Why does $a^x=e^{\ln(a^x)}$? Why does $a^x=e^{\ln(a^x)}$? 
I know if I actually evaluate this I get
\begin{align}
\ln(a^x)&=\ln[e^{\ln(a^x)}] \\ x\ln(a)&=x\ln(a)
\underbrace{\ln(e)}_{1} \\ \therefore x\ln(a)&=x\ln(a)
\end{align}
But I want to know how I arrive to $a^x=e^{\ln(a^x)}$ without evaluating the equality. 
 A: The function $\ln:(0,\infty)\to\mathbb R$ is by definition the inverse of exponential function $\mathbb R\to(0,\infty)$.
So: $$y=e^{\ln(y)}\text{ for every }y\in(0,\infty)$$
Now substitute $y=a^x$.
A: By definition, $\ln{a^x}$ equals a number that when $e$ is raised to that number, you get $a^x$. So, let's say that $\ln{a^x}=y$. Then, $e^y=a^x$. But we know what $y$ is in that last expression. It's equal to  $\ln{a^x}$. Thus, $e^{\ln{a^x}}=a^x$.
A: Depending on how you define $\ln x$, one possible way is to define $f^{-1} (x) = \ln x$ as the inverse function of $f(x) = e^x$.
Then: $$e^{\ln(a^x)} =  f\Big(f^{-1} \left(a^x \right) \Big) = a^x$$.
A: Proof 1: Show the equality from the right to the left. It's simply
$$\mathrm{e}^{\ln(a^x)}=\mathrm{e}^{x\ln(a)}=\mathrm{e}^{\ln(a)x}=(\mathrm{e}^{\ln(a)})^x=a^x$$
provided $a>0$ and $x\in\mathbb{R}$.
Proof 2: Let $a,b>0$ and $a\neq 1$ and consider $a^x=b$. Then you can define $x=\log_a(b)$, in case of $a=\mathrm{e}$ you can abbreviate $\ln(b)=\log_{\mathrm{e}}(b)$. In terms of $\ln(\cdot)$ the equation $a^x=b$ is rewritten $\mathrm{e}^{\ln(b)}=b$. Now just replace $b$ with $a^x$. 
