6log\(_x2\) - 3log\(_x3\) = 3log\(_50.2\)
= log\(_x2^6\) - 3log\(_x3^3\) = log\(_5(0.2)^3\)
= logx(64/27) = log5(1/5)3
logx(64/27) = log5(1/125)
let logx(64/27) = y
∴xy = 64/27
and log5(1/125) = y
∴ 5y = 1/125
5y = 125-1
5y = 5-3
∴ y = -3
substitute y = -3 in xy = 64/27
implies x-3 = 64/27
1/x3 = 64/27
64x3 = 27
x3 = 27/64
x3 = 3√27/64
x = 3/4