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Section 4.3 Rules of Logarithms

Before Starting this Section, Review

1 Definition of logarithm (Section 4.2 , page 445)
2 Basic properties of logarithms (Section 4.2 , page 447)
3 Rules of exponents (Section 4.1 , page 425)

Objectives

1 Learn the rules of logarithms.
2 Estimate a large number.
3 Change the base of a logarithm.
4 Apply logarithms to growth and decay.

The Boy King Tut

Today the most famous pharaoh of Ancient Egypt is King Nebkheperure Tutankhamun, popularly called “King Tut.” He was only 9 years old when he became a pharaoh. In 2005, a team of Egyptian scientists headed by Dr. Zahi Hawass determined that the pharaoh was 19 years old when he died (around 1346 b.c.).

Tutankhamun was a short-lived boy king who, unlike the great Egyptian kings Khufu (builder of the Great Pyramid), Amenhotep III (builder of temples throughout Egypt), and Ramesses II (prolific builder and usurper), accomplished nothing significant during his reign. In fact, little was known about him prior to Howard Carter’s discovery of his tomb (and the amazing treasures it held) in the Valley of the Kings on November 4, 1922. Carter, with his benefactor Lord Carnarvon at his side, entered the tomb’s burial chamber on November 26, 1922. Lord Carnarvon died seven weeks after entering the burial chamber, giving rise to the theory of the “curse” of King Tut.

Work on the tomb continued until 1933. The tomb contained a pristine mummy of an Egyptian king, lying intact in his original burial furniture. He was accompanied by a small slice of the royal world of the pharaohs: golden chariots, statues of gold and ebony, a fleet of miniature ships to accommodate his trip to the hereafter, his throne of gold, bottles of perfume, precious jewelry, and more. The “Treasures of Tutankhamun” exhibition, first shown at the British Museum in London in 1972, traveled to many countries, including the United States.

In Example 8, we discuss the age of a work of art found in King Tut’s tomb.

Rules of Logarithms

1 Learn the rules of logarithms.

Recall the basic properties of logarithms from Section 4.2 for $a > 0, a \neq 1$ :

$\log_{a} a = 1$ (7)

$\log_{a} 1 = 0$ (8)

$\log_{a} a^{x} = x, x real$ (9)

$a^{\log_{a} x} = x, x > 0$ (10)

In this section, we will discuss some important rules of logarithms that are helpful in calculations, simplifications, and applications.

Rules of Logarithms

Let M, N, and a be positive real numbers with $a \neq 1$ and let r be any real number.

Rule

Description

Examples

1. Product Rule:

$\log_{a} (M N) = \log_{a} M + \log_{a} N$

The logarithm of the product of two (or more) numbers is the sum of the logarithms of the numbers.

$\ln (5 \cdot 7) = \ln 5 + \ln 7$

$\log (3 x) = \log 3 + \log x$

$\log_{2} (5 \cdot 17) = \log_{2} 5 + \log_{2} 17$

2. Quotient Rule:

$\log_{a} (\frac{M}{N}) = \log_{a} M - \log_{a} N$

The logarithm of the quotient of two numbers is the difference of the logarithms of the numbers.

$\ln \frac{5}{7} = \ln 5 - \ln 7$

$\log \frac{5}{x} = \log 5 - \log x$

3. Power Rule:

$\log_{a} M^{r} = r \log_{a} M$

The logarithm of a number to the power r is r times the logarithm of the number.

$\ln 5^{7} = 7 \ln 5$

$\log 5^{3 / 2} = \frac{3}{2} \log 5$

$\log_{2} 7^{- 3} = - 3 \log_{2} 7$

Rules 1, 2, and 3 follow from the corresponding rules of the exponents:

$\begin{array}{rcll} a^{x} \cdot a^{y} & = & a^{x + y} & Product rule \\ \frac{a^{x}}{a^{y}} & = & a^{x - y} & Quotient rule \\ {(a^{x})}^{r} & = & a^{x r} & Power rule \end{array}$

For example, to prove the product rule for logarithms, we let

$\log_{a} M = x and \log_{a} N = y .$

The corresponding exponential forms of these equations are

$M = a^{x}^{} and N = a^{y} .$

Then

$\begin{array}{rcll} M N & = & a^{x} \cdot a^{y} \\ M N & = & a^{x + y} & Product rule of exponents \\ \log_{a} M N & = & x + y & Logarithmic form \\ \log_{a} M N & = & \log_{a} M + \log_{a} N & Replace x with \log_{a} M and y with \log_{a} N . \end{array}$

This proves the product rule of logarithms. In Exercise 126, you are asked to prove the quotient rule and the power rule similarly by using the corresponding rules for exponents. The following table compares the rules of exponents and logarithms.

Compare the Rules of Exponents and Logarithms

Exponents	Logarithms
1. $a^{x} \cdot a^{y} = a^{x + y}$	$\log_{a} x y = \log_{a} x + \log_{a} y$
2. $\frac{a^{x}}{a^{y}} = a^{x - y}$	$\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$
3. ${(a^{x})}^{y} = a^{x y}$	$\log_{a} x^{y} = y \log_{a} x$

Example 1 Using Rules of Logarithms to Evaluate Expressions

Given that $\log_{5} z = 3$ and $\log_{5} y = 2$ , evaluate each expression.

