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Section 7.3 Logarithms

In this section, we introduce a new mathematical tool called a logarithm, which will help us solve exponential equations.

Subsection A Logarithm is an Exponent

Suppose that a colony of bacteria doubles in size every day. If the colony starts with 50 bacteria, how long will it be before there are 800 bacteria? We answer questions of this type by writing and solving an exponential equation. The function

\begin{equation*} P(t) = 50 \cdot 2^t \end{equation*}

gives the number of bacteria present on day \(t\text{,}\) so we must solve the equation

\begin{equation*} 800 = 50 \cdot 2^t \end{equation*}

We are looking for an unknown exponent on base 2. Dividing both sides by 50 yields

\begin{equation*} 16 = 2^t \end{equation*}

This equation asks the question:

To what power must we raise \(2\) in order to get \(16\text{?}\)

Because \(2^4 = 16\text{,}\) we see that the solution of the equation is 4. You can check that\(t=4\) solves the original problem:

\begin{equation*} P(\alert{4}) = 50 \cdot 2^{\alert{4}} = 800 \end{equation*}

The unknown exponent that solves the equation \(16=2^t\) is called the base \(2\) logarithm of \(16\text{.}\) The exponent in this case is \(4\text{,}\) and we write this fact as

\begin{equation*} \log_{2}16 = 4 \end{equation*}

So, we solve an exponential equation by computing a logarithm. We make the following definition.

Definition 7.39. Definition of Logarithm.
(logarithm)

For \(b\gt 0, b\ne 1\text{,}\) the base \(b\) logarithm of \(x\), written \(\log_{b} x\text{,}\) is the exponent to which \(b\) must be raised in order to yield \(x\text{.}\)

Example 7.40.

Compute the logarithms.

  1. \(\log_3 9 = 2\) because \(3^2=9\)

  2. \(\log_5 125 = 3\) because \(5^3=125\)

  3. \(\log_4 \dfrac{1}{16} = -2\) because \(4^{-2}=\dfrac{1}{16}\)

  4. \(\log_5 \sqrt{5} = \dfrac{1}{2}\) because \(5^{1/2} = \sqrt{5}\)

The positive constant \(b\) is called the base of the exponential function.

From the definition of a logarithm and the examples above, we see that the following two statements are equivalent.

Logarithms and Exponents: Conversion Equations.

If \(b \gt 0\text{,}\) \(b\ne 1\text{,}\) and \(x \gt 0\text{,}\)

\begin{equation*} \blert{y = \log_b x}~~~ \text{ if and only if }~~~ \blert{ x = b^y} \end{equation*}

This equivalence tells us that the logarithm, \(y\text{,}\) is the same as the exponent in \(x = b^y\text{.}\) We see again that a logarithm is an exponent; it is the exponent to which \(b\) must be raised to yield \(x\text{.}\)

The conversion equations allow us to convert from logarithmic to exponential form, or vice versa. You should memorize the conversion equations, because we will use them frequently.

As special cases of the equivalence above, we can compute the following useful logarithms.

For any base \(b \gt 0, b\ne 1\text{,}\)

Some Useful Logarithms.

For any base \(b \gt 0, b\ne 1\text{,}\)

\begin{align*} \log_b b \amp = 1~~~ \text{ because } ~~~b^1 = b\\ \log_b 1 \amp = 0 ~~~ \text{ because } ~~~b^0 = 1\\ \log_b{b^x} \amp = x~~~ \text{ because } ~~~b^x = b^x \end{align*}
Example 7.42.

