Why do you reject negative base solution for Logs? $log_x64=2$ 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 ?

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.
