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

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

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

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