  1. \(\displaystyle \log_{2}{2} = 1\)

  2. \(\displaystyle \log_{5}{1} = 0\)

  3. \(\displaystyle \log_{3}{3^4} = 4\)

Subsection Logs and Exponential Equations

We use logarithms to solve exponential equations, just as we use square roots to solve quadratic equations. Consider the two equations

\begin{equation*} x^2 = 25 ~~~~ \text{ and } ~~~~ 2^x = 8 \end{equation*}

We solve the first equation by taking a square root, and we solve the second equation by computing a logarithm:

\begin{equation*} x = \pm\sqrt{25} = \pm 5 ~~~~ \text{ and } ~~~~ x = \log_{2}{8} = 3 \end{equation*}

The operation of taking a base \(b\) logarithm is the inverse operation for raising the base \(b\) to a power, just as extracting square roots is the inverse of squaring a number.

Every exponential equation can be rewritten in logarithmic form by using the conversion equations. Thus,

\begin{equation*} 3 = \log_{2}{8}~~~~ \text{ and }~~~~ 8 = 2^3 \end{equation*}

are equivalent statements, just as

\begin{equation*} 5 = \sqrt{25}~~~~ \text{ and }~~~~ 25 = 5^2 \end{equation*}

are equivalent statements. Rewriting an equation in logarithmic form is a basic strategy for finding its solution.

Example 7.44.

Rewrite each equation in logarithmic form.

  1. \(\displaystyle 2^{-1} = \dfrac{1}{2}\)

  2. \(\displaystyle a^{1/5} = 2.8\)

  3. \(\displaystyle 6^{1.5} = T\)

  4. \(\displaystyle M^v = 3K\)

Solution

First identify the base \(b\text{,}\) and then the exponent or logarithm \(y\text{.}\) Use the conversion equations to rewrite \(b^y = x\) in the form \(\log_{b}{x} = y\text{.}\)

  1. The base is \(2\) and the exponent is \(-1\text{.}\) Thus, \(\log_{2}{\dfrac{1}{2}}= -1\text{.}\)

  2. The base is \(a\) and the exponent is \(\dfrac{1}{5}\text{.}\) Thus, \(\log_{a}{2.8} = \dfrac{1}{5}\text{.}\)

  3. The base is \(6\) and the exponent is \(1.5\text{.}\) Thus, \(\log_{6}{T} = 1.5\text{.}\)

  4. The base is \(M\) and the exponent is \(v\text{.}\) Thus, \(\log_{M}{3K} = v\text{.}\)

Subsection Approximating Logarithms

Now let's consider computing logarithms that are not obvious by inspection. Suppose we would like to solve the equation

\begin{equation*} 2^x = 26 \end{equation*}

The solution of this equation is \(x = \log_{2}{26}\text{,}\) but can we find a decimal approximation for this value? There is no integer power of 2 that equals 26, because

\begin{equation*} 2^4 = 16~~~~~ \text{and }~~~~ 2^5 = 32 \end{equation*}

So \(\log_{2}{26}\) must be between 4 and 5. We can use trial and error to find the value of \(\log_{2}{26}\) to the nearest tenth. Use your calculator to make a table of values for \(y = 2^x\text{,}\) starting with \(x = 4\) and using increments of 0.1.

\(x\) \(2^x\) \(x\) \(2^x\)
\(4\) \(2^4=16\) \(4.5\) \(2^{4.5}=22.627\)
\(4.1\) \(2^{4.1}=17.148\) \(4.6\) \(2^{4.6}=24.251\)
\(4.2\) \(2^{4.2}=18.379\) \(\alert{4.7}\) \(2^{\alert{4.7}}=25.992\)
\(4.3\) \(2^{4.3}=19.698\) \(\alert{4.8}\) \(2^{\alert{4.8}}=27.858\)
\(4.4\) \(2^{4.4}=21.112\) \(4.9\) \(2^{4.9}=29.857\)

From the table we see that 26 is between \(2^{4.7}\) and \(2^{4.8}\text{,}\) and is closer to \(2^{4.7}\text{.}\) To the nearest tenth, \(\log_{2}{26} \approx 4.7\text{.}\)

Trial and error can be a time-consuming process. In the Example below, we illustrate a graphical method for estimating the value of a logarithm.

