Subsection 10.1.1 Logarithms and Exponents
Before we look at logarithmic functions, let’s quickly review exponents and logs. For a particular base, let’s say 5, taking a logarithm is the opposite operation for raising to a power. For example, if we raise base 5 to a power of \(\alert{2}\text{,}\) we get
\begin{equation*}
5^\alert{2} = 25 ~~~~ \text{and thus} ~~~~ \log_5 {(25)} = \alert{2}
\end{equation*}
We can say that a logarithm is actually an exponent. Asking for the log base 5 of 25 is asking "what power of 5, or what exponent on base 5 will give me 25?"
Example 10.1.1.
Write each logarithmic equation as an equivalent exponential equation.
\(\displaystyle \log_3 {(81)} = 4\)
\(\displaystyle \log_b {(32)} = 5\)
Solution.
The logarithm asks "To what power must I raise 3 to get 81?" The base is 3 and the logarithm (or exponent) is 4, so \(3^4 = 81\text{.}\)
The logarithm asks "To what power must I raise \(b\) to get 32?" The base is \(b\) and the logarithm (or exponent) is 5, so \(b^5 = 32\text{.}\)
For a more thorough review of logarithms you can refer to
Section 7.3.1.
Checkpoint 10.1.2. QuickCheck 1.
Which of these could you use to estimate the value of \(\log_5 {(378)}\) ?
Find multiples of 5.
Find the fifth root of 378.
Find powers of 5.
Divide 378 by 5.
Now we’ll consider functions defined in terms of logarithms, or logarithmic functions. For example,
\begin{equation*}
f(x)=\log_2 {(x)}
\end{equation*}
is a logarithmic function. In order to understand logarithmic functions better, we first investigate how they are related to more familiar functions, the exponential functions.
Subsection 10.1.2 Inverse of a Function
You know that raising to the \(n\)th power and taking \(n\)th roots are inverse operations. For example, if we first cube a number and then take the cube root of the result, we return to the original number.
\begin{align*}
x \amp = \alert{5}~~~~~~\rightarrow~~~~~~x^3=125~~~~~~\rightarrow~~~~~~\sqrt[3]{x^3}=\sqrt[3]{125}=\alert{5}\\
\amp \blert{\text{Cube the number}}~~~~~~~~\blert{\text{Take the cube root}}~~~~~~~~\blert{\text{Original number}}
\end{align*}
Now consider the functions that describe those operations: The graphs of \(f(x)=x^3\) and \(g(x)=\sqrt[3]{x}\) are related in an interesting way, as shown below.
If we place a mirror along the line \(y=x\text{,}\) each graph is the reflection of the other. Or imagine folding the grid along the line \(y=x\text{,}\) and one graph will match the other exactly. We say that the graphs are "symmetric about the line \(y=x\text{.}\)"
The two functions \(f(x)=x^3\) and \(g(x)=\sqrt[3]{x}\) are called inverse functions. Look at the tables of values for the two functions.
| \(x\) |
\(-2\) |
\(-1\) |
\(\dfrac{-1}{2}\) |
\(0\) |
\(\dfrac{1}{2}\) |
\(1\) |
\(2\) |
| \(f(x)=x^3\) |
\(-8\) |
\(-1\) |
\(\dfrac{-1}{8}\) |
\(0\) |
\(\dfrac{1}{8}\) |
\(1\) |
\(8\) |
| \(x\) |
\(-8\) |
\(-1\) |
\(\dfrac{-1}{8}\) |
\(0\) |
\(\dfrac{1}{8}\) |
\(1\) |
\(8\) |
| \(g(x)=\sqrt[3]{x}\) |
\(-2\) |
\(-1\) |
\(\dfrac{-1}{2}\) |
\(0\) |
\(\dfrac{1}{2}\) |
\(1\) |
\(2\) |
By interchanging the rows in the table for \(f(x)\text{,}\) we get the table for \(g(x)\text{.}\) This makes sense when we recall that each function undoes the effect of the other. In fact, we define the cube root function by
\begin{equation*}
y=\sqrt[3]{x}~~~~\text{if and only if}~~~~x=y^3
\end{equation*}
In other words, if we interchange the variables in the cubing function, \(y=x^3\text{,}\) to get \(x=y^3\text{,}\) we have an equivalent formula for the cube root function, \(y=\sqrt[3]{x}\text{.}\)
Let’s look at inverse operations in terms of functions. If we apply a function and then its inverse to an input, we return to that input. Use \(f(x)=x^3\) and \(g(x)=\sqrt[3]{x}\text{,}\) for example, and start by applying the innermost function.
