Question 991939

{{{log(2 ,(2x-1))=log(4 ,(3x^2-4x+2))}}}......change to base ten

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

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

{{{log((2x-1))/log(2 )=log((3x^2-4x+2))/2log(2) }}}....cross multiply

{{{2cross(log(2))*log((2x-1))=log((3x^2-4x+2))*cross(log(2 )) }}}

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

{{{log((2x-1)^2)=log((3x^2-4x+2)) }}}........if log same, we have

{{{(2x-1)^2=3x^2-4x+2 }}}......solve for {{{x}}}

{{{4x^2-4x+1=3x^2-4x+2 }}}

{{{4x^2-3x^2-4x+4x+1-2=0 }}}

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

{{{(x-1)(x+1)=0 }}}

solutions:

{{{(x-1)=0 }}}=>{{{x=1}}}

{{{(x+1)=0 }}}=>{{{x=-1}}}....disregard negative solution,

so, your solution is {{{highlight(x=1)}}}