How would you find the inverse function of $f(x)=e^{x/2}$?
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$\begingroup$ 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) $\endgroup$– Christopher A. WongOct 30, 2012 at 21:43
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$\begingroup$ Real-valued.. $f(x) = e^{g(x)} \iff \log f(x) = g(x).$ $\endgroup$– user2468Oct 30, 2012 at 21:47
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2 Answers
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$$y = \exp(x/2) \implies \log_e(y) = \log_e(\exp(x/2)) = x/2 \implies x = 2 \log_e(y)$$
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$\begingroup$ @AdiDani: No, it's not. $\log_e$ is the same thing as $\ln$. $\endgroup$ Oct 31, 2012 at 7:26
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$\begingroup$ @Hans Lundmarkt. It is wrong in sense that $x=2\log_e(y)$ is not inverse but needs to interchange x and y to $y=2\log_e(x)$ $\endgroup$– Adi DaniOct 31, 2012 at 7:54
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$\begingroup$ @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. $\endgroup$ Oct 31, 2012 at 8:12
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$$f(x)=y=e^{\frac{x}{2}}$$ inverse is $x=e^{\frac{y}{2}}\iff\ln x=\ln e^{\frac{y}{2}}\iff \ln x =\frac{y}{2}\iff y=2\ln x$
