Fact 13.2.1.
The domain of a logarithm is where the inside of the log is greater than 0. The vertical asymptote is where the inside of the log is equal to 0.
Section 13.2 Graphs of Logarithmic Functions
We already learned about how to find the domain of a function in Section 8.4. In this section, we are just adding another "problem" to the list of things to deal with: logarithms. Other than that, we do the same thing we did before.
Example 13.2.2.
Suppose we want to find the domain and asymptote of \(f(x)=3\log_6(8-5x)+7\text{.}\)
Remember that the domain only cares about where there is a potential problem. So, it only cares about what is inside the log. Since the inside of the log has to be greater than 0, we set \(8-5x\gt 0\) and solve.
\begin{align*}
8-5x \amp\gt 0\\
-5x \amp\gt -8\\
x \amp \lt \frac{8}{5}
\end{align*}
Notice that in the last step, we had to switch the direction of the inequality because we divided by a negative number. Now, we just have to turn that into interval notation to get our domain: \(\left(-\infty,\frac{8}{5}\right)\text{.}\)
To find the equation of the asymptote, we have to set the inside equal to 0 and solve for \(x\text{.}\)
\begin{align*}
8-5x \amp= 0\\
-5x \amp= -8\\
x \amp = \frac{8}{5}
\end{align*}
Note that logs always have a vertical asymptote, but they never have a horizontal asymptote. Therefore, our answer is the entire equation \(x=\frac{8}{5}\text{,}\) not just the number \(\frac{8}{5}\text{.}\)
Checkpoint 13.2.3.
Find the domain and the equation of the asymptote for \(f(x)=-2\log_3(2x-7)+5\text{.}\)
Answer.
Solution.
Remember that the domain only cares about where there is a potential problem. So, it only cares about what is inside the log. Since the inside of the log has to be greater than 0, we set \(2x-7\gt 0\) and solve.
\begin{align*}
2x-7 \amp\gt 0\\
2x \amp\gt 7\\
x \amp \gt \frac{7}{2}
\end{align*}
Now, we just have to turn that into interval notation to get our domain: \(\left(\frac{7}{2},\infty\right)\text{.}\)
To find the equation of the asymptote, we have to set the inside equal to 0 and solve for \(x\text{.}\)
\begin{align*}
2x-7 \amp= 0\\
2xx \amp= 7\\
x \amp = \frac{7}{2}
\end{align*}
Note that logs always have a vertical asymptote, but they never have a horizontal asymptote. Therefore, our answer is the entire equation \(x=\frac{7}{2}\text{,}\) not just the number \(\frac{7}{2}\text{.}\)