$\log_{5} (y z)$
$\log_{5} (125 y^{7})$
$\log_{5} \sqrt{\frac{z}{y}}$
$\log_{5} (z^{1 / 30} y^{5})$

Solution

$\begin{array}{rcll} \log_{5} (y z) & = & \log_{5} y + \log_{5} z & Product rule \\ = & 2 + 3 = 5 & Use the given values and simplify . \end{array}$
$\begin{array}{rcll} \log_{5} (125 y^{7}) & = & \log_{5} 125 + \log_{5} y^{7} & Product rule \\ = & \log_{5} 5^{3} + \log_{5} y^{7} & 125 = 5^{3} \\ = & 3 + 7 \log_{5} y & Power rule and \log_{a} a = 1 \\ = & 3 + 7 (2) = 17 & Use the given values and simplify . \end{array}$
$\begin{array}{rcll} \log_{5} \sqrt{\frac{z}{y}} & = & \log_{5} {(\frac{z}{y})}^{1 / 2} & Rewrite radical as exponent . \\ = & \frac{1}{2} \log_{5} \frac{z}{y} & Power rule \\ = & \frac{1}{2} (\log_{5} z - \log_{5} y) & Quotient rule \\ = & \frac{1}{2} (3 - 2) = \frac{1}{2} & Use the given values and simplify . \end{array}$
$\begin{array}{rcll} \log_{5} (z^{1 / 30} y^{5}) & = & \log_{5} z^{1 / 30} + \log_{5} y^{5} & Product rule \\ = & \frac{1}{30} \log_{5} z + 5 \log_{5} y & Power rule \\ = & \frac{1}{30} (3) + 5 (2) & Use the given values . \\ = & 0.1 + 10 = 10.1 & Simplify . \end{array}$

Practice Problem 1

Evaluate each expression for the given values of logarithms in Example 1 .
1. $\log_{5} (\frac{y}{z})$
2. $\log_{5} (y^{2} z^{3})$

In many applications in more advanced mathematics courses, the rules of logarithms are used in both directions; that is, the rules are read from left to right and from right to left. For example, by the product rule of logarithms, we have

$\log_{2} 3 x = \log_{2} 3 + \log_{2} x .$

The expression $\log_{2} 3 + \log_{2} x$ is the expanded form of $\log_{2} 3 x,$ while $\log_{2} 3 x$ is the condensed form, or the single logarithmic form of $\log_{2} 3 + \log_{2} x .$

Example 2 Writing Expressions in Expanded Form

Write each expression in expanded form.

$\log_{2} \frac{x^{2} {(x - 1)}^{3}}{{(2 x + 1)}^{4}}$
$\log_{c} \sqrt{x^{3} y^{2} z^{5}}$

Solution

$\begin{array}{rcll} \log_{2} \frac{x^{2} {(x - 1)}^{3}}{{(2 x + 1)}^{4}} & = & \log_{2} x^{2} {(x - 1)}^{3} - \log_{2} {(2 x + 1)}^{4} & Quotient rule \\ = & \log_{2} x^{2} + \log_{2} {(x - 1)}^{3} - \log_{2} {(2 x + 1)}^{4} & Product rule \\ = & 2 \log_{2} x + 3 \log_{2} (x - 1) - 4 \log_{2} (2 x + 1) & Power rule \end{array}$
$\begin{array}{rcll} \log_{c} \sqrt{x^{3} y^{2} z^{5}} & = & \log_{c} {(x^{3} y^{2} z^{5})}^{1 / 2} & \sqrt{a} = a^{1 / 2} \\ = & \frac{1}{2} \log_{c} (x^{3} y^{2} z^{5}) & Power rule \\ = & \frac{1}{2} [\log_{c} x^{3} + \log_{c} y^{2} + \log_{c} z^{5}] & Product rule \\ = & \frac{1}{2} [3 \log_{c} x + 2 \log_{c} y + 5 \log_{c} z] & Power rule \\ = & \frac{3}{2} \log_{c} x + \log_{c} y + \frac{5}{2} \log_{c} z & Distributive property \end{array}$

Practice Problem 2

Write each expression in expanded form.
1. $\ln \frac{2 x - 1}{x + 4}$
2. $\log \sqrt{\frac{4 x y}{z}}$

Example 3 Writing Expressions in Condensed Form

Write each expression in condensed form.

$\log 3 x - \log 4 y$
$2 \ln x + \frac{1}{2} \ln (x^{2} + 1)$
$2 \log_{2} 5 + \log_{2} 9 - \log_{2} 75$
$\frac{1}{3} [\ln x + \ln (x + 1) - \ln (x^{2} + 1)]$

