Show that $\ln (x) \leq x-1 $ Show that $\ln (x) \leq x-1 $
I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that $e^x$ is the inverse? I mean, if $e^{x-1} \geq x$ then would the statement be proved?

Yes, one can use $$\tag1e^x\ge 1+x,$$ which holds for all $x\in\mathbb R$ (and can be dubbed the most useful inequality involving the exponential function). This again can be shown in several ways.
If you defined $e^x$ as limit $\lim_{n\to\infty}\left(1+\frac xn\right)^n$, then $(1)$ follows from Bernoullis inequality: $(1+t)^n>1+nt$ if $t>-1$ and $n>0$.
To show that $\ln(x)\le x-1$ for all $x>0$, just substitute $\ln x$ for $x$ in $(1)$.
