If $x \gt0$, then $\log(1+x) \lt x$.
My attempt at the proof thus far...
Let $f(x) = x-\log(1+x)$, then $f'(x) = 1-\frac{1}{1+x}$ for some $\mu \in (0,x)$ (since $x>0$)
MVT give us $f(x) = f(a) + f'(\mu)(x-a)$
So plug in our values we get:
$$x-\log(1+x) = 0+(1-\frac{1}{1+\mu})(x-0)$$
which we can reduce to
$$\log(1+x)=\frac{x}{1+\mu}$$
Now For $x\leq1$, then $0\lt\mu\lt1$
S0 $0\lt \frac{1}{1+\mu}\lt 1$, thus $\log(1+x)\lt x$
If $x>1$, then....
So I can see clearly that if $x>1$ is plugged into $\frac{x}{1+\mu}$
then $\log(1+x)<x$, but I am not sure of how to prove this.
I would appreciate tips hints or proof completions.
You're almost there. Have a look at your statement $$\log(1 + x) = \frac{x}{1+\mu}.$$ No matter what $\mu$ is, $1 + \mu$ is bigger than $1$, hence $$\frac{1}{1+\mu} < 1.$$ Multiplying this final inequality through by $x$ gives you the result you want. In words, dividing $x$ by something bigger than $1$ gives you something smaller than $x$.