When n is an odd integer, f(n)=3^n and when n is an even int

$$When\ n=odd\ integer.\ Then,\ f\left(n\right)=3^n$$
$$When\ n=even\ integer.\ Then,\ f\left(n\right)=4^{\frac{n}{2}}$$
$$What\ is\ the\ value\ of\ f\ \left(2^5\right)?$$
$$f\ \left(2^5\right)=f\left(32\right)$$
$$when\ n=32\ and\ 32\ is\ an\ even\ integer.$$
$$f\left(n\right)=4^{\frac{n}{2}}$$
$$f\left(n\right)=4^{\frac{n}{2}}=4^{\frac{32}{2}}$$
Using the fractional index law in indices
$$a^{\frac{x}{y}}=\left(y\sqrt{a}\right)^x$$
$$4^{\frac{32}{2}}=\left(2\sqrt{4}\right)^{32}\ =2^{32}$$

Final answer = Option E

When n is an odd integer, f(n)=3^n and when n is an even integer, f(n)=4^(n/2). What is the value of f(2^5)=?

A. 2^5
B. 3^11
C. 3^12
D. 3^13
E. 2^32

\[f\left( n \right) = \left\{ \begin{gathered}
{3^n}\,\,\,,\,\,\,n\,\,{\text{odd}} \hfill \\
{4^{n/2}}\,\,\,,\,\,\,n\,\,{\text{even}} \hfill \\
\end{gathered} \right.\]
\[? = f\left( {{2^5}} \right)\]
\[{2^5} = 32\,\,{\text{is}}\,\,{\text{even}}\,\,\,\,\, \Rightarrow \,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,\,{{\text{4}}^{{\text{32/2}}}} = {\left( {{2^2}} \right)^{16}} = {2^{32}}\]

The above follows the notations and rationale taught in the GMATH method.

=>
Since n = 2^5 is an even integer, f(2^5) = 4^2^{5/2}=4^2^4=(2^2)^{16}=2^{32}.

Therefore, the answer is E.
Answer: E