Question

Дано $log _{6} 15=a$; $log _{12} 18=b$
Найти $log _{25} 24$

Answer 1

  $log_{6}15=a\ log_{12}18=b\\ log_{25}24=frac{log_{5}24}{2} = frac{2log_{5}2+log_{5}6}{2}\\ log_{12}18=frac{2log_{6}3+log_{6}2}{2log_{6}2+log_{6}3}=b\ log_{6}15=log_{6}3+log_{6}5=a\\ log_{6}2=frac{log_{5}2}{log_{5}6}\\ log_{12}18=frac{2log_{6}3+frac{log_{5}2}{log_{5}6}}{2*frac{log_{5}2}{log_{5}6}+log_{6}3}=b\ log_{6}15=log_{6}3+frac{1}{log_{5}6} = a$
 
 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 
 $log_{25}{24}=frac{5-b}{2ab+2a-4b+2}$