We have the inequality $$\large\large \log _{\frac{1}{2}}x+\log _{\frac{1}{3}}x>\log _{\frac{1}{6}}\left(2x-1\right)$$
Here is my work:
$$\large-\log_2x-\log_3x>-\log_6(2x-1)$$
$$\large\log_2x+\log_3x<\log_6(2x-1)$$
$$\frac{\log x}{\log2}+\frac{\log x}{\log 3}<\frac{\log(2x-1)}{\log(2)+\log(3)}$$
Multiplying both sides of the equation by $\log2+\log3$ ,
$$\log(x)(2+\log_23+\log_32)<\log(2x-1)$$
$\log x=0$ (i.e $x=1$) is not a solution. Hence we can divide both sides by $\log x$ to obtain
$$2+\log_23+\log_32<\log_x{(2x-1)}\qquad\text{For $\quad 1<x$}$$
$$\qquad2+\log_23+\log_32>\log_x{(2x-1)}\qquad\text{For $\quad \frac12<x<1$}$$
But I don't know how to continue from here.
Best Answer
We have that
$$\log(x)(2+\log_23+\log_32)<\log(2x-1) $$
$$\iff \log(2x-1)-\log(x)(2+\log_23+\log_32)>0$$
and by $f(x)=\log(2x-1)-\log(x)(2+\log_23+\log_32)$ with $f(1)=0$ we have
$$f'(x)=\frac 2{2x-1}-\frac{2+\log_23+\log_32}x=0 \iff x_M=\frac{2+\log_23+\log_32}{2(1+\log_23+\log_32)}\approx 0.65548$$
with $f(x_M)>0$ and since
$$\lim_{x \to \frac12^+} f(x) = -\infty$$
by IVT another root exists $x_0 \in \left(\frac12, x_M\right)$ and then