Showing that $f(x)=e^{x-5}-\ln(x-4)-1$ has only one real root

Question

There is a simple problem. I need to prove that there is just one $x$ such that $f(x)=0$.

$$f(x)=e^{x-5}-\ln(x-4)-1$$

I realized that $f(5)=0$ but how can I prove that it is the only one?

Answer 1

$\textbf{Hint}:$ Use second derivative test to show $x=5$ is a global $\textit{minimum}$.