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\sqrt27/64 x = 3/4