Solution

$\log 3 x - \log 4 y = \log (\frac{3 x}{4 y}) Quotient rule$
$\begin{array}{rcll} 2 \ln x + \frac{1}{2} \ln (x^{2} + 1) & = & \ln x^{2} + \ln {(x^{2} + 1)}^{1 / 2} & Power rule \\ = & \ln (x^{2} \sqrt{x^{2} + 1}) & Product rule; \sqrt{a} = a^{1 / 2} \end{array}$
$\begin{array}{rcl} 2 \log_{2} 5 + \log_{2} 9 - \log_{2} 75 & = & \log_{2} 5^{2} + \log_{2} 9 - \log_{2} 75 & Power rule \\ = & \log_{2} (25 \cdot 9) - \log_{2} 75 & 5^{2} = 25; Product rule \\ = & \log_{2} \frac{25 \cdot 9}{75} & Quotient rule \\ = & \log_{2} 3 & \frac{25 \cdot 9}{75} = \frac{9}{3} = 3 \end{array}$
$\begin{array}{rcl} \frac{1}{3} [\ln x + \ln (x + 1) - \ln (x^{2} + 1)] & = & \frac{1}{3} [\ln x (x + 1) - \ln (x^{2} + 1)] \begin{array}{l} Product \\ rule \end{array} \\ = & \frac{1}{3} \ln [\frac{x (x + 1)}{x^{2} + 1}] Quotient rule \\ = & \ln \sqrt[3]{\frac{x (x + 1)}{x^{2} + 1}} Power rule; a^{1 / 3} = \sqrt[3]{a} \end{array}$

Practice Problem 3

Write the expression in condensed form:

$\frac{1}{2} [\log (x + 1) + \log (x - 1)] .$

Be careful when using the rules of logarithms to simplify expressions. For example, there is no property that allows you to rewrite $\log_{a} (x + y) .$

In general,

$\log_{a} (x + y) \neq \log_{a} x + \log_{a} y .$ (11)

To illustrate this statement, let $x = 100, y = 10,$ and $a = 10 .$ Then the value of the left side of (11) is

$\log (100 + 10) = \log (110) \approx 2.0414 . Use a calculator .$

The value of the right side of (11) is

$\begin{array}{rcll} \log 100 + \log 10 & = & \log 10^{2} + \log 10 \\ = & 2 \log 10 + \log 10 & Power rule \\ = & 2 (1) + 1 = 3. & \log 10 = \log_{10} 10 = 1 \end{array}$

Therefore, $\log (100 + 10) \neq \log 100 + \log 10 .$

Warning

To avoid common errors, be aware that in general,

$\begin{array}{l} \log_{a} (M + N) \neq \log_{a} M + \log_{a} N . \\ \log_{a} (M - N) \neq \log_{a} M - \log_{a} N . \\ (\log_{a} M) (\log_{a} N) \neq \log_{a} M N . \\ \log_{a} (\frac{M}{N}) \neq \frac{\log_{a} M}{\log_{a} N} . \\ \frac{\log_{a} M}{\log_{a} N} \neq \log_{a} M - \log_{a} N . \\ {(\log_{a} M)}^{r} \neq r \log_{a} M . \end{array}$

Number of Digits

2 Estimate a large number.

In estimating the magnitude of a positive number K, we look for the number of digits needed to express K. If K has a fractional part, then “number of digits” means the number of digits to the left of the decimal point. For example, 357.29 and 231.4796 are viewed as 3-digit numbers in this context.

A number K written in the form $K = s \times 10^{n}$ , where $1 \leq s < 10$ and n is an integer, is said to be in scientific notation. For example, $432.1 = 4.321 \times 10^{2}$ and $0.56 = 5.6 \times 10^{- 1}$ are both in scientific notation.

The exponent n represents the magnitude of K and is closely related to the common logarithm. In fact, we have

$\begin{array}{rcll} \log K & = & \log (s \times 10^{n}) & Take log of both sides of K = s \times 10^{n} . \\ = & \log s + \log 10^{n} & Product rule \\ = & \log s + n \log 10 & Power rule \\ = & \log s + n & \log 10 = \log_{10} 10 = 1 \end{array}$

$\begin{array}{rcccll} Because 1 & \leq & s & < & 10, we have \\ \log 1 & \leq & \log s & < & \log 10 & y = \log_{10} x is an increasing function . \\ 0 & \leq & \log s & < & 1 & \log 1 = 0 and \log 10 = 1 \end{array}$

This says that the exponent n in $K = s \times 10^{n}$ is the largest integer less than or equal to log K. We note that a number m is a 3-digit number if

$\begin{array}{l} 100 = 10^{2} & \leq & m & < & 10^{3} = 1000 \\ or & \log 10^{2} & \leq & \log m & < & \log 10^{3} & y = \log_{10} x is an increasing function . \\ 2 & \leq & \log m & < & 3 & \log 10^{x} = \log_{10} 10^{x} = x \end{array}$

Similarly, a number K has $(n + 1)$ digits if $n \leq \log K < n + 1 .$

For example, the number 8765.43, which is a 4-digit number, has its common logarithm between 3 and 4. Conversely, if log N is approximately 57.3 then N is a 58-digit number.

Example 4 Estimating a Large Number

Write an estimate of the number $K = e^{700}$ in scientific notation.

Solution

$\begin{array}{rcl} K & = & e^{700} \\ \log K & = & \log (e^{700}) \\ = & 700 \log e \\ \approx & 304.0061373 Use a calculator . \end{array}$

Because log K lies between the integers 304 and 305, the number K requires 305 digits to the left of the decimal point. Also, by definition of the common logarithm, we have

$\begin{array}{rcll} K & \approx & 10^{304.0061373} & Exponential form \\ = & 10^{0.0061373} \times 10^{304} & a^{m + n} = a^{m} \times a^{n} \\ \approx & 1.014232 \times 10^{304} & Use a calculator . \end{array}$

Practice Problem 4

Repeat Example 4 for $K = {(234)}^{567} .$

Change of Base

3 Change the base of a logarithm.

Calculators usually come with two types of log keys: a key for natural logarithms (base e, $LN$ ) and a key for common logarithms (base 10, $LOG$ ). These two types of logarithms are frequently used in applications. Sometimes, however, we need to evaluate logarithms for bases other than e and 10. The change-of-base formula helps us evaluate these logarithms.

