How do I prove that a^(log x with the base of a) = x? All I know is that it does equal x.
The logarithm function, let's call it g(x), $\displaystyle g(x) = \log_{a} x$, is defined as the inverse function of $\displaystyle f(x) = a^x$.
From the definition of an inverse function we know if must be true that
$\displaystyle f(f^{-1}(x))=x$
So applying that definition here
$\displaystyle f(g(x)) = x$
$\displaystyle f(\log_{a} x) = x$
$\displaystyle a^{\log_{a} x} = x$
And there you have it.