Example 7.47.

Approximate \(\log_{3}{7}\) to the nearest hundredth.

Solution

If \(\log_{3}{7}=x\text{,}\) then \(3^x = 7\text{.}\) We will use the graph of \(y = 3^x\) to approximate a solution to \(3^x = 7\text{.}\)

We graph \(Y_1 =3\)^ X and \(Y_2 = 7\) in the standard window (ZOOM 6) to obtain the graph shown below. Next we activate the intersect feature to find that the two graphs intersect at the point \((1.7712437, 7)\text{.}\) Because this point lies on the graph of \(y = 3^x\) , we know that

\begin{equation*} 3^{1.7712437} \approx 7~~~\text{, or }~~~ \log_{3}{7} \approx 1.7712437 \end{equation*}

To the nearest hundredth, \(\log_{3}{7} \approx 1.77\text{.}\)

GC intersection of expnential curve and horizontal line

Subsection Base 10 Logarithms

Some logarithms are used so frequently in applications that their values are programmed into scientific and graphing calculators. These are the base 10 logarithms, such as

\begin{equation*} \log_{10}{1000} = 3 ~~~\text{ and }~~~ \log_{10}{0.01} = -2 \end{equation*}

Base 10 logarithms are called common logarithms, and the subscript 10 is often omitted, so that \(\log x\) is understood to mean \(\log_{10}{x}\text{.}\)

To evaluate a base 10 logarithm, we use the LOG key on a calculator. Many logarithms are irrational numbers, and the calculator gives as many digits as its display allows. We can then round off to the desired accuracy.

Example 7.50.

Approximate the following logarithms to 2 decimal places.

  1. \(\displaystyle \log{6.5}\)

  2. \(\displaystyle \log{256}\)

Solution

  1. The keying sequence LOG \(6.5\) )ENTER produces the display

    \(\log {(6.5)}\) \(\) \(\)
    \(\) \(\) \(.812913566\)

    so \(\log {6.5}\approx 0.81\text{.}\)

  2. The keying sequence LOG \(256\) ) ENTER yields \(2.408239965\text{,}\) so \(\log {256} \approx 2.41\text{.}\)

Note 7.51.

We can check the approximations found in Example 7.50 with our conversion equations. Remember that a logarithm is an exponent, and in this example the base is 10. We find that

\begin{align*} \amp\amp 10^{0.81} \amp\approx 6.45654229\\ \text{and} \amp\amp 10^{2.41} \amp\approx 257.0395783 \end{align*}

so our approximations are reasonable, although you can see that rounding a logarithm to 2 decimal places does lose some accuracy.

For this reason, rounding logarithms to 4 decimal places is customary.

Subsection Solving Exponential Equations

We can now solve any exponential equation with base 10.

Example 7.54.

Solve the equation \(~~38 = 95 - 15 \cdot 10^{0.4x}\)

Solution

First, we isolate the power of 10: We subtract 95 from both sides of the equation and divide by \(-15\) to obtain

\begin{align*} -57 \amp = -15 \cdot 10^{0.4x} \amp\amp \blert{\text{Divide by }-15.}\\ 3.8 \amp = 10^{0.4x} \end{align*}

Next, we convert the equation to logarithmic form as

\begin{equation*} \log_{10}{3.8} = 0.4x \end{equation*}

Solving for \(x\) yields

\begin{equation*} \frac{\log_{10}{3.8}}{0.4}= x \end{equation*}

We can evaluate this expression on the calculator by entering

LOG \(3.8\) ) ÷ \(0.4\) ENTER

which yields \(1.449458992\text{.}\) Thus, to four decimal places, \(x \approx 1.4495\text{.}\)

Caution 7.55.

Do not omit the parenthesis when entering the expression in Example 5. Without the parenthesis, you are calculating \(\log_{10}{\dfrac{3.8}{0.4}}\text{.}\) You can check that this is not the same as \(\dfrac{\log_{10}{3.8}}{0.4}\text{.}\)

To solve exponential equations involving powers of 10, we can use the following steps.

