Question 987574
.
Your equation is 


{{{log(4,(x^2))}}} = {{{log(2, (2x-1))}}}.


It is not about using a calculator.  It is about solving the equation.


The domain is  2x - 1 > 0,  or  x > {{{1/2}}}. 


First,  you have  {{{log(4,(x^2))}}} = {{{1/2}}}{{{log(2,(x^2))}}},  since  {{{4}}} = {{{2^2}}}. 


Substitute it into the equation.  You will have


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


{{{log(2,(x^2))}}} = {{{2}}}{{{log(2, (2x-1))}}}, 


{{{log(2,(x^2))}}} = {{{2}}}{{{log(2, (2x-1))^2}}}.


Hence, 


{{{x^2}}} = {{{(2x-1)^2}}},


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


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


{{{x[1,2]}}} = {{{(4 +- sqrt(16 -12))/6}}} = {{{(4 +- 2)/6}}}.


{{{x[1]}}} = {{{(4+2)/6}}} = {{{1}}}.     It is in the domain.


{{{x[2]}}} = {{{(4-2)/6}}} = {{{1/3}}}.     It is outside the domain.


<B>Answer</B>. &nbsp;There is the unique solution &nbsp;{{{x}}} = {{{1}}}.