Prove that $\log_23>\log_35>\log_47$. 
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 $\log_23>\log_35>\log_47$.
$f(x)=\log_x(2x-1)$
$f(x)=\dfrac{\ln(2x-1)}{\ln x}$
$f'(x)=\dfrac{2x\ln x-(x-1)\ln(2x-1)}{(\ln x)^2x(2x-1)}$
 A: 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.
A: You might need to do your derivative again. The numerator of the derivative of $\frac{\ln(2x-1)}{\ln x}$ obtained by the quotient rule is
$$\frac1{2x-1}\ln x-\frac{2x-1}x$$
which we can upper-bound as
$$\frac1{2x-1}\ln x-\frac{2x-1}x\le\frac x{2x-1}-\frac{2x-1}x=\frac{-3x^2+4x-1}{x(2x-1)}$$
The denominator of this last fraction is positive for $x>\frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2\le x\le 4$, the interval of interest to the question, and $\log_x(2x-1)$ is decreasing over $(2,4)$, so $\log_23>\log_35>\log_47$.
A: Again $g(x)=2x\ln x-(2x-1)\ln(2x-1)$ is a decreasing function in $(1,\infty)$ because $g'(x)<0$ then from $x>1$ we have 
$$2x\ln x-(2x-1)\ln(2x-1)<0$$
then $f'(x)>0$.
