TITLE: Where is the mistake in this argument that $(\sqrt8)^{\sqrt 7} >(\sqrt7)^{\sqrt 8}$?
QUESTION [5 upvotes]: I posted an answer in this question to prove that $(\sqrt8)^{\sqrt 7}<(\sqrt7)^{\sqrt 8}$
I started with 
$$f(x)=\frac{\ln x}{x}$$
$$f'(x)=\frac{1-\ln x}{x^2}$$
$$f'(x)>0 : x\in)0,e($$  $$f'(x)<0 : x\in)e,+\infty($$
so $f(1/7)>f(1/8)$, hence $-8\ln 8<-7\ln7$
so
$$\begin{align}
8\ln 8 & > 7\ln7\\\\
0.5\cdot 8\ln 8 & >0.5\cdot 7\ln7\\\\
8\ln \sqrt8 & > 7\ln\sqrt7\\\\
\ln (\sqrt8)^{\sqrt 7}& >\ln(\sqrt7)^{\sqrt 8}\\\\
(\sqrt8)^{\sqrt 7} &>(\sqrt7)^{\sqrt 8}
\end{align}$$
but where is the mistake with my answer, because $(\sqrt8)^{\sqrt 7}<(\sqrt7)^{\sqrt 8}$?!

REPLY [5 votes]: In your answer, you made the following comment:

@BarryCipra $8\ln\sqrt8>7\ln\sqrt7$, so $1/7∗1/8∗8\ln\sqrt8>1/7∗1/8∗7\ln\sqrt7$, so $1/7\ln\sqrt8>1/8\ln\sqrt7$ and note that $a\ln b=\ln b^a$.

It is true that:
$$
\dfrac{1}{7}\ln\sqrt8 > \dfrac{1}{8}\ln\sqrt7
$$
However, if you apply the power rule for logs, you instead obtain:
$$
\ln\sqrt[7]{\sqrt8} > \ln\sqrt[8]{\sqrt7}
$$
which is true, but not quite what we wanted.