Find $x$, such that $\log_{64^{\frac{1}{3}}}4^x=1$.
This problem involves logarithms and exponents, so I will try to rewrite them using the same base if possible. I notice that $64^{\frac{1}{3}}=4$, so I can simplify the base of the logarithm to $\log_44^x=1$. Now I have a logarithm with base 4 and argument 4^x, which means I can use the property $\log_b b^a=a$. This gives me $x=1$, which is the value of $x$ that satisfies the equation. 