\begin{equation*}
g(f(5)) = g(5^3) = g(125) = \sqrt[3]{125} = 5
\end{equation*}
Without the function notation we are simply saying that \(\sqrt[3]{x^3} = x\text{.}\) But this is an example of the more general rule that if \(f(x)\) and \(g(x)\) are inverse functions, then \(g(f(x))= x\) and \(f(g(x)) = x\text{.}\)
Checkpoint 10.1.3. Quickcheck 2.
Simplify each expression without a calculator.
\(\displaystyle \sqrt[4]{17^4}\)
\(\displaystyle (\sqrt{213})^2\)
\(\displaystyle (\sqrt[5]{b})^5\)
Subsection 10.1.3 Inverse of an Exponential Function
Because \(f(x) = x^3\) and \(g(x) = \sqrt[3]{x}\) are inverse functions, we saw that
\begin{equation*}
y=\sqrt[3]{x}~~~~\text{if and only if}~~~~x=y^3
\end{equation*}
A similar rule relates the operations of raising a base \(b\) to a power and taking a base \(b\) logarithm. You may recall from Chapter 7 that
\begin{equation*}
y = \log_b {(x)}~~~~\text{if and only if}~~~~x=b^y
\end{equation*}
So the function \(g(x)=\log_b {(x)}\) is the inverse function for \(f(x)=b^x\text{.}\) Each function undoes the effect of the other. For example, let \(f(x)=2^x\) and \(g(x)=\log_2 {(x)}\text{.}\) Start with \(x=3\text{,}\) apply \(f\text{,}\) and then apply \(g\) to the result.
\begin{equation*}
x = \alert{3}~~~~~~\rightarrow~~~~~~f(3)=2^3=8~~~~~~\rightarrow~~~~~~g(8)=\log_2{(8)} = \alert{3}
\end{equation*}
We return to the original number, 3.
And because \(f(x)=b^x\) and \(g(x)=\log_b {(x)}\) are inverse functions, we can write these operations in one expression as
\begin{equation*}
g(f(3)) = \log_2{(2^{\alert{3}})} = \alert{3}
\end{equation*}
(Remember the order of operations: do what’s inside of parentheses first, to get \(\log_2{(8)}\text{.}\)) Because the log and the exponential are inverse functions, this identity holds for any value of \(x\) and any base \(b \gt 0\text{.}\) That is
\begin{equation*}
\blert{\log_b{(b^{x})} = x}
\end{equation*}
We can also apply the two functions in the opposite order, so that
\begin{equation*}
f(g(8)) = 2^{\log_2{(\alert{8})}} = \alert{8}
\end{equation*}
(We compute the exponent, \(\log_2{(8)} = 3\text{,}\) first, and then compute the power \(2^3\text{.}\)) And in general
\begin{equation*}
\blert{b^{\log_b{(x)}} = x}
\end{equation*}
Example 10.1.4.
For this Example, recall that \(\log {(x)}\text{,}\) the "common logarithm," means \(\log_{10} {(x)}\text{.}\) Simplify each expression.
\(\displaystyle \log {(10^{3.4})}\)
\(\displaystyle 10^{\log{(0.6)}}\)
Solution.
Because \(\log {(x)}\) and \(10^x\) are inverse functions, \(\log {(10^{3.4})} = 3.4\text{.}\) You can verify the result on your calculator by first computing \(10^{3.4}\) and then taking the log of the result.
Again, because \(\log {(x)}\) and \(10^x\) are inverse functions, \(10^{\log{(0.6)}} = 0.6\text{.}\) You can verify the result on your calculator by first computing \(\log {(0.6)}\) and then computing \(10\) to the result.
Example 10.1.5.