Steps for Solving Exponential Equations.
  1. Isolate the power on one side of the equation.
  2. Rewrite the equation in logarithmic form.
  3. Use a calculator, if necessary, to evaluate the logarithm.
  4. Solve for the variable.

Subsection Application to Exponential Models

We have seen that exponential functions are used to describe some applications of growth and decay, \(P(t) = P_0b^t\text{.}\) There are two common questions that arise in connection with exponential models:

  1. Given a value of \(t\text{,}\) what is the corresponding value of \(P(t)\text{?}\)

  2. Given a value of \(P(t)\text{,}\) what is find the corresponding value of \(t\text{?}\)

To answer the first question, we evaluate the function \(P(t)\) at the appropriate value. To answer the second question, we must solve an exponential equation, and this usually involves logarithms.

Example 7.58.

The value of a large tractor originally worth $30,000 depreciates exponentially according to the formula

\begin{equation*} V(t) = 30,000(10)^{-0.04t} \end{equation*}

where \(t\) is in years. When will the tractor be worth half its original value?

Solution

We want to find the value of \(t\) for which \(V(t) = 15,000\text{.}\) That is, we want to solve the equation

\begin{align*} 15,000 \amp = 30,000(10)^{-0.04t} \amp \amp \blert{\text{Divide both sides by 30,000.}}\\ 0.5 \amp = 10^{-0.04t} \end{align*}

Once we have isolated the power, we convert the equation to logarithmic form.

\begin{align*} \log_{10}{0.5} \amp = -0.04t \amp \amp \blert{\text{Divide both sides by } -0.04.}\\ \frac{\log_{10}{0.5}}{-0.04} \amp = t \end{align*}

To evaluate this expression, we key in

LOG \(0.5\) ) ÷ (-) \(0.04\) ENTER

to find \(t \approx 7.525749892\text{.}\) The tractor will be worth $15,000 in approximately \(7\frac{1}{2}\) years.

At this stage, it seems we will only be able to solve exponential equations in which the base is 10. However, we will see in future sections how the properties of logarithms enable us to solve exponential equations with any base.

Exercises Problem Set 7.3

Warm Up
1.

There were 300 lizards on a small island in 2000, and since then the population has been growing by 15% each year.

  1. Write a formula for the size of the lizard population \(t\) years after 2000.
  2. Write an equation to calculate how long it will take the lizard population to reach 500.
  3. Use trial and error to estimate the solution to the equation.
2.

Complete each table. Then use the tables to approximate each logarithm between two integers.

  1. \(2^{-5}\) \(\hphantom{0000}\)
    \(2^{-4}\) \(\hphantom{0000}\)
    \(2^{-3}\) \(\hphantom{0000}\)
    \(2^{-2}\) \(\hphantom{0000}\)
    \(2^{-1}\) \(\hphantom{0000}\)
    \(2^0\) \(\hphantom{0000}\)
    \(2^1\) \(\hphantom{0000}\)
    \(2^2\) \(\hphantom{0000}\)
    \(2^3\) \(\hphantom{0000}\)
    \(2^4\) \(\hphantom{0000}\)
    \(2^5\) \(\hphantom{0000}\)
    \begin{gather*} \underline{\hspace{1.363636363636364em}} \lt \log_2{12} \lt \underline{\hspace{1.363636363636364em}}\\ \underline{\hspace{1.363636363636364em}} \lt \log_2{\dfrac{1}{6}} \lt \underline{\hspace{1.363636363636364em}}\\ \underline{\hspace{1.363636363636364em}} \lt \log_2{0.02} \lt \underline{\hspace{1.363636363636364em}} \end{gather*}
  2. \(10^{-5}\) \(\hphantom{0000}\)
    \(10^{-4}\) \(\hphantom{0000}\)
    \(10^{-3}\) \(\hphantom{0000}\)
    \(10^{-2}\) \(\hphantom{0000}\)
    \(10^{-1}\) \(\hphantom{0000}\)
    \(10^0\) \(\hphantom{0000}\)
    \(10^1\) \(\hphantom{0000}\)
    \(10^2\) \(\hphantom{0000}\)
    \(10^3\) \(\hphantom{0000}\)
    \(10^4\) \(\hphantom{0000}\)
    \(10^5\) \(\hphantom{0000}\)
    \begin{gather*} \underline{\hspace{1.363636363636364em}} \lt \log_{10}{296} \lt \underline{\hspace{1.363636363636364em}}\\ \underline{\hspace{1.363636363636364em}} \lt \log_{10}{0.095} \lt \underline{\hspace{1.363636363636364em}}\vphantom{\dfrac{1}{6}}\\ \underline{\hspace{1.363636363636364em}} \lt \log_{10}{3.8} \lt \underline{\hspace{1.363636363636364em}} \end{gather*}
Skills Practice

