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Section 15.2 Graphs of Logarithmic Functions

Remember that exponential functions had a horizontal asymptote. Since logs and exponentials are inverses of each other, and inverses are all about swapping, logs are going to have a vertical asymptote.

We already learned about how to find the domain of a function in Section 3.4. 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 15.8.

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 15.9.

Find the domain and the equation of the asymptote for \(f(x)=-2\log_3(2x-7)+5\text{.}\)

Answer.

The domain is \(\left(\frac{7}{2},\infty\right)\text{,}\) and the equation for the vertical asymptote is \(x=\frac{7}{2}\text{.}\) (You must inclue the entire equation, not just the number \(\frac{7}{2}\text{.}\))

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 \(\frac{7}{2}\text{.}\)