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 $\in (a,b)$.

Hint: Use the mean value theorem.