Suppose we are given $\log_{b} x$ and we want to find an equivalent expression in terms of logarithms with base a. We let

$u = \log_{b} x .$ (12)

In exponential form, we have

$x = b^{u} .$ (13)

Then

$\begin{array}{rcll} \log_{a} x & = & \log_{a} b^{u} & Take \log_{a} of each side of (13) . \\ \log_{a} x & = & u \log_{a} b & Power rule \\ u & = & \frac{\log_{a} x}{\log_{a} b} & Solve for u . \\ \log_{b} x & = & \frac{\log_{a} x}{\log_{a} b} (14) & Substitute for u from (12) . \end{array}$

Equation (14) is the change-of-base formula. By choosing $a = 10$ and $a = e,$ we can state the change-of-base formula as follows:

Example 5 Using a Change of Base to Compute Logarithms

Compute $\log_{5} 13$ by changing to

common logarithms and
natural logarithms.

Solution

$\begin{array}{rcll} \log_{5} 13 & = & \frac{\log 13}{\log 5} & Change to base 10. \\ \approx & 1.59369 & Use a calculator . \end{array}$
$\begin{array}{rcll} \log_{5} 13 & = & \frac{\ln 13}{\ln 5} & Change to base e . \\ \approx & 1.59369 & Use a calculator . \end{array}$

Practice Problem 5

Find $\log_{3} 15 .$

Example 6 Modeling with Logarithmic Functions

Find the following.

An equation of the form $y = c + b \log_{} x$ whose graph contains the points (2, 3) and $(4, - 5) .$
An equation of the form $y = c + b \log_{a} x$ whose graph contains the points (2, 3) and $(4, - 5)$ where a, b, and c are integers.

Solution

Substitute (2, 3) and $(4, - 5)$ in the equation $y = c + b \log x$ to obtain:

$3 = c + b \log 2$ (15)

$- 5 = c + b \log 4$ (16)

Find b: Subtract equation (16) from equation (15) to obtain

$\begin{array}{l} 8 = b \log 2 - b \log 4 = b (\log 2 - \log 4) = b (\log \frac{2}{4}) = b \log (\frac{1}{2}) \\ 8 = b \log (\frac{1}{2}) = b \log 2^{- 1} = - b \log 2 \\ b = - \frac{8}{\log 2} Solve for b . \end{array}$

Find c:

$\begin{array}{rcll} c & = & 3 - b \log 2 & From equation (15) \\ = & 3 - (- \frac{8}{\log 2}) \log 2 & Replace b with - \frac{8}{\log 2} \\ = & 3 + 8 = 11 & Simplify . \\ So, y & = & 11 - \frac{8}{\log 2} \log x & Substitute for b and c . \end{array}$
Use the change-of-base formula to find a, b, and c with integer values.

$\begin{array}{rcll} y = 11 - \frac{8}{\log 2} (\log x) & = & 11 - 8 (\frac{\log x}{\log 2}) & Answer to part a \\ y & = & 11 - 8 \log_{2} x & \log_{2} x = \frac{\log x}{\log 2} \end{array}$

Then $a = 2, b = - 8,$ and $c = 11 .$ (Answers may vary.)

Practice Problem 6

Repeat Example 6 for the graph that contains the points (3, 3) and (9, 1).

Side Note

The change-of-base formula can be used to produce a variety of integer values for a and b in part b of Example 6. Because $8 \log_{2} x = 4 \log_{4} x$ we could also chose $a = 4, b = - 4 .$

Growth and Decay

4 Apply logarithms to growth and decay.

We express the exponential growth (or decay) formula (page 437) in an equivalent logarithmic form.

Growth and Decay Formula

Exponential Form	Logarithmic Form
$A (t) = A_{0} e^{k t}$	$\ln (\frac{A (t)}{A_{0}}) = k t$

$\begin{array}{rcl} where A (t) & = & the amount of substance (or population) at time t, \\ A_{0} & = & A (0) is the initial amount, and \\ t & = & time . \end{array}$

Growth occurs when $k > 0$ and decay occurs when $k < 0 .$

Half-Life

The half-life of any quantity whose value decreases with time is the time required for the quantity to decay to half its initial value.

Radioactive substances undergo exponential decay. Krypton-85, for example, has a half-life of about ten years; this means that any initial quantity of krypton-85 will dissipate or decay to half that amount in about ten years.

If h is the half-life of a substance undergoing exponential decay at the rate k, then any mass $A_{0}$ decays to $\frac{1}{2} A_{0}$ in time h. See Figure 4.18. This fact leads to a half-life formula.

Figure 4.18

Time `h` is the half-life for the substance with this decay curve

$\begin{array}{rcll} \ln (\frac{\frac{1}{2} A_{0}}{A_{0}}) & = & k h & t = h and A (h) = \frac{1}{2} A_{0} in the logarithmic form \\ \ln (\frac{1}{2}) & = & k h & Simplify by dividing out A_{0.} \\ - \ln 2 & = & k h & \ln (\frac{1}{2}) = \ln (2^{- 1}) = - \ln 2 \\ h & = & - \frac{\ln 2}{k} & Solve for h . \end{array}$

Example 7 Finding the Half-Life of a Substance

In an experiment, 18 grams of the radioactive element sodium-24 decayed to 6 grams in 24 hours. Find its half-life to the nearest hour.

