Showing that $\ln(b)-\ln(a)=\frac 1x \cdot (b-a)$ has one solution $x \in (\sqrt{ab}, {a+b\over2})$ for $0 < a < b$ 
For $0<a<b$, show that $\ln(b)-\ln(a)=\frac 1x \cdot (b-a)$ has one solution $x \in (\sqrt{ab}, {a+b\over2})$.

I guess that this is an application of the Lagrange theorem, but I'm unsure how to deal with $a+b\over2$ and $\sqrt{ab}$ since Lagrange's theorem offers a solution $\in (a,b)$.
 A: You could show the relations
$$ \frac{2}{a+b}(b-a) < \log(b)-\log(a) < \frac{1}{\sqrt{ab}}(b-a)$$
directly. (Exactly how does this solve your problem?)
To prove the first inequality, note that it is equivalent to
$$ 2\frac{\frac{b}{a}-1}{\frac{b}{a}+1} < \log\left(\frac{b}{a}\right). $$
Now set $x:= \frac{b}{a}$. We get
$$2(x-1) < x\log(x) + \log(x)$$
or
$$ x\log(x) + \log(x) - 2x < -2,$$
which needs to be true for all $x>1$, since by assumption, $b>a>0$. Let's denote the LHS by $f$. To prove the inequality above, we calculate the derivative of $f$ w.r.t. $x$:
$$f'(x) = \frac{d}{dx}\left(x\log(x)+\log(x) - 2x \right) = \frac{1}{x}-1,$$
which is zero if and only if $x=1$ (where $f'$ also changes sign from $+$ to $-$). Consequently $f$ attains its only maximum at $x=1$, so we get
$$f(x) \leq f(1) = -2 \quad \forall x\in\mathbb{R}^+$$
and the condition $x>1$ makes that inequality strict, as required.
The second inequality can be proven by an analogue approach. I'll leave it up to you to give it a try.
A: Hint: Use the mean value theorem.
