# SOLUTION: Hi guys, im stumped with this one: 2ln2 - ln(x-1) = ln(2x) solve for 'x'. cheers =)

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 Click here to see ALL problems on logarithm Question 186597: Hi guys, im stumped with this one: 2ln2 - ln(x-1) = ln(2x) solve for 'x'. cheers =)Found 2 solutions by nerdybill, josmiceli:Answer by nerdybill(6951)   (Show Source): You can put this solution on YOUR website!You will need to apply "log rules". Review them at: http://www.purplemath.com/modules/logrules.htm . 2ln2 - ln(x-1) = ln(2x) 2ln(2/(x-1)) = ln(2x) ln(2/(x-1))^2 = ln(2x) (2/(x-1))^2 = 2x 2^2/(x-1)^2 = 2x 4/(x-1)^2 = 2x 2/(x-1)^2 = x 2 = x(x-1)^2 2 = x(x-1)(x-1) 2 = x(x^2-x-x+1) 2 = x(x^2-2x+1) 2 = x^3-2x^2+x 0 = x^3-2x^2+x-2 Factor (by grouping) on the right: 0 = (x^3-2x^2)+(x-2) 0 = x^2(x-2)+(x-2) 0 = (x-2)(x^2+1) . Setting each term (in parenthesis to zero) we get: x = {2, i} Answer by josmiceli(9661)   (Show Source): You can put this solution on YOUR website!In general: and check: OK The negative root doesn't work since raising to a power can't result in a negative