How would you find the inverse function of $f(x)=e^{x/2}$?
Inverse function of $f(x)=e^{x/2}$
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$\begingroup$
calculus
functions
inverse
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0This 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) – 2012-10-30
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0*Real-valued..* $f(x) = e^{g(x)} \iff \log f(x) = g(x).$ – 2012-10-30
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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0this is wrong abswer – 2012-10-30
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0@AdiDani: No, it's not. $\log_e$ is the same thing as $\ln$. – 2012-10-31
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0@Hans Lundmarkt. It is wrong in sense that $x=2\log_e(y)$ is not inverse but needs to interchange x and y to $y=2\log_e(x)$ – 2012-10-31
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0@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. – 2012-10-31
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$$f(x)=y=e^{\frac{x}{2}}$$ inverse is $x=e^{\frac{y}{2}}\iff\ln x=\ln e^{\frac{y}{2}}\iff \ln x =\frac{y}{2}\iff y=2\ln x$