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 initial value of the population is \(P_0=50\text{,}\) and the growth factor is \(2\text{.}\) Thus, the function
\begin{equation*}
P(t) = 50 \cdot 2^t
\end{equation*}
gives the number of bacteria present on day \(t\text{,}\) and 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?"
Because \(2^4 = 16\text{,}\) we see that the solution of the equation is \(4\text{.}\) 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*}
In other words, we solved an exponential equation by computing a logarithm. We make the following definition.
Definition 7.3.1. Definition of Logarithm.
For \(b\gt 0~\) and \(~b\ne 1\text{,}\)
the base \(b\) logarithm of \(x\) is the exponent to which \(b\) must be raised in order to yield \(x\text{.}\)
We write the logarithm as \(\log_{b} {(x)}\text{.}\)
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.
Here are some examples of logarithms.
\(\log_3 {(9)} = \alert{2}~~~\) because \(~~~3^\alert{2}=9\)
\(\log_5 {(125)} = \alert{3}~~~\) because \(~~~5^\alert{3}=125\)
\(\log_4 {\left(\dfrac{1}{16}\right)} = \alert{-2}~~~\) because \(~~~4^{\alert{-2}}=\dfrac{1}{16}\)
\(\log_5 {\left(\sqrt{5}\right)} = \alert{\dfrac{1}{2}}~~~\) because \(~~~5^\alert{1/2} = \sqrt{5}\) \(~\alert{\text{[TK]}}~~\)
We see that in each example the logarithm is the exponent we need on the given base.
Checkpoint 7.3.4. Practice 1.
Find each logarithm.
\(\log_{3}{(81)}~~~\) (What exponent on 3 gives me 81?)
\(\log_{10}{\left(\dfrac{1}{1000}\right)}~~~\) (What exponent on 10 gives me \(\frac{1}{1000}\text{?}\))
Answer.
\(\displaystyle 4\)
\(\displaystyle -3\)
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 \(x\text{.}\)
For example, to convert the equation \(3 = \log_5 {(125)}\) to exponential form, we note that the base is \(b=5\) and the logarithm is \(y=3\text{,}\) so the exponent on base 5 will be 3, like this: \(125 = 5^3\text{.}\)
The conversion equations allow us to convert from logarithmic to exponential form, or vice versa. It will help you to become familiar with the conversion, because we will use it frequently.
As special cases of the equivalence above, we can compute the following useful logarithms.
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 7.3.5.
Here are some more examples. Once again, we are looking for the exponent on the given base \(b\) to get \(x\text{.}\)
\(\log_{2}{(2)} = 1~~\) because \(~~2^1=2\)
\(\log_{5}{(1)} = 0~~\) because \(~~5^0=1\)
\(\log_{3}{(3^4)} = 4~~\) because, well, \(~3^4=3^4\)
Checkpoint 7.3.6. QuickCheck 1.
\(\log_6{(216)}=\) because \(~~6^\fillinmath{XX} = 216\)
\(\log_8{(1)} =\) because \(~~8^\fillinmath{XX}= 1\)
\(\log_9{(9^5)} =\)
If \(\log_w{(87)} = p,\) then \(87 =\)
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 7.3.7.
Checkpoint 7.3.8. Practice 2.
Rewrite each equation in logarithmic form.
\(\displaystyle 8^{-1/3} = \dfrac{1}{2}\)
\(\displaystyle V=S^{3/2}\)
Answer.
\(\displaystyle \log_8 {\left(\dfrac{1}{2} \right)}=\dfrac{-1}{3} \)
\(\displaystyle \log_S {(V)} = \dfrac{3}{2}\)
We can solve an exponential equation by converting the equation to logarithmic form.
Example 7.3.9.
Solve each equation by converting to logarithmic form.
\(~\alert{\text{[TK]}}\)
\(\displaystyle 4^x = 64\)
\(\displaystyle 4^x = 60\)
Solution.
In this case we can see by inspection that the solution is \(3\text{,}\) or \(x = \log_4 {(64)}.\)
The log form is \(x = \log_4 {(60)}\text{,}\) which is not easy to evaluate without a calculator.
Checkpoint 7.3.10. Practice 3.
Solve each equation by converting to logarithmic form.
\(\displaystyle 18^x = 324\)
\(\displaystyle 5^x = 46\)
Answer.
\(\displaystyle x = \log_18 {(324)} = 2 \)
\(\displaystyle x = \log_5{(46)}\)
Checkpoint 7.3.11. QuickCheck 2.
Answer the question or complete the statement.
What is the inverse operation for squaring a number?
What is the inverse operation for raising to a power?
Every exponential equation can be rewritten in logarithmic form by using the .
A logarithm is an unknown .
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.
| \(\hphantom{00} x \hphantom{00}\) |
\(2^x\) |
|
\(\hphantom{00} x \hphantom{00}\) |
\(2^x\) |
| \(4\) |
\(2^4=16\) |
|
\(4.5\) |
\(\hphantom{00} 2^{4.5}=22.627 \hphantom{00} \) |
| \(4.1\) |
\(\hphantom{00} 2^{4.1}=17.148 \hphantom{00}\) |
|
\(\hphantom{00} 4.6 \hphantom{00}\) |
\(2^{4.6}=24.251\) |
| \(4.2\) |
\(2^{4.2}=18.379\) |
|
\(\alert{4.7}\) |
\(2^{\alert{4.7}}=25.992\) |
| \(4.3\) |
\(2^{4.3}=19.698\) |
|
\(\alert{4.8}\) |
\(2^{\alert{4.8}}=27.858\) |
| \(4.4\) |
\(2^{4.4}=21.112\) |
|
\(4.9\) |
\(2^{4.9}=29.857\) |
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 next Example we illustrate a graphical method for estimating the value of a logarithm.
Example 7.3.12.
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 same window 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 7.3.13. Practice 4.
Rewrite the equation \(3^x = 90\) in logarithmic form.
Use a graph to approximate the solution to the equation in part (a). Round your answer to three decimal places.
Answer.
\(\displaystyle \log_8{(90)}=x \)
\(\displaystyle x\approx 4.096 \)
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 7.3.14. QuickCheck 3.
If \(\log {(x)} = 2.5\text{,}\) which statement(s) about \(x\) are true?
\(\displaystyle x=25\)
\(x\) is between 20 and 30
\(x\) is between 100 and 1000
\(\displaystyle x=\dfrac{1}{2.5}=0.4\)
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 7.3.15.
Approximate the following logarithms to 2 decimal places.
\(\displaystyle \log{(6.5)}\)
\(\displaystyle \log{(256)}\)
Solution.
The keying sequence LOG \(6.5\) )ENTER yields \(.812913566\text{,}\) 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 7.3.17. Practice 5.
Approximate the logarithms to four decimal places.
\(\displaystyle \log {(0.2)}\)
\(\displaystyle \log {(846,000)}\)
Answer.
\(\displaystyle -0.6990\)
\(\displaystyle 5.9274\)
Checkpoint 7.3.18. QuickCheck 4.
Decide whether each statement is true or false.
The LOG key on a calculator computes logarithms base 2.
\(\log_5 {(100)}\) is a number between 2 and 3.
Rounding a logarithm to two decimal places gives very accurate results.
The value of \(\log_4 {(392)}\) is also the solution of \(4^x = 392\text{.}\)
