Why do you reject negative base solution for Logs?

Question

$log_x64=2$ translates to $x^2=64$

This solves to $x=\pm8$

Why do you reject the solution of $x=-8$ ?
Doesn't it successfully check?

$log_{-8}64=2$ means "The exponent for -8 to get 64 is 2" which is a true statement, no ?

Answer 1

$\log_{-8}x$ would be an inverse function of $(-8)^x$ but this function does not behave well at all. What would be $(-8)^π$ for example?

Answer 2

Because $\log$ is the inverse of $\exp$ with base $-8$. Think about the domain of the function $x\to(-8)^x$. Is it defined for $x=1/2$ or in general for fractions $m/(2n)$? So in my opinion negative bases are avoided to be able to define the exponential everywhere.