Say we have $y=e^x$, we then apply to both sides $\ln$ and hence get: $\ln y=x$....apparently from there we get to $y=\ln x$ somehow..although I cannot think of any logarithm law which should yield this..?

The aim of this operation was to find the inverse function of $y=e^x$.

  • $\begingroup$ "apparently from there we get to $y=lan x$", Can I ask where does it say that?... Also, I assume than $lan$ is $\log$... $\endgroup$ – Cure Nov 10 '14 at 19:57
  • $\begingroup$ Keep in mind that the function and it's inverse are different functions. See mathuprising.comlu.com/inverse-functions.html for more details. $\endgroup$ – John Joy Nov 10 '14 at 23:29

We start with $y=e^x$. Taking the $\ln$ of both sides gives us $\ln y=\ln e^x$.

Using the logarithm property that $\ln a^b=b\ln a$, we simplify to:

$\ln y=\ln e^x\implies \ln y=x(\ln e)\implies \ln y=x(1)=x$

It should not be $y=\ln x$

We can see that $\ln y=x$ does not mean that $y=\ln x$ in general with a counterexample. Consider $y=e$, and $x=1$. In the first equation those values work, but in the second equation $e\neq\ln1=0$

To find the inverse of $e^x$:

We start with $y=e^x$. We switch the $x$ and the $y$, giving us $x=e^y$. Now we solve for $y$. Taking the $\ln$ of both sides gives us, $\ln x=y\ln e=y$.

Hence the inverse of $f(x)=e^x$ is $f^{-1}(x)=\ln x$


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