I have a screenshot of the solution for a $\log$ equation below, but I don't understand how the exponent of $2$ got turned into $\dfrac12$. I thought it's supposed to stay a $2$ when you transfer it to the front of the $\log$?
$$\begin{align} \sqrt{x}&=2^{\log_4(x)}&&\log_4(x)=\log_{2^2}(x)\\ &=2^{\log_{2^2}(x)}&&\log_{2^2}(x)=\frac12\log_{2}(x)\\ &=2^{\frac12\log_{2}(x)}\\ &=\sqrt{2^{\log_{2}(x)}}&&a^{\log_a(b)}=b\\ &=\sqrt{x} \end{align}$$
$\log_a(b)$ is by definition the unique number $p$ such that $a^p = b$.
So $\log{a^b}(c)$ is by definition the unique number $p$ such that $(a^b)^p = c$. We can write $(a^b)^p = a^{bp}$ by the rules of exponentiation. So $a^{bp} = c$ and so $bp = \log_a(c)$ as the latter is the unique exponent that will give $c$ (for base $a$). But recall that $p = \log_{a^b}(c)$ from the starting definition.
So (with dividing both sides by $b$):
$$b\log_{a^b}(c) = \log_a(c) \implies \log_{a^b}(c) = \frac{1}{b}\log_a(c)$$
Now you can see where the $\frac{1}{b}$ comes from..
OTOH, in the identity $\log_a(c^d) = d\log_a(c)$ we move the $d$ from the exponent in the argument to the front and it stays the same, while in your case above we move the $2$ in the exponent of the base $2^2$ to the front. Quite a different thing.
Let’s say we have a logarithmic equation.
$$\log_b x = y \longleftrightarrow b^y = x$$
Suppose $b$ can be written as an exponent.
$$b = a^c$$
Rewriting the original equation this way, we get:
$$\log_{a^c} x = y \longleftrightarrow(a^c)^y = x$$
$$\implies a^{cy} = x \longleftrightarrow \log_a x = cy$$
$$\implies y = \frac{1}{c}\cdot\log_a x$$
Recall that at the beginning, we said $\log_b x = y$. Therefore, we reach a conclusion.
$$\boxed{\log_{a^c} x = \frac{1}{c}\cdot\log_a x}$$
You confused this with the following identity, which is a different thing altogether.
$$\log_b x^a = a\cdot\log_b x$$
