Evaluate $\log_2 (4^2)$.
$\log_24=\boxed{2}$, so $\log_2(4^2) = \log_2((2^2)^2) = \log_2 (2^4) = \boxed{4}$
4
Level 1
Algebra
Problem: Evaluate $\log_2 (4^2)$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate $\log_2 (4^2)$, first recognize that $4$ can be expressed as a power of $2$. Specifically, $4 = 2^2$. Therefore, $4^2$ can be rewritten using this relationship: 

$4^2 = (2^2)^2$. 

Applying the power of a power property of exponents, we have:

$(2^2)^2 = 2^{2 \cdot 2} = 2^4$.

Next, we substitute this back into the logarithmic expression:

$\log_2 (4^2) = \log_2 (2^4)$.

Using the property of logarithms that states $\log_b (b^x) = x$, we can simplify:

$\log_2 (2^4) = 4$.

Thus, the final answer is $4$.<|END_OF_LATENT|> The final answer is $4$.