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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
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2 Answers 2

up vote 4 down vote accepted

$$y = \exp(x/2) \implies \log_e(y) = \log_e(\exp(x/2)) = x/2 \implies x = 2 \log_e(y)$$

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this is wrong abswer –  Adi Dani Oct 30 '12 at 21:57
    
@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
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$$f(x)=y=e^{\frac{x}{2}}$$ inverse is $x=e^{\frac{y}{2}}\iff\ln x=\ln e^{\frac{y}{2}}\iff \ln x =\frac{y}{2}\iff y=2\ln x$

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