Show that $T(2^n) = \Theta(3^n)$
First of all, try to solve the recurrence relation using the master method. 

Since $f(n) = O(n^{\log_2 3 - \epsilon})$ for some $\epsilon$, this is the first case of the master method. Therefore $$T(n) = \Theta(n^{\log_2 3})$$

If you replace $n$ by $2^n$, you will get your answer. i.e. 
$$T(2^n) = \Theta(2^{n\log_2 3}) = \Theta(3^n)$$