Subsection 7.3.1 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 7.3.1. Definition of 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 \(x\text{.}\)
Some logarithms, like some square roots, are easy to evaluate, while others require a calculator. We’ll start with the easy ones.
Example 7.3.2.
Compute the logarithms.
\(\log_3 9 = 2\) because \(3^2=9\)
\(\log_5 125 = 3\) because \(5^3=125\)
\(\log_4 \dfrac{1}{16} = 2\) because \(4^{2}=\dfrac{1}{16}\)
\(\log_5 \sqrt{5} = \dfrac{1}{2}\) because \(5^{1/2} = \sqrt{5}\)
Checkpoint 7.3.3. Practice 1.
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 7.3.4.
\(\displaystyle \log_{2}{2} = 1\)
\(\displaystyle \log_{5}{1} = 0\)
\(\displaystyle \log_{3}{3^4} = 4\)
Checkpoint 7.3.5. QuickCheck 1.
Subsection 7.3.2 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 7.3.6.
Checkpoint 7.3.7. Practice 2.
Checkpoint 7.3.8. QuickCheck 2.
Subsection 7.3.3 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 timeconsuming process. In the
Example below, we illustrate a graphical method for estimating the value of a logarithm.
Example 7.3.9.
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, \(\log_{3}{7} \approx 1.77\text{.}\)
Checkpoint 7.3.10. Practice 3.
Subsection 7.3.4 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 \(\log_{10}{x}\text{.}\)
Checkpoint 7.3.11. QuickCheck 3.
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 7.3.12.
Approximate the following logarithms to 2 decimal places.
\(\displaystyle \log{6.5}\)
\(\displaystyle \log{256}\)
Solution.

The keying sequence LOG \(6.5\) )ENTER produces the display
\(\log {(6.5)}\) 
\(\) 
\(\) 
\(\) 
\(\) 
\(.812913566\) 
so \(\log {6.5}\approx 0.81\text{.}\)
The keying sequence LOG \(256\) ) ENTER yields \(2.408239965\text{,}\) so \(\log {256} \approx 2.41\text{.}\)
Checkpoint 7.3.14. Practice 4.
Checkpoint 7.3.15. QuickCheck 4.
Subsection 7.3.6 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:
Given a value of \(t\text{,}\) what is the corresponding value of \(P(t)\text{?}\)
Given a value of \(P(t)\text{,}\) what is 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 7.3.20.
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.
Checkpoint 7.3.21. QuickCheck 6.
Checkpoint 7.3.22. Practice 6.
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.