Выразите log√3(корень 6 степени из 1.8) через а, если log0.2(27)=a ПОМОГИТЕ!!!!!!

1)
$log_{0.2}27=a \\ \\ log{ \frac{1}{5} }(3^3)=a \\ \\ 3log_{5^{-1}}3=a \\ \\ -3log_{5}3=a \\ \\ log_{5}3=- \frac{a}{3}$

2)
$log_{ \sqrt{3} } \sqrt[6]{1.8}=log_{3^{ \frac{1}{2} }}(1.8^{ \frac{1}{6} })= \frac{ \frac{1}{6} }{ \frac{1}{2} }log_{3}1.8= \frac{1}{3}log_{3}(3*0.6)= \\ \\ = \frac{1}{3}(log_{3}3+log_{3}0.6)= \frac{1}{3}(1+log_{3}0.6)= \\ \\ = \frac{1}{3}(1+log_{3}( \frac{3}{5} ))= \frac{1}{3}(1+log_{3}3-log_{3}5)= \\ \\ = \frac{1}{3}(1+1- \frac{1}{log_{5}3} )= \frac{1}{3}(2- \frac{1}{log_{5}3} )= \\ \\ = \frac{2}{3}- \frac{1}{3log_{5}3}= \frac{2}{3}- \frac{1}{3*(- \frac{a}{3} )}= \\ \\$
$= \frac{2}{3}+ \frac{1}{a}= \frac{2a+3}{3a}$