 < b < 2

(iii) f(x) = b^x, < b < 1

(iv) f(x) = b^x, 0 < b <

18. In Figure 6.R.2, fill in the blanks for the coordinates of the points on each graph.

Figure 6.R.2 Graphs for Problem 18

In Problems 19 and 20, find the slope of the secant line L given in each figure.

19.

Figure 6.R.3 Graph for Problem 19

20.

Figure 6.R.4 Graph for Problem 20

In Problems 21-26, match each of the following functions with one of the given graphs.

(i) y = ln (x - 2)

(ii) y = 2 - ln x

(iii) y = 2 + ln(x + 2)

(iv) y = -2 - ln(x + 2)

(v) y = -ln(2x)

(vi) y = 2 + ln(-x + 2)

21.

Figure 6.R.5 Graph for Problem 21

22.

Figure 6.R.6 Graph for Problem 22

23.

Figure 6.R.7Graph for Problem 23

24.

Figure 6.R.8Graph for Problem 24

25.

Figure 6.R.9Graph for Problem 25

26.

Figure 6.R.10Graph for Problem 26

In Problems 27 and 28, in words describe the graph of the function f in terms of a transformation of the graph of y = ln x.

27. f(x) = ln ex

28. f(x) = ln x³

29. Find a function f(x) = Ae^kx if (0, 5) and (6, 1) are points on the graph of f.

30. Find a function f(x) = A10^kx if f(3) = 8 and f(0) =

31. Find a function f(x) = a + b^x, 0 < b < 1, if f(1) = 5.5 and the graph of f has a horizontal asymptote y = 5.

32. Find a function f(x) = a + log₃(x - c) if f(11) = 10 and the graph of f has a vertical asymptote x = 2.

33. Doubling Time If the initial number of bacteria present in a culture doubles after 9 hours, how long will it take for the number of bacteria in the culture to double again?

34. Got Bait? A commercial fishing lake is stocked with 10,000 fingerlings. Find a model P(t) = P₀e^kt for the fish population of the lake at time t if the owner of the lake estimates that there will be 5000 fish left after 6 months. After how many months does the model predict that there will be 1000 fish left?

35. Radioactive Decay Tritium, an isotope of hydrogen, has a half-life of 12.5 years. How much of an initial quantity of this element remains after 50 years?

36. Old Bones It is found that 97% of C-14 has been lost in a human skeleton found at an archeological site. What is the approximate age of the skeleton?

37. Wishful Thinking A person facing retirement invests $650,000 in a savings account. She wants the account to be worth $1,000,000 in 10 years. What annual rate r of interest compounded continuously will achieve this dream?

38. Light Intensity According to the Beer-Lambert-Bouguer law, the intensity I (measured in lumens) of a vertical beam of light passing through a transparent substance decreases according to the exponential function I(x) = I₀e^kx, k < 0, where I₀ is the intensity of the incident beam and x is the depth measured in meters. If the intensity of light 1 meter below the surface of water is 30% of I₀, what is the intensity 3 meters below the surface?

39. The graph of the function y = ae^-be-cx is called a Gompertz curve. Solve for x in terms of the other symbols.

40. If a > 0and b > 0, then show that log _a² b² = log _a b.

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*In biology the doubling time is sometimes referred to as the generation time.