Question 317778
Let {{{y=log(64,(4))}}}. Now convert to exponential form to get {{{64^y=4}}}. We can rewrite 64 as {{{4^3}}} and 4 as {{{4^1}}} which means that {{{(4^3)^y=4^(3y)}}} and {{{4^(3y)=4^1}}}. Since the bases are equal, the exponents must be equal. So {{{3y=1}}}. Solve for 'y' to get {{{y=1/3}}}. This then means that {{{log(64,(4))=1/3}}}