Дано ; Найти

$log_{6}15=a\\ log_{12}18=b\\\\ log_{25}24=\frac{log_{5}24}{2} = \frac{2log_{5}2+log_{5}6}{2}\\\\ log_{12}18=\frac{2log_{6}3+log_{6}2}{2log_{6}2+log_{6}3}=b\\ log_{6}15=log_{6}3+log_{6}5=a\\\\ log_{6}2=\frac{log_{5}2}{log_{5}6}\\\\ log_{12}18=\frac{2log_{6}3+\frac{log_{5}2}{log_{5}6}}{2*\frac{log_{5}2}{log_{5}6}+log_{6}3}=b\\ log_{6}15=log_{6}3+\frac{1}{log_{5}6} = a\\\\ \\$

откуда $log_{6}15=a\\ log_{12}18=b\\\\ log_{25}24=\frac{log_{5}24}{2} = \frac{2log_{5}2+log_{5}6}{2}\\\\ log_{12}18=\frac{2log_{6}3+log_{6}2}{2log_{6}2+log_{6}3}=b\\ log_{6}15=log_{6}3+log_{6}5=a\\\\ log_{6}2=\frac{log_{5}2}{log_{5}6}\\\\ log_{12}18=\frac{2log_{6}3+\frac{log_{5}2}{log_{5}6}}{2*\frac{log_{5}2}{log_{5}6}+log_{6}3}=b\\ log_{6}15=log_{6}3+\frac{1}{log_{5}6} = a\\\\ \\ log_{6}3=\frac{log_{5}3}{log_{5}6}\\$
заменяя $log_{5}3=y\\ log_{5}2=x$
и решая систему

$\frac{y+1}{x+y}=a\\ \frac{x+2y}{2x+y}=b$
подставляя
получим
$log_{25}{24}=\frac{5-b}{2ab+2a-4b+2}$