Question 133258
{{{(9^28 - 9^27) / 8 = 3^x }}} Start with the given equation



{{{(9^(27+1) - 9^27) / 8 = 3^x }}} Rewrite {{{9^28}}} into {{{9^(27+1)}}}




{{{(9^27*9^1- 9^27) / 8 = 3^x }}} Break up {{{9^(27+1)}}} into {{{9^27*9^1}}} through the identity {{{x^(y+z)=x^y*x^z}}}



{{{(9^27(9^1- 1)) / 8 = 3^x }}} Factor out the GCF {{{9^27}}}



{{{(9^27(9- 1)) / 8 = 3^x }}} Evaluate {{{9^1}}} to get 9



{{{(9^27(8)) / 8 = 3^x }}} Subtract



{{{(9^27cross(8)) / cross(8) = 3^x }}} Cancel like terms



{{{9^27 = 3^x }}} Simplify



{{{(3^2)^27 = 3^x }}} Rewrite {{{9}}} as {{{3^2}}}



{{{3^(2*27) = 3^x }}} Multiply the exponents. Remember, {{{(x^y)^z=x^(y*z)}}}



{{{3^54 = 3^x }}} Multiply



Since the bases are equal, the exponents are equal. So this means that {{{x=54}}}