Why is it not correct to apply as follows the rule : $\log(a^n) = n\times \log(a) $?

Question

The equation I want to solve is : $2\log(x)= \log(25)$ ( Source : Blitzer , College Algebra, 4.4, #79).

Symbolab calculator gives the solution $x=5$, after having divided both sides by $2$ and having simplified the RHS.

Why isn't it correct to apply the rule : $\log(a^n) = n\times \log(a) $ , on the LHS, in the following way?

$2\log(x)= \log(25)$

$\rightarrow \log (x^2) = \log (25)$

$\rightarrow x^2 = 25 $

$\rightarrow x =5$ OR $ x = -5$.

Answer 1

You have the existence condition on $\log x$, that is $x>0$; so the solution $x=-5$ is not allowed.