problem in realizing "AEP-like limit" I have a problem in realizing the solution of this problem:

As I marked in the picture, I cannot understand how $\lim(p(X_1,X_2,\dots,X_n))^{1/n}$ changed to a base $2$ logarithm and then again it changed to what you see in the picture. Would you please explain how and why it happened?
 A: call $Y = p(X_1, \cdots, X_n)^{1/n}$.
By definition : $Y = 2^{log_2(Y)}$.
And $log_2(Y) = log_2(p(X_1, \cdots, X_n)^{1/n}) = \frac{1}{n} log_2(p(X_1, \cdots, X_n))$.
$X_1, \cdots, X_n$ are idependent thus $p(X_1, \cdots, X_n)) = \prod_i p(X_i) \implies log_2(p(X_1, \cdots, X_n)) = log_2(\prod_i p(X_i)) = \sum_i log_2(p(X_i))$.
Sum up all of these facts with remark that the $lim$ operator taken over $n$, you have what happens in the solution.
A: Just to clarify the above solution a bit more,
\begin{align}
\lim_{n\to\infty}p(x^n)^{\frac{1}{n}} & = \lim_{n\to\infty}2^{\frac{1}{n}\log p(x^n)}\\
& = \lim_{n\to\infty}2^{\frac{1}{n}\sum_{k=1}^n\log p(x_k)} \\
& = 2^{\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\log p(x_k)} \\
& = 2^{-H(P)}
\end{align}
where the second line uses the fact that $x_i's$ are i.i.d., the third line uses the continuity of the map $t\mapsto 2^t$, and the last line follows due to strong law of large numbers that:
$$
\frac{1}{n}\sum_{k=1}^n \log p(x_k) \xrightarrow{\text{a.s.}}\mathbb{E}[\log p(x)] = -H(X)
$$
as $n\to\infty$, almost surely.
