TITLE: Prove that $\log_23>\log_35>\log_47$.
QUESTION [6 upvotes]: Using calculus, prove that $\log_23>\log_35>\log_47$ .

My try :
If $\log_x(2x-1)$ is decreasing function then we can say that $\log_23>\log_35>\log_47$.
$f(x)=\log_x(2x-1)$
$f(x)=\dfrac{\ln(2x-1)}{\ln x}$
$f'(x)=\dfrac{2x\ln x-(x-1)\ln(2x-1)}{(\ln x)^2x(2x-1)}$

REPLY [4 votes]: From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x \log x  < (2x-1) \log (2x-1).\tag{$*$}$$
The derivative of $g(x) = (2x-1) \log (2x-1) - 2 x \log x$ is $$g'(x) = 2 \log(2x-1) +  2 - 2 (\log x + 1) = 2 \log (2 - \frac{1}{x}),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.