How to prove that $a^x$ is crescent for $a>1$ and decrescent if $0<a<1$ using logarithms?

Question

I need to prove that $f(x) = a^x$ is crescent when $a>1$, decrescent when $0<a<1$ and constant when $a=1$.

I just know that $$a^x = e^{\log a^x} = e^{x\log a} = (e^{\log a})^x$$

but I don't think it helps, because $e^{\log a}$ can be $>1$ or $<1$ depending on the value of $a$, and I'd arrive at the same problem. Could somebody help me?