Cfrgtkky

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 $log$?

$$begin{align}
sqrt{x}&=2^{log_4(x)}&&log_4(x)=log_{2^2}(x)\
&=2^{log_{2^2}(x)}&&log_{2^2}(x)=frac12log_{2}(x)\
&=2^{frac12log_{2}(x)}\
&=sqrt{2^{log_{2}(x)}}&&a^{log_a(b)}=b\
&=sqrt{x}
end{align}$$

$log_a(b)$ 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 $b$):

$$blog_{a^b}(c) = log_a(c) implies log_{a^b}(c) = frac{1}{b}log_a(c)$$

Now you can see where the $frac{1}{b}$ comes from..

OTOH, in the identity $log_a(c^d) = dlog_a(c)$ 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:
$$log_{a^c} x = y longleftrightarrow(a^c)^y = x$$
$$implies a^{cy} = x longleftrightarrow log_a x = cy$$
$$implies y = frac{1}{c}cdotlog_a x$$
Recall that at the beginning, we said $log_b x = y$. Therefore, we reach a conclusion.
$$boxed{log_{a^c} x = frac{1}{c}cdotlog_a x}$$
You confused this with the following identity, which is a different thing altogether.
$$log_b x^a = acdotlog_b x$$