Section 10.5 Logarithmic Functions
The inverse of an exponential function is a new function known as a logarithm. Logarithms are studied in detail in advanced algebra, here we will take an introductory look at how logarithms work. When working with radicals we found that there were two ways to write radicals. The expression \(\displaystyle \sqrt[\leftroot{-3}\uproot{3}m]{a^n} \) could be written as \(\displaystyle a^{\frac{n}{m}}\text{.}\) Each form has its advantages; thus, we need to be comfortable using both the radical form and the rational exponent form. Similarly, an exponent can be written in two forms, each with its own advantages. The first form we are very familiar with, \(b^x = a\text{,}\) where \(b\) is the base, \(a\) can be thought of as our answer, and \(x\) is the exponent. The second way to write this is with a logarithm, \(\log_b a = x \text{.}\) The word “log” tells us that we are in this new form. The variables all still mean the same thing. \(b\) is still the base, \(a\) can still be thought of as our answer.
Using this idea the problem \(5^2 = 25\) could also be written as \(\log_5 25 = 2 \text{.}\) Both mean the same thing, both are still the same exponent problem, but just as roots can be written in radical form or rational exponent form, both our forms have their own advantages. The most important thing to be comfortable doing with logarithms and exponents is to be able to switch back and forth between the two forms. This is what is shown in the next few examples.
Example 10.5.1.
Write each exponential equation in logarithmic form.
Example 10.5.2.
Write each logarithmic equation in exponential form
Once we are comfortable switching between logarithmic and exponential form, we are able to evaluate and solve logarithmic expressions and equations. We will first evaluate logarithmic expressions. An easy way to evaluate a logarithm is to set the logarithm equal to \(x\) and change it into an exponential equation.
Example 10.5.3.
Example 10.5.4.
Example 10.5.5.
World View Note: Dutch mathematician Adriaan Vlacq published a text in \(1628\) which listed logarithms calculated out from \(1\) to \(100,000\text{!}\)
Solving equations with logarithms is done in a very similar way, we simply will change the equation into exponential form and try to solve the resulting equation.
Example 10.5.6.
Example 10.5.7.
Example 10.5.8.
There is one base on a logarithm that gets used more often than any other base, base \(10\text{.}\) Similar to square roots not writing the common index of \(2\) in the radical, we don’t write the common base of \(10\) in the logarithm. So if we are working on a problem with no base written we will always assume that base is base \(10\text{.}\)
Example 10.5.9.
This lesson has introduced the idea of logarithms, changing between logs and exponents, evaluating logarithms, and solving basic logarithmic equations. In an advanced algebra course, logarithms will be studied in much greater detail.
Exercises Exercises – Logarithmic Functions
Exercise Group.
Rewrite each equation in exponential form.
1.
\(\log_9 81=2 \)
2.
\(\log_b a=-16 \)
3.
\(\log_7 \frac{1}{49}=-2 \)
4.
\(\log_{16} 256=2 \)
5.
\(\log_{13} 169 = 2 \)
6.
\(\log_{11} 1 = 0 \)
Exercise Group.
Rewrite each equation in logarithmic form.
7.
\(8^0=1 \)
8.
\(17^{-2}=\frac{1}{289} \)
9.
\(15^2=225 \)
10.
\(144^{\frac{1}{2}} = 12 \)
11.
\(64^{\frac{1}{6}}=2 \)
12.
\(19^2=361 \)
Exercise Group.
Evaluate each expression.
13.
\(\log_{125} 5 \)
14.
\(\log_5 125 \)
15.
\(\log_{343} \frac{1}{7} \)
16.
\(\log_7 1 \)
17.
\(\log_4 16 \)
18.
\(\log_4 \frac{1}{64} \)
19.
\(\log_6 36 \)
20.
\(\log_{36} 6 \)
21.
\(\log_2 64 \)
22.
\(\log_3 243 \)
Exercise Group.
Solve each equation.
23.
\(\log_5 x=1 \)
24.
\(\log_8 k=3 \)
25.
\(\log_2 x = -2 \)
26.
\(\log n = 3 \)
27.
\(\log_{11} k=2 \)
28.
\(\log_4 p=4 \)
29.
\(\log_9 (n+9)=4 \)
30.
\(\log_{11} (x-4) = -1 \)
31.
\(\log_5 (-3m)=3 \)
32.
\(\log_2 (-8r)=1 \)
33.
\(\log_{11} (x+5) = -1 \)
34.
\(\log_7 -3n = 4 \)
35.
\(\log_4 (6b+4)=0 \)
36.
\(\log_{11} (10v+1) = -1 \)
37.
\(\log_5 (-10x+4)=4 \)
38.
\(\log_9 (7-6x)=-2 \)
39.
\(\log_2 (10-5a)=3 \)
40.
\(\log_8 (3k-1)=1 \)