Solution

$\begin{array}{rcll} \ln (\frac{A (t)}{A_{0}}) & = & k t & Logarithmic growth and decay model \\ \ln (\frac{6}{18}) & = & k (24) & A_{0} = 18, A (24) = 6 \\ \ln (\frac{1}{3}) & = & 24 k & Simplify . \\ - \ln 3 & = & 24 k & \ln (\frac{1}{3}) = \ln (3^{- 1}) = - \ln 3 \\ k & = & - \frac{\ln 3}{24} & Solve for k . \end{array}$

To find the half-life of sodium-24, we use the formula

$\begin{array}{l} h = - \frac{\ln 2}{k} & Half-life formula \\ h = - \frac{\ln 2}{(- \frac{\ln 3}{24})} & Replace k with - \frac{\ln 3}{24} . \\ h = \frac{24 \ln 2}{\ln 3} \approx 15 hours & Simplify; use a calculator . \end{array}$

Practice Problem 7

In an experiment, 100 grams of the radioactive element strontium-90 decayed to 66 grams in 15 years. Find its half-life to the nearest year.

Willard Frank Libby (1908–1980)

Libby was born in Grand Valley, Colorado. In 1927, he entered the University of California at Berkeley, where he earned his BSc and PhD degrees in 1931 and 1933, respectively. He then became an instructor in the Department of Chemistry at the University of California (Berkeley) in 1933. He was awarded a Guggenheim Memorial Foundation Fellowship in 1941, but the fellowship was interrupted when Libby went to Columbia University on the Manhattan District Project upon America’s entry into World War II.

Libby was a physical chemist and a specialist in radiochemistry, particularly in hot-atom chemistry, tracer techniques, and isotope tracer work. He became well known at the University of Chicago for his work on natural carbon-14 (a radiocarbon) and its use in dating archaeological artifacts and on natural tritium and its use in hydrology and geophysics. He was awarded the Nobel Prize in Chemistry in 1960.

Radiocarbon Dating

Ordinary carbon, called carbon-12 $({}^{12}C),$ is stable and does not decay. However, carbon-14 $({}^{14}C)$ is a form of carbon that decays radioactively with a half-life of 5700 years. Sunlight constantly produces carbon-14 in Earth’s atmosphere. When a living organism breathes or eats, it absorbs carbon-12 and carbon-14. After the organism dies, no more carbon-14 is absorbed; so the age of its remains can be calculated by determining how much carbon-14 has decayed. The method of radiocarbon dating was developed by the American scientist W. F. Libby.

Example 8 King Tut’s Treasure

In 1960, a group of specialists from the British Museum in London investigated whether a piece of art containing organic material found in Tutankhamun’s tomb had been made during his reign or (as some historians claimed) whether it belonged to an earlier period. We know that King Tut died in 1346 b.c. and ruled Egypt for ten years. What percent of the amount of carbon-14 originally contained in the object should be present in 1960 if the object was made during Tutankhamun’s reign?

Solution

Because the half-life h of carbon-14 is approximately 5700 years, we have

$\begin{array}{rcll} 5700 & = & - \frac{\ln 2}{k} & Half-life formula with h = 5700 \\ k & = & - \frac{\ln 2}{5700} & Solve for k . \\ \approx & - 0.0001216 & Use a calculator . \end{array}$

Substituting this value of k in the exponential form, we obtain

$A (t) = A_{0} e^{- 0.0001216 t}$ (17)

The time t that elapsed between King Tut’s death and 1960 is

$\begin{array}{rcll} t & = & 1960 + 1346 & 1346 B .C . to A .D . 1960 \\ = & 3306 \\ A (3306) & = & A_{0} e^{- 0.0001216 (3306)} & Replace t with 3306 in (17) . \\ \approx & 0.66897 A_{0} & Use a calculator . \end{array}$

Therefore, the percent of the original amount of carbon-14 remaining in the object (after 3306 years) is 66.897%.

Because King Tut began ruling Egypt ten years before he died, the time $t_{1}$ that elapsed from the beginning of his reign to 1960 is 3316 years.

$\begin{array}{rcll} A (3316) & = & A_{0} e^{- 0.0001216 (3316)} & Replace t with 3316 in (17) . \\ \approx & 0.66816 A_{0} & Use a calculator . \end{array}$

Therefore, a piece of art made during King Tut’s reign would have between 66.816% and 66.897% of carbon-14 remaining in 1960.

Practice Problem 8

Repeat Example 8 , assuming that the art object was made 194 years before King Tut’s death.

