Question 937829
{{{ 2log((x-1))= log((x+1))}}}

{{{ log((x-1)^2)= log((x+1))}}} if log same, we have

{{{(x-1)^2= x+1}}} ...solve for {{{x}}}

{{{x^2-2x+1= x+1}}} 

{{{x^2-2x+1-x-1=0}}} 

{{{x^2-3x=0}}} 

{{{x(x-3)=0}}} 

solutions:

{{{x=0}}} and {{{x=3}}}

since you are given {{{ 2log((x-1))}}} we cannot use solution {{{x=0}}} because we get {{{ 2log((0-1))=2log((-1))}}} and {{{log(10,(-1))}}} is undefined

so, your answer is {{{x=3}}}