For Problems 3-8, use the definition to find each logarithm without using a calculator.

3.

  1. \(\displaystyle \log_7{49}\)

  2. \(\displaystyle \log_5{625} \)

4.

  1. \(\displaystyle \log_4{4}\)

  2. \(\displaystyle \log_6{1} \)

5.

  1. \(\displaystyle \log_3{\sqrt{3}}\)

  2. \(\displaystyle \log_3{\dfrac{1}{3}}\)

6.

  1. \(\displaystyle \log_9{9^{-6}}\)

  2. \(\displaystyle \log_8{8^5}\)

7.

  1. \(\displaystyle \log_{10}{0.1}\)

  2. \(\displaystyle \log_{10}{10,000}\)

8.

  1. \(\displaystyle \log_{10}{0.001}\)

  2. \(\displaystyle \log_{10}{1000}\)

For Problems 9 and 10, without using a calculator, estimate the log between two integers.

9.

  1. \(\displaystyle x=\log_{10}{5678}\)

  2. \(\displaystyle y=\log_{10}{0.25}\)

10.

  1. \(\displaystyle \log_{10}{137,624}\)

  2. \(\displaystyle \log_{10}{0.009}\)

For Problems 11–16, solve. Round your answers to hundredths.

11.

\(10^{-3x}=5\)

12.

\(25 \cdot 10^{0.2x} = 80\)

13.

\(12.2 = 2(10^{1.4x})-11.6\)

14.

\(16.1 = 28.2 - 4 \left(10^{-0.7x}\right)\)

15.

\(3\left(10^{-1.5x}\right)-14.7=17.1 \)

16.

\(80\left(1-10^{-0.2x}\right) =65 \)

For Problems 17 and 18, rewrite each equation in logarithmic form.

17.

  1. \(\displaystyle t^{3/2} = 16\)

  2. \(\displaystyle 0.8^{1.2} = M\)

18.

  1. \(\displaystyle 3.7^{2.5} = Q\)

  2. \(\displaystyle 3^{-0.2t} = 2N_0\)

For Problems 19 and 20, rewrite each equation in exponential form.

19.

  1. \(\displaystyle \log_{16}{256} = w\)

  2. \(\displaystyle \log_b{9} = -2\)

20.

  1. \(\displaystyle \log_{10}{a} = -2.3\)

  2. \(\displaystyle \log_4 {36} = 2q-1\)

21.

Solve each equation, writing your answer as a logarithm. Then use trial and error to approximate the logarithm to tenths.

  1. \(\displaystyle 4^x = 2.5\)
  2. \(\displaystyle 2^x = 0.2\)

For Problems 22–24, use a graph to approximate each logarithm to the nearest hundredth. (Hint: Graph an appropriate function \(y=b^x\text{.}\))

22.

\(\log_3 {67.9}\)

23.

