Question 1154521

 Let 
{{{f(x)= pi^2 }}}-> constant
{{{g(x) = e^x}}}


Then solve the following equation and give answer(s) in exact form:


{{{f(x-1)-2g(2x-1)=0}}}...........{{{f(x-1)= pi^2 }}},{{{g(2x-1) = e^(2x-1)}}}


{{{pi^2-2(e^(2x-1))=0}}}


{{{pi^2=2(e^(2x-1))}}}


{{{pi^2/2=e^(2x-1)}}}...take log of both sides


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


{{{log(pi^2/2)=(2x-1)log(e)}}}......{{{log(e)=1}}}


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


{{{2log(pi)-log(2)+1=2x}}}


{{{x=2log(pi)/2-log(2)/2+1/2}}}


{{{x=log(pi)-log(2)/2+1/2}}}