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