\(f(x)=\log_6 {(x)}~\) and \(g(x)\) is its inverse function. \(~\alert{\text{[TK]}}\)
What is \(~g\text{?}\)
Evaluate and simplify \(~f\left(g(4)\right)\)
Evaluate and simplify \(~g\left(f(5)\right)\)
Solution.
The inverse of a logartihmic function is the exponential function with the same base, so \(~g(x) = 6^x\text{.}\)
First compute \(g(4) = 6^4\text{.}\) Then \(~f(g(4)) = f(6^4) = \log_6 {(6^4)} = 4\text{.}\)
First compute \(f(5) = \log_6 {(5)}\text{.}\) Then \(~g(\log_6 {(5)}) = 6^{\log_6 {(5)}} = 5\)
Checkpoint 10.1.6. Practice 1.
Simplify each expression.
\(\displaystyle \log_4 {(4^6)}\)
\(\displaystyle 4^{\log_4 {(64)}}\)
\(\displaystyle \log_{10} {(10^6)}\)
\(\displaystyle 10^{\log_{10} {(1000)}}\)
Answer.
\(\displaystyle 6\)
\(\displaystyle 64\)
\(\displaystyle 6\)
\(\displaystyle 1000\)
Subsection 10.1.4 Graphs of Logarithmic Functions
What does the graph of a log function look like? We can use exponential functions to help us.
Example 10.1.7.
Graph \(g(x) = \log_2{(x)}\)
Solution.
We can make a table of values for \(g(x) = \log_2{(x)}\) by interchanging the columns in a table for \(f(x) = 2^x\text{.}\)
| \(x\) |
\(f(x)=2^x\) |
| \(-2\) |
\(\dfrac{1}{4}\) |
| \(-1\) |
\(\dfrac{1}{2}\) |
| \(0\) |
\(1\) |
| \(1\) |
\(2\) |
| \(2\) |
\(4\) |
| \(3\) |
\(8\) |
| \(~x~\) |
\(g(x)=\log_2{(x)}\) |
| \(\dfrac{1}{4}\) |
\(-2\) |
| \(\dfrac{1}{2}\) |
\(-1\) |
| \(1\) |
\(0\) |
| \(2\) |
\(1\) |
| \(4\) |
\(2\) |
| \(8\) |
\(3\) |
We plot the points for each function, connecting them with smooth curves, as shown below. You can see that the two graphs are symmetric about the line \(y=x\text{.}\)
The same technique works for graphing a log function with any base.
Checkpoint 10.1.8. Practice 2.
Graph the function \(f(x)=10^x\) and its inverse function \(g(x)=\log_{10}{(x)}\) on the same axes.
In addition, the logarithmic function has the following properties.
Properties of Log Functions.
For any base \(b \gt 0, b \ne 1\text{:}\)
The logarithmic function \(y = \log_b {(x)}\) is defined for positive \(x\) only.
The \(x\)-intercept of its graph is \((1,0)\text{.}\)
The graph has a vertical asymptote at \(x=0\text{.}\)
The graphs of \(y = \log_b {(x)}\) and \(y=b^x\) are symmetric about the line \(y=x\text{.}\)
You can also see that while an exponential growth function increases very rapidly for positive input values, its inverse, the logarithmic function, grows extremely slowly.
Checkpoint 10.1.10. QuickCheck 3.
What is the \(y\)-intercept of the graph of \(y=\log_5 {(x)}\) ?
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
\(\displaystyle (0,5)\)
There is none.
Subsection 10.1.5 Using Logarithmic Functions
We can use the LOG key on a calculator to evaluate the function \(f(x) = \log_{10}{(x)}\text{.}\)
Example 10.1.11.
Let \(f(x) = \log_{10}{(x)}\text{.}\) Evaluate the following expressions.
\(\displaystyle f(35)\)
\(\displaystyle f(-8)\)
\(\displaystyle 2 f(16) + 1\)
Solution.
\(\displaystyle f (35) = \log_{10}{(35)}\approx 1.544\)
\(f(-8)\text{,}\) or \(\log_{10}{(-8)}\text{,}\) is undefined.
\(\displaystyle 2 f(16) + 1 = 2(\log_{10}{(16)}) + 1 \approx 2(1.204) + 1 = 3.408\)
Checkpoint 10.1.12. Practice 3.
