Question 239806
{{{2 logx= log((8x-160))+1}}}
{{{logx^2= log((8x-160))+1}}}
{{{logx^2 - log((8x-160))= 1}}}
{{{log((x^2/(8x-160))) = 1}}}
{{{x^2/(8x-160) = 10^1}}}
{{{x^2 /(8x-160) = 10}}}
{{{x^2 = 80x - 1600}}}
{{{x^2 - 80x + 1600=0}}}
(((x-40)(x-40)=0}}}
x=40

Check your answer with a calculator. Does 2 log(40) = log(320-160) + 1 ???