$$\log_{1\over31} (4x-5)^2 > \log_{1\over31} (5x-7)^2$$
$\begin{cases} x\neq {7\over 5} \\ x\neq {5\over 4} \\ (4x-5)^2<(5x-7)^2 \end{cases}$
$(4x-5)^2-(5x-7)^2<0, \quad (4x-5-(5x-7))(4x-5+5x-7)<0, \quad (2-x)(x-{4\over3})<0, \quad x \in \left(-\infty; {4\over3}\right) \cup \left(2; +\infty\right)$
$x \in \left(-\infty; {5\over4}\right)\cup\left({5\over4};{4\over3}\right)\cup\left(2;+\infty\right)$
But right answer is $x \in \left({4\over3};{7\over5}\right) \cup \left({7\over5}; 2\right)$
$\log_b$ with basis less than $1$ is decreasing. Therefore the equality is equivalent to $$(4x-5)^2>(5x-7)^2\iff 9x^2-30x+24=3(3x^2-10x+8)>0.$$ Use the Rational root theorem to find $2$ is a root, hence by Vieta's relations, $4/3$ is the other root. The solutions are the $x$ outide of the interval of the roots: $$(-\infty,4/3)\cup(2,+\infty)\smallsetminus\{5/4,7/5\}=(-\infty,5/4)\cup(5/4,4/3))\cup(2,+\infty).$$