Evaluate \(~H(t)=20 \log_{10}{\left(\dfrac{t}{0.5}\right)}~\) for \(t=2\text{.}\)
Answer.
\(20 \log_{10}{(4)} \approx 12.04\)
Recall that the subscript 10 is often omitted in applications that use common or base 10 logarithms, as illustrated in the following Example.
Example 10.1.13.
Evaluate the expression
\begin{equation*}
~~T=\dfrac{1}{k}\log{\left(\dfrac{M_f}{M_0}+1\right)}~
\end{equation*}
for \(k=0.028, ~M_f=1832\text{,}\) and \(M_0=15.3\text{.}\)
Solution.
Follow the order of operations and calculate
\begin{align*}
T \amp =\dfrac{1}{0.028}\log{\left(\dfrac{1832}{15.3}+1\right)} = \dfrac{\log{(120.739)}}{0.028}\\
\amp \approx \dfrac{2.082}{0.028} \approx 74.35
\end{align*}
Checkpoint 10.1.14. Practice 4.
The formula
\begin{equation*}
T = \dfrac{\log {(2)} \cdot t_i}{3 \log{(D_f /D_0)}}
\end{equation*}
is used by X-ray technicians to calculate the doubling time of a malignant tumor. \(D_0\) is the diameter of the tumor when first detected, \(D_f\) is its diameter at the next reading, and \(t_i\) is the time interval between readings, in days.
Calculate the doubling time of the following tumor: its diameter when first detected was 1 cm, and 7 days later its diameter was 1.05 cm.
Logarithmic functions are useful for modeling increasing functions that slow down as the input increases.
Example 10.1.15.
Life expectancy at birth is the average number of years a newborn child is expected to live. In 1900, the average life expectancy at birth in the U.S. was 47.3 years, and in 1910 it had risen to 50.0 years. During rest of the twentieth century, life expectancy was modeled by the formula
\begin{equation*}
L(x)=19.13 + 28.34 \log {(x)}
\end{equation*}
where \(x\) is the number of years after 1900.
Graph the life expectancy function for the years 1910 to 2000.
The life expectancy in 1950 was 68.2 years. What does the function \(L(x)\) predict for life expectancy in 1950?
According to the model, how much did life expectancy increase between 1920 and 1930? How much did it increase between 1990 and 2000?
Solution.
-
We can make a table of values for \(L(x)\) and plot points to obtain the graph below, which also shows the actual data points for life expectancy for the decades from 1910 to 2000.
| \(x\) |
\(10\) |
\(20\) |
\(30\) |
\(40\) |
\(50\) |
\(60\) |
\(70\) |
\(80\) |
\(90\) |
\(100\) |
| \(L(x)\) |
\(47.5\) |
\(56.1\) |
\(61.1\) |
\(64.6\) |
\(67.4\) |
\(69.6\) |
\(71.5\) |
\(73.1\) |
\(74.6\) |
\(75.9\) |
We substitute \(x=50\) into the function to find
\begin{equation*}
L(50) = 19.13 + 28.34 \log {(50)} \approx 67.4
\end{equation*}
The function predicts a life expectancy of 67.4 years in 1950.
Between 1920 and 1930, life expectancy increased from 56.1 to 61.1, or 5 years. Between 1990 and 2000 it increased from 74.6 to 75.9, or 1.3 years.
In the previous Example, we see that although life expectancy has been increasing over time, it has been slowing down or leveling off. In fact, life expectancy in the US actually declined slightly from 78.94 in 2013 to 78.81 in 2018. (What factors may have contributed to this decline?) By 2024 it had rebounded to 79.25. It remains to be seen how well the model predicts life expectancy in the 21st century.
Checkpoint 10.1.16. Practice 5.
The CDC (Centers for Disease Control and Prevention) provides Growth Charts for the average height and weight of children from age 2 to 20. The average height of girl children is given in centimeters by
\begin{equation*}
H(t) = 49.29+91.3 \log {(t)}
\end{equation*}
where \(t\) is age in years.