Section 4.3 Exercises

Concepts and Vocabulary

$\log_{a} M N = \underline{} + \underline{}$ .
$\underline{} = \log_{a} M - \log_{a} N .$
$\log_{a} M^{r} = \underline{}$ .
The change-of-base formula using base e is $\log_{a} M = \underline{}$ .
True or False. $\log_{a} (u + v) = \log_{a} u + \log_{a} v .$
True or False. $\log u - \log v = \frac{\log u}{\log v} .$
True or False. $\ln (2 \sqrt{x}) = \frac{1}{2} \ln (2 x) .$
True or False. $\log \frac{x}{10} = \log x - 1 .$

Building Skills

In Exercises 9–20, given that $\log x = 2, \log y = 3, \log 2 \approx 0.3,$ and $\log 3 \approx 0.48,$ evaluate each expression without using a calculator.

log 6
log 4
log 5 $[H i n t : 5 = \frac{10}{2}]$
log (3x)
$\log (\frac{2}{x})$
${\log x}^{2}$
$\log (2 x^{2} y)$
$\log x y^{3}$
$\log \sqrt[3]{x^{2} y^{4}}$
$\log (\log x^{2})$
$\log \sqrt[3]{48}$
$\log_{2} 3$

In Exercises 21–44, write each expression in expanded form. Assume that all expressions containing variables represent positive numbers.

$\ln [x (x - 1)]$
$\ln \frac{x - 2}{x + 3}$
$\ln \frac{x + 1}{{(x - 2)}^{2}}$
$\ln \frac{x (x + 1)}{{(x - 1)}^{2}}$
$\log_{a} \sqrt{x} y^{3}$
$\log_{a} \frac{3 x^{2}}{\sqrt{y}}$
$\log_{a} \sqrt[3]{\frac{x}{y}}$
$\log_{a} \sqrt[3]{\frac{x^{2}}{y^{5}}}$
$\log_{2} \sqrt[4]{\frac{x y^{2}}{8}}$
$\log \sqrt[3]{\frac{x^{2} y}{100}}$
$\log \frac{\sqrt{x^{2} + 1}}{x + 3}$
$\log \frac{x^{2}}{\sqrt{x + 1}}$
$\log_{3} \frac{(x - 1) (x + 1)}{x^{2} - 4}$
$\log_{4} {(\frac{x^{2} - 9}{x^{2} - 6 x + 8})}^{2 / 3}$
$\log_{b} x^{2} y^{3} z$
$\log_{b} \sqrt{x y z}$
$\ln [\frac{x \sqrt{x - 1}}{x^{2} + 2}]$
$\ln [\frac{\sqrt{x - 2} \sqrt[3]{x + 1}}{x^{2} + 3}]$
$\ln [\frac{{(x + 1)}^{2}}{(x - 3) \sqrt{x + 4}}]$
$\ln [\frac{2 x + 3}{{(x + 4)}^{2} {(x - 3)}^{4}}]$
$\ln [(x + 1) \sqrt{\frac{x^{2} + 2}{x^{2} + 5}}]$
$\ln [\frac{\sqrt[3]{2 x + 1} (x + 1)}{{(x - 1)}^{2} (3 x + 2)}]$
$\ln [\frac{x^{3} {(3 x + 1)}^{4}}{\sqrt{x^{2} + 1} {(x + 2)}^{- 5} {(x - 3)}^{2}}]$
$\ln [\frac{{(x + 1)}^{1 / 2} {(x^{2} - 2)}^{2 / 5}}{{(2 x - 1)}^{3 / 2} {(x^{2} + 2)}^{- 4 / 5}}]$

In Exercises 45–60, write each expression in condensed form.

$\log_{2} x + \log_{2} 7$
$\log_{2} x - \log_{2} 3$
$\log (3 x + 2) - \log x$
$2 \log x + \log y$
$\frac{1}{2} \ln x + 2 \ln y$
$3 \log x + 4 \log y$
$\frac{1}{2} \log x - \log y + \log z$
$\frac{1}{2} (\log x + \log y)$
$\frac{1}{5} (\log_{2} z + 2 \log_{2} y)$
$\frac{1}{3} (\log x - 2 \log y + 3 \log z)$
$\ln x + 2 \ln y + 3 \ln z$
$2 \ln x - 3 \ln y + 4 \ln z$
$3 \log x - \log y + \frac{1}{2} \log (x^{2} + 4)$
$\log x + \log (x^{2} + 1) - \log (x - 1) - \log (2 x^{2} + 3)$
$2 \ln x - \frac{1}{2} \ln (x^{2} + 1)$
$2 \ln x + \frac{1}{2} \ln (x^{2} - 1) - \frac{1}{2} \ln (x^{2} + 1)$

In Exercises 61–68, write an estimate of each number in scientific notation.

$e^{500}$
$π^{650}$
$324^{756}$
$723^{416}$
Which is larger, $234^{567}$ or $567^{234}$ ?
Which is larger, $4321^{8765}$ or $8765^{4321}$ ?
Find the number of digits in $17^{200} \cdot 53^{67} .$
Find the number of digits in $67^{200} \div 23^{150} .$

In Exercises 69–76, use the change-of-base formula and a calculator to evaluate each logarithm.

$\log_{2} 5$
$\log_{4} 11$
$\log_{1 / 2} 3$
$\log_{\sqrt{3}} 12.5$
$\log_{\sqrt{5}} \sqrt{17}$
$\log_{15} 123$
$\log_{2} 7 + \log_{4} 3$
$\log_{2} 9 - \log_{\sqrt{2}} 5$

In Exercises 77–84, find the value of each expression without using a calculator.

$\log_{3} \sqrt{3}$
$\log_{1 / 4} 4$
$\log_{3} (\log_{2} 8)$
$2^{\log_{2} 2}$
$5^{2 \log_{5} 3 + \log_{5} 2}$
$e^{3 \ln 2 - 2 \ln 3}$
$\log 4 + 2 \log 5$
$\log_{2} 160 - \log_{2} 5$

In Exercises 85–92, find an equation of the form $y = c + b \log_{a} x$ (a, b, and c integers) whose graph contains the two given points.

