Inverse function of $f(x)=e^{x/2}$

How would you find the inverse function of $f(x)=e^{x/2}$?

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This is the same as setting $y = e^{x/2}$, and then solving for $x$ in terms of $y$. Are you familiar with how that is done? (Hint: logarithms) – Christopher A. Wong Oct 30 '12 at 21:43
Real-valued.. $f(x) = e^{g(x)} \iff \log f(x) = g(x).$ – user2468 Oct 30 '12 at 21:47

$$y = \exp(x/2) \implies \log_e(y) = \log_e(\exp(x/2)) = x/2 \implies x = 2 \log_e(y)$$
@AdiDani: No, it's not. $\log_e$ is the same thing as $\ln$. – Hans Lundmark Oct 31 '12 at 7:26
@Hans Lundmarkt. It is wrong in sense that $x=2\log_e(y)$ is not inverse but needs to interchange x and y to $y=2\log_e(x)$ – Adi Dani Oct 31 '12 at 7:54
@AdiDani: The answer gives $x=f^{-1}(y)$ instead of $y=f^{-1}(x)$. So what? It's the same function $f^{-1}$ no matter what variables you use for writing it. – Hans Lundmark Oct 31 '12 at 8:12
$$f(x)=y=e^{\frac{x}{2}}$$ inverse is $x=e^{\frac{y}{2}}\iff\ln x=\ln e^{\frac{y}{2}}\iff \ln x =\frac{y}{2}\iff y=2\ln x$