Question 133287
{{{9^(y/5)=27}}} Start with the given equation



{{{(3^2)^(y/5)=27}}} Rewrite {{{9}}} as {{{3^2}}}



{{{(3^2)^(y/5)=3^3}}}  Rewrite {{{27}}} as {{{3^3}}}




{{{3^(2*(y/5))=3^3}}} Multiply the exponents. Remember, {{{(x^y)^z=x^(y*z)}}}



{{{3^(2y/5)=3^3}}} Multiply 



{{{2y/5=3}}} Since the bases are equal, this means that the exponents are equal



{{{2y=15}}} Multiply both sides by 5



{{{y=15/2}}} Divide both sides by 2



So our answer is


{{{y=15/2}}} or {{{y=7.5}}}