(10, 1) and (1, 2)
(4, 10) and (2, 12)
(e, 1) and (1, 2)
(3, 1) and (9, 2)
(5, 4) and (25, 7)
(4, 3) and (8, 5)
(2, 4) and (4, 9)
(1, 1) and (5, 7)

In Exercises 93–96, determine the half-life of each substance that decays from $A_{0}$ to A in time t.

$A_{0} = 50, A = 23, t = 12 years$
$A_{0} = 200, A = 65, t = 10 years$
$A_{0} = 10.3, A = 3.8, t = 15 hours$
$A_{0} = 20.8, A = 12.3, t = 40 minutes$

Applying the Concepts

In Exercises 97–100, assume the exponential growth model $A (t) = A_{0} e^{k t}$ or $\ln (\frac{A (t)}{A_{0}}) = k t$ and a world population of 7 billion in 2011.

Rate of population growth. If the world adds about 90 million people in 2012, what is the rate of growth?
Population. Use the rate of growth from Exercise 97 to estimate the year in which the world population will be
1. 12 billion.
2. 20 billion.
Population growth. If the population must stay below 20 billion during the next 100 years, what is the maximum acceptable annual rate of growth? (Round your answer to six decimal places.)
Population decline. If the world population must decline below 5 billion during the next 25 years, what must be the annual rate of decline? (Round your answer to six decimal places.)
Continuous compounding. If $1000 is deposited in a bank at 10% interest compounded continuously, how many years will it take for the money to grow to $3500?
Continuous compounding. At what rate of interest compounded continuously will an investment double in six years?
Half-life. Iodine-131, a radioactive substance that is effective in locating brain tumors, has a half-life of only eight days. A hospital purchased 20 grams of the substance but had to wait five days before it could be used. How much of the substance was left after five days?
Half-life. Plutonium-241 has a half-life of 13 years. A laboratory purchased 10 grams of the substance but did not use it for two years. How much of the substance was left after two years?
Half-life. Tritium is used in nuclear weapons to increase their power. It decays at the rate of 5.5% per year. Calculate the half-life of tritium.
Half-life. Sixty grams of magnesium-28 decayed into 7.5 grams after 63 hours. What is the half-life of magnesium-28?

Effective medicine dosage. Use the following model in Exercises 107–110: The concentration C(t) of a drug administered to a patient intravenously jumps to its highest level almost immediately. The concentration subsequently decays exponentially according to the law

$C (t) = C_{0} e^{- k t}, or equivalently, \ln (\frac{C (t)}{C_{0}}) = - k t,$

where $C (0) = C_{0} .$ The physician administering the drug would like to have

$m < C (t) < M,$

where m is the concentration below which the drug is ineffective and M is the concentration above which the drug is dangerous.

Drug concentration. Suppose that immediately after a certain drug is administered, the concentration is 5 milligrams per milliliter. Ten hours later the concentration drops to 1.5 milligrams per milliliter. Determine the value of k for this drug.
Drug concentration. For the drug of Exercise 107, suppose the maximum safe level is $M = 12$ milligrams per milliliter and the minimum effective level is $m = 3$ milligrams per milliliter. If the initial concentration is M, find the maximum possible time between doses of this drug.
Half-life. The half-life of Valium (diazepam) is 36 hours. Suppose a patient receives 16 milligrams of the drug at 8 a.m. How much Valium is in the patient’s blood at 4 p.m. the same day?
Half-life. The half-life of aspirin in the blood is 12 hours. Estimate the time required for the aspirin to decay to 90% of the original amount ingested.

In Exercises 111–114, use the following amortization formula:

$P = \frac{r \cdot M}{1 - {(1 + \frac{r}{n})}^{- n t}} \div n$

where $P = the$ payment, $r = the$ annual interest rate, $M = the$ mortgage amount, $t = the$ number of years, and $n = the$ number of payments per year.

Finding P. What is the monthly payment on a mortgage of $120,000 with a 6% interest rate for 20 years? How much interest will be paid over 20 years?
Finding t. The Garcia family wanted to take out a mortgage for $150,000 at 8% interest with monthly payments. The family can afford monthly payments of $1200. How long would they have to make payments to pay off their mortgage, and how much interest would they be paying?
Finding M. The First National Bank offers Andy an 8.5% interest rate on a 30-year mortgage to be paid back in monthly payments. The most Andy can afford to pay in monthly payments is $850. What mortgage amount can Andy afford?
Finding r. Lisa is 40 years old and needs to take out a mortgage of $120,000. She can afford monthly payments up to $850. She wants her mortgage paid off when she retires at age 65. What interest rate on the mortgage will satisfy all her requirements? [Hint: Start with $r = 7 %$ and compute the monthly payment, increasing r by 0.001 each time.]

