Выразите log2 (3) через "а", если а = log6 (108). Помогите решить

Делаем так log6(108) = log2 (108) / log2 (6) = log2 (36*3) / log2 (2*3) = = (log2 (36) + log2 (3)) / (log2 (2) + log2 (3)) = = (2log2 (2) + 2log2 (3)+ log2 (3)) / (1 + log2 (3) = = (2 + 3log2 (3)) / (1 + log2 (3)) a = (2 + 3log2 (3)) / (1 + log2 (3)) 2 + 3log2 (3) = a + alog2 (3) 3log2 (3) - alog2 (3) = a - 2 log2(3)*(3 - a) = a - 2 log2 (3) = (a - 2) / (3 -a)

log6 (108)=log6(36*3)=2+log6(3)=2+log2(3)/log2(6)=2+(log2(3)+1-1)/(log2(3)+1)=3-1/(log2(3)+1); 3-1/(log2(3)+1)=a 3-a=1/(log2(3) log2(3)=1/(3-a)-1=(a-2)/(3-a)