Выразите log√3(6√a) через b, если log a(27)=b

Решите задачу:

$\log_{a}27=b\\a=27^ \frac{1}{b} = 3^ \frac{3}{b} \\\\\log _{ \sqrt{3}}\sqrt[6]{a} =\log_{ 3^\frac{1}{2} }a^\frac{1}{6}=\log_{3^ \frac{1}{2}} 3^ {\frac{3}{b}* \frac{1}{6}} = \\ = \log_{3^ \frac{1}{2}}3^ \frac{1}{2b} = 2\log_{3}3^ \frac{1}{2b} = \frac{1}{b} \log_{3}3 = \frac{1}{b}$

$a = \log_{\frac{1}{5}}27 = \frac{1}{\log_{27}\frac{1}{5}} = \frac{3}{\log_{3}\frac{1}{5}}{}\\\\ \log_{3^\frac{1}{2}}(\frac{9}{5})^\frac{1}{6} = \frac{1}{3}\log_{3}\frac{9}{5} =\frac{1}{3}\log_{3}9+\frac{1}{3}\log_{3}\frac{1}{5} = \\\\ = \frac{2}{3} +\frac{1}{3}*\frac{3}{a} = \frac{2}{3} +\frac{1}{a}$