Question 74477
I have added parentheses to the equation which I think were missing:

      (3^x*9^(x+20))/(27^(2x+5))=81^(x+5)


Reduce all the numbers in the unknown terms to powers of 3 and simplify

{{{(3^x*(3^2)^(x+20))/((3^3)^(2x+5))=(3^4)^(x+5)}}}   or   {{{(3^(3x+40))/(3^(6x+15))=3^(4x+20)}}}


Multiply both sides by the denominator on the left side

       {{{3^(3x+40)=3^(10x+35)}}}


Since bases are the same, exponents are equal

        {{{3x+40=10x+35}}}    so {{{x=5/7}}}