Graph the height function for \(2 \le t \le 20\text{.}\)
-
Use the height function to complete the table.
| \(t\) |
\(2\) |
\(5\) |
\(10\) |
\(15\) |
\(20\) |
| \(H(t)\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
-
How much is a girl’s height expected to increase between the ages of 5 and 10?
Between the ages of 15 and 20?
Answer.
| \(t\) |
\(2\) |
\(5\) |
\(10\) |
\(15\) |
\(20\) |
| \(H(t)\) |
\(77\) |
\(113\) |
\(141\) |
\(157\) |
\(168\) |
28 cm, 11 cm
Checkpoint 10.1.17. QuickCheck 4.
Which function grows most slowly for large values of \(x\) ?
\(\displaystyle f(x)=2x\)
\(\displaystyle f(x)=\log_2{(x)}\)
\(\displaystyle f(x)=\sqrt{x}\)
\(\displaystyle f(x)=\dfrac{x}{2}\)
Subsection 10.1.6 Solving Logarithmic Equations
A logarithmic equation is one in which the variable appears inside of a logarithm. For example,
\begin{equation*}
\log_4 {(x)} = 3
\end{equation*}
is a log equation.
If there is only one log involved, we can use the conversion equations to write the equation in exponential form. (Recall the conversion equations from
Section 7.3.1.)
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*}
Example 10.1.18.
Solve \(~\log_4{(2x-8)}=3\text{.}\)
Solution.
We use the conversion equations to write the equation in exponential form.
\begin{align*}
2x-8 \amp = 4^3 \amp \amp \blert{\text{Evaluate the power.}}\\
2x-8 \amp = 64 \amp \amp \blert{\text{Solve for}~x.}\\
x \amp = 36
\end{align*}
The solution is 36.
Checkpoint 10.1.19. Practice 6.
Solve \(~\log_5{(5-3x)} = 3\)
If an equation contains more than one log, we must first combine any expressions involving logs into a single logarithm.
Example 10.1.20.
Solve \(~~\log_{10}{(x + 1)} + \log_{10}{(x - 2)}= 1\text{.}\)
Solution.
\begin{equation*}
\log_{10}{(x + 1)(x - 2)} = 1
\end{equation*}
Once the left-hand side is expressed as a single logarithm, we can rewrite the equation in exponential form as
\begin{equation*}
(x + 1)(x - 2) = 10^1
\end{equation*}
Simplifying the right side gives us a quadratic equation to solve.
\begin{align*}
x^2 - x - 2 \amp = 10 \amp \amp \blert{\text{Subtract 10 from both sides.}}\\
x^2 - x - 12 \amp = 0 \amp \amp \blert{\text{Factor the left side.}}\\
(x - 4)(x + 3) \amp= 0 \amp \amp \blert{\text{Apply the zero-factor principle.}}
\end{align*}
We find \(x = 4\) or \(x = -3\text{.}\) But let us check both of these values in the original equation.
For \(x=4\) we have
\begin{equation*}
\log_{10}{(4 + 1)(4 - 2)} = \log_{10}{(5 \cdot 2)} = \log_{10}{(10)} = 1
\end{equation*}
so \(x=4\) is a solution.
But \(x = -3\) is not a solution of the original equation, because neither \(\log_{10}{(x + 1)}\) nor \(\log_{10}{(x - 2)}\) is defined for \(x = -3\text{.}\) (Remember that we cannot take a logarithm of a negative number or zero.) We say that the apparent solution \(x=-3\) is extraneous, and the only solution of the original equation is \(4\text{.}\)
Extraneous solutions can arise whenever we solve a logarithmic equation, especially if there is more than one apparent solution. Therefore, we should always check that a possible solution does not cause one of the logarithms to be undefined.
Checkpoint 10.1.21. Practice 7.
Solve \(~\log_{10}{(x)} + \log_{10}{(2)}= 3\)
Hint:
Rewrite the left side as a single logarithm.
Rewrite the equation in exponential form.
Solve for \(x\text{.}\)
Check for extraneous solutions.
Checkpoint 10.1.22. QuickCheck 5.
Fill in the blanks to complete each statement.
We cannot take a logarithm of .
After solving a logarithmic equation, we must check for .
If an equation contains more than one log, we must first combine them into .
If there is only one log involved, we rewrite the equation in form.