\(\log_5{86.3}\)

24.

\(\log_{10}{50}\)

Applications
25.

In 2015, Summit City used 4.2 million kilowatt-hours of electricity, and the demand for electricity has increased every year according to the formula

\begin{equation*} E(t) = 4.2(10^{0.0261t}) \end{equation*}
  1. When will Summit City need 10 million kilowatt-hours annually?
  2. What is the annual growth rate in the demand for electricity?
26.

Radium, a radioactive element, was used to paint watches and instrument dials until its serious health effects were discovered in the 1920's. Radium-226 is the most stable isotope of radium and decays according to the formula

\begin{equation*} R(t)=R_0 (10^{-1.0177t}) \end{equation*}

where \(t\) is in years.

  1. How long will it take for one gram of radium-226 to decay to one-half gram?
  2. What is the annual decay rate for radium-226?

The atmospheric pressure decreases with altitude above the surface of the Earth, according to the function

\begin{equation*} P(h) = 30 (10)^{-0.09h} \end{equation*}

where \(h\) is altitude given in miles, and \(P\) is atmospheric pressure in inches of mercury. Graph this function in the window

\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 9.4\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} = 30 \end{align*}

Solve Problems 27 and 28 algebraically, then verify with your graph.

27.

  1. The elevation of Mount Everest, the highest mountain in the world, is 29,028 feet. What is the atmospheric pressure at the top?

    Hint: 1 mile = 5280 feet

  2. The atmospheric pressure at the top of Mount McKinley, the highest peak in the United States, is 13.51 inches of mercury. Estimate the elevation of Mount McKinley.
28.
  1. What is the atmospheric pressure at sea level (\(h=0\))?
  2. Find the height above sea level at which the atmospheric pressure is equal to one-half the pressure at sea level.
29.

From 1950 to 2000, the population of the state of Nevada increased according to the formula

\begin{equation*} P(t) = 162,000 (10)^{0.02191t} \end{equation*}

where \(t\) is measured in years since 1950.

  1. What was the population in 2000?

  2. What was the annual percent growth rate from 1950 to 2000?

  3. According to this model, when would the population of Nevada reach 3,000,000? (In fact the population of Nevada was 3,139,658 in 2020.)

30.

From 1970 to 1990, the population of the state of California increased according to the formula

\begin{equation*} P(t) = 19,971,000 (10)^{0.0088t} \end{equation*}

where \(t\) is measured in years since 1970.

  1. What was the population in 1990?

  2. What was the annual percent growth rate from 1970 to 1990?

  3. According to this model, when would the population of California reach 40,000,000? (In fact the population of California was 39,747,267 in 2020.)

Equations Practice: For Problems 31 and 32, solve the equation and check your solutions.

31.

  1. \(\displaystyle x^2=16 \)

  2. \(\displaystyle 2^x = 16\)

  3. \(\displaystyle x^{2/3}=16 \)

  4. \(\displaystyle x^3 = 16\)

32.

  1. \(\displaystyle 2^{3x-2}=16 \)

  2. \(\displaystyle (3x-2)^2 = 16\)

  3. \(\displaystyle x^2-6x = 16 \)

  4. \(\displaystyle x^2-3x = 16\)

33.

  1. Complete the table.

    \(x\) \(x^2\) \(\log_{10}{x}\) \(\log_{10}{x^2}\)
    \(1\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(2\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(6\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  2. Do you notice a relationship between \(\log_{10}{x}\) and \(\log_{10}{x^2}\text{?}\) State the relationship as an equation.
34.

  1. Complete the table.

    \(x\) \(\dfrac{1}{x}\) \(\log_{10}{x}\) \(\log_{10}{\dfrac{1}{x}}\)
    \(1\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(2\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(6\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  2. Do you notice a relationship between \(\log_{10}{x}\) and \(\log_{10}{\dfrac{1}{x}}\text{?}\) State the relationship as an equation.