Выразите log(3√2)((18^1/3)/(12^1/2)),если log(9)6=a

Так как

$log_{9}6=\frac{log_{2}6}{log_{2}9}=\frac{log_{2}2+log_{2}3}{log_{2}3^2}=\frac{1+log_{2}3}{2log_{2}3}$

$a=\frac{1+log_{2}3}{2log_{2}3};\\ \\log_{2}3=\frac{1}{2a-1};\\ \\log_{18}12=\frac{log_{2}12}{log_{2}18}=\frac{log_{2}3+log_{2}4}{log_{2}2+log_{2}9}=\frac{\frac{1}{2a-1}+2 }{1+2\cdot\frac{1}{2a-1} } = \frac{4a-1}{2a+1}$

$log_{3\sqrt{2} } \frac{18^{\frac{1}{3}}}{12^{\frac{1}{2} }}=log_{\sqrt{18} } (18^{\frac{1}{3}})-log_{\sqrt{18} } {12^{\frac{1}{2} }}=log_{18^{\frac{1}{2} } } 18^{\frac{1}{3}}-log_{18^{\frac{1}{2} } }12^{\frac{1}{2} }=\\ \\=\frac{1}{3}\cdot\frac{1}{\frac{1}{2} }-log_{18}12=\frac{2}{3}-\frac{4a-1}{2a+1}=\frac{4a+2-12a+3}{3(2a+1)}=\frac{5-8a}{6a+3}$