Log5(15)=a;5-основ.
Log12(24)=b;12-осн.
Log125(45)-?

$a=log_515=log_5(5cdot 3)=log_55+log_53=1+log_53; to ; log_53=a-1; ; b=log_{12}24=log_{12}(12cdot 2)=log_{12}12+log_{12}2=1+frac{1}{log_212}==1+frac{1}{log_2(2^2cdot 3)}=1+frac{1}{2log_22+log_23}=1+frac{1}{2+log_23} ;; to frac{1}{2+log_23}=b-1; to ; 2+log_23=frac{1}{b-1}; to ; log_23=frac{1}{b-1}-2=frac{3-2b}{b-1};log_{125}45=log_{5^3}(5cdot 3^2)=frac{1}{3}cdot log_5(5cdot 3^2)=frac{1}{3}cdot (log_55+2log_53)=$

$=frac{1}{3}cdot (1+2log_53)=$ $frac{1}{3}cdot (1+2cdot (a-1))=frac{1}{3}cdot (1+2a-2)=$

$=frac{1}{3}cdot (2a-1)=frac{2a-1}{3}$