Suppose $L:(0,\infty)\rightarrow \mathbb{R}$ is differentiable, that $L(1)=0$, and that $L'(x)=1/x$ for all $x>0$.
Show that $L(2^n)>n/2$ for all $n\in N$.
Based off of a hint I am to use the mean value theorem in $[1,2],\dots,[2^{n-1},2^n]$.
The case $n=1$ is trivial. However, I am trying to show that $\frac{L(2^{k-1})-L(2^k)}{2^{k-1}-2^k}>\frac{k}{2}$ implies $\frac{L(2^{k})-L(2^{k+1})}{2^{k}-2^{k+1}}>\frac{k+1}{2}$ and I am hitting dead ends.
If you add $\frac{1}{2}$ to both sides of $\frac{L(2^{k-1})-L(2^k)}{2^{k-1}-2^k}>\frac{k}{2}$ you get $\frac{2(L(2^{k-1})-L(2^k))+(2^{k-1}-2^k)}{2^{k}-2^{k+1}}>\frac{k+1}{2}$
This is almost where I want to be, but not sure where to go from here? or maybe I went completely the wrong direction? I am sure it has something to do with expanding on the inequality so that $\frac{L(2^{k})-L(2^{k+1})}{2^{k}-2^{k+1}}>\frac{2(L(2^{k-1})-L(2^k))+(2^{k-1}-2^k)}{2^{k}-2^{k+1}}>\frac{k+1}{2}$ but I am unsure.
MVT says
$$ \frac{f(y) - f(x)}{y-x} = f'(s) $$
for some $s$ between $x$ and $y$. So in your case
$$ \frac{L(2^k) - L(2^{k-1})}{2^k - 2^{k-1}} = \frac{1}{s} $$
For some $s$ between $2^{k-1}$ and $2^k$. So $1/s > 1/2^k$ Also note $2^k - 2^{k-1} = 2^{k-1}$, so
$$ \frac{L(2^k) - L(2^{k-1})}{2^{k-1}} > \frac{1}{2^k} $$
Multiply by $2^{k-1}$ and use that ( by induction ) $L(2^{k-1}) > (k-1)/2$ and you get
$$ L(2^k) > \frac{1}{2} + L(2^{k-1}) > \frac{k}{2} $$