Question 1028182
{{{log((2x-1)^2)-log(3x-2x^2)=log((4x-3)/(x))}}}


{{{log((2x-1)^2/(3x-2x^2))=log((4x-3)/x)}}}


Temporary change in variables saying as {{{log((A))=log((B))}}}.



{{{(2x-1)^2/(3x-2x^2)=(4x-3)/x}}}


{{{x(2x-1)^2=(3x-2x^2)(4x-3)}}}


{{{x(4x^2-4x+1)=12x^2-9x-8x^3+6x^2}}}


{{{4x^3-4x^2+x=18x^2-9x-8x^3}}}


{{{12x^3-22x^2+10x=0}}}


{{{6x^3-11x^2+5x=0}}}


{{{x(6x^2-11x+5)=0}}}
You are not interested in x=0, because you cannot have log of negative values as would happen in the original equation.  Put attention instead on the remaining roots of the quadratic factor.


Discrim,  {{{(-11)^2-4*6*5=121-6*20=1}}}
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{{{x=(11+- 1)/12}}}

{{{highlight(system(x=5/6,or,x=1))}}}