Beyond the Basics

Show that $\log_{b} (\sqrt{x^{2} + 1} - x) + \log_{b} (\sqrt{x^{2} + 1} + x) = 0 .$
Show that $\log_{b} (\sqrt{x + 1} + \sqrt{x}) = - \log_{b} (\sqrt{x + 1} - \sqrt{x}) .$
Show that $(\log_{b} a) (\log_{a} b) = 1 .$
Show that $\log \frac{a}{b} + \log \frac{b}{a} = 0 .$
Write $\log_{2} x = \frac{\ln x}{\ln 2}$ by the change-of-base formula. Then use a graphing calculator to sketch the graph of $y = \log_{2} x .$
Sketch the graph of $y = \log_{5} (x + 3) .$
Simplify each expression.
1. $\log (\frac{a}{b}) + \log (\frac{b}{a}) + \log (\frac{c}{a}) + \log (\frac{a}{c})$
2. $\log (\frac{a^{2}}{b c}) + \log (\frac{b^{2}}{c a}) + \log (\frac{c^{2}}{a b})$
3. $\log_{2} 3 \cdot \log_{3} 4$
4. $\log_{a} b \cdot \log_{b} c \cdot \log_{c} a$
Find the number of digits in N if $\log_{2} (\log_{2} N) = 4 .$
Find the domain of $f (x) = \log_{4} (\log_{5} (\log_{3} (18 x - x^{2} - 77)))$
Let $f (x) = \log_{a} x .$ Show that
1. $f (\frac{1}{x}) = - f (x) = \log_{1 / a} x .$
2. $\frac{f (x + h) - f (x)}{h} = \log_{a} {(1 + \frac{h}{x})}^{1 / h}, h \neq 0 .$
State whether each of the following is true or false for all permissible values of the variable.
1. $\log (x + 2) = \log x + \log 2$
2. $\log 2 x = \log x + \log 2$
3. $\log_{2} 3 x = 3 \log_{2} x$
4. ${(\log_{3} 2 x)}^{4} = 4 \log_{3} 2 x$
5. $\log_{5} x^{2} = 2 \log_{5} | x |$
6. $\log \frac{x}{3} = \log x - \log 3$
7. $\log \frac{x}{4} = \frac{\log x}{\log 4}$
8. $\ln (\frac{1}{x}) = - \ln x$
9. $\log_{2} x^{2} = (\log_{2} x) (\log_{2} x)$
10. $\log | 10 x | = 1 + \log | x |$
1. Prove the quotient rule for logarithms:
  
  $\log_{a} \frac{M}{N} = \log_{a} M - \log_{a} N .$
2. Prove the power rule for logarithms: $\log_{a} M^{r} = r \log_{a} M .$
If $\log (\frac{a + b}{3}) = \frac{1}{2} (\log a + \log b),$ show that $a^{2} + b^{2} - 7 a b = 0 .$
Evaluate $\frac{1}{1 + \log_{c} a b} + \frac{1}{1 + \log_{a} b c} + \frac{1}{1 + \log_{b} c a} .$

[Hint: Convert to common logarithms and then simplify.]

Critical Thinking / Discussion / Writing

What went wrong? Find the error in the following argument:

$\begin{array}{rcl} 3 & < & 4 \\ 3 \log \frac{1}{2} & < & 4 \log \frac{1}{2} \\ \log {(\frac{1}{2})}^{3} & < & \log {(\frac{1}{2})}^{4} \\ \log \frac{1}{8} & < & \log \frac{1}{16} \\ \frac{1}{8} & < & \frac{1}{16} \\ 2 & < & 1 \end{array}$
By the power rule for logarithms, ${\log x}^{2} = 2 \log x .$ However, the graphs of $f (x) = {\log x}^{2}$ and $g (x) = 2 \log x$ are not identical. Explain why.
Prime Number. As of 2008, the largest known prime number was $2^{43112609} - 1 .$ How many digits does this prime number have?

[Hint: First, argue that if $p = 2^{m} - 1$ is a prime number, then p and $2^{m}$ have the same number of digits.]
Find the domain of $g (x) = \frac{1}{\log (1 - x)} + \frac{1}{\ln (x + 2)} .$

Getting Ready for the Next Section

In Exercises 133–136, simplify each expression and write your result without exponents.

$11 \cdot 3^{0}$
$- 4 \cdot 5^{x} \cdot 5^{- x}$
$4^{x} \cdot 2^{- 2 x + 1}$
${(7^{x})}^{2} \cdot {(7^{2})}^{- x}$

In Exercises 137–140, let $t = 5^{x}$ and write each expression in the form $a t^{2} + b t + c = 0,$ where a, b, and c are real numbers.

$5^{2 x} - 5^{x} = - 1$
$3 \cdot 5^{2 x} - 2 \cdot 5^{x} = 7$
$\frac{5^{x} + 3 \cdot 5^{- x}}{5^{x}} = \frac{1}{4}$
$\frac{5^{- x} + 5^{x}}{5^{- x} - 5^{x}} = \frac{1}{2}$

In Exercises 141–144, solve for x.

$2 x - (11 + x) = 8 x + (7 + 2 x)$
$4 x - 1 + (6 x - 2) = 3 - (5 x + 1)$
$x^{2} + 3 x - 1 = 3$
$2 x^{2} - 7 x = 3 x + 48$

In Exercises 145–148, solve for x.

$2 x < 7 + x$
$5 x - 2 \leq 19 - (1 - x)$
$12 x > 30 - 3 x$
$4 x - 17 \geq 6 - (5 + 2 x)$

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Table of Contents for Section 4.3 Rules of Logarithms

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Table of Contents for
Section 4.3 Rules of Logarithms