Question 755750

{{{(1/36)^(x-1) =1296^(2x)}}}.......write {{{(1/36)^x-1}}} as {{{36^(1-x)}}} and {{{1296}}} as {{{36^2}}}

{{{36^(1-x) =(36^2)^(2x)}}} .........since {{{(36^2)^(2x)=(36)^(4x)}}} we have


{{{36^(1-x) =36^(4x)}}}..since bases same, exponents are same too; so, we have


{{{1-x =4x}}}....solve for {{{x}}}


{{{1=4x+x}}}


{{{1=5x}}}


{{{1/5=x}}}