If 6logx2 – 3logx3 = 3log50.2, find x.

Explanation

6log_x2 - 3log_x3 = 3log₅0.2
= logx²⁶ - 3logx³³ = log5^(0.2)3
= logx(64/27) = log5(1/5)³
logx(64/27) = log5(1/125)
let logx(64/27) = y
∴x^y = 64/27
and log5(1/125) = y
∴ 5^y = 1/125
5^y = 125^-1
5^y = 5^-3
∴ y = -3
substitute y = -3 in x^y = 64/27
implies x^-3 = 64/27
1/x³ = 64/27
64x³ = 27
x³ = 27/64
x³ = ³√27/64
x = 3/4