# [SOLVED]6.1.202 AP Calculus Inverse of e^x

#### karush

##### Well-known member
If $f^{-1}(x)$ is the inverse of $f(x)=e^x$, then $f^{-1}(x)=$

$a. \ln\dfrac{2}{x}$
$b. \ln \dfrac{x}{2}$
$c. \dfrac{1}{2}\ln x$
$d. \sqrt{\ln x}$
$e. \ln(2-x)$

ok, it looks slam dunk but also kinda ???

my initial step was
$y=e^x$ inverse $\displaystyle x=e^y$
isolate
$\ln{x} = y$

the overleaf pdf of this project is here .... lots of placeholders...

https://drive.google.com/open?id=1WyjkfLAzhs4qF3RYOgSJrllP4hoKC5d4

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
The inverse of $f(x)=e^x$ is $f^{-1}(x) = \ln{x}$

... there is an obvious mistake in the answer choices.

Maybe a typo? $f(x) = e^{2x}$ ???

#### karush

##### Well-known member

well graphing it looks like its (c)

so how???

#### skeeter

##### Well-known member
MHB Math Helper
the graph is close, but no cigar.

$f(1)=e \implies f^{-1}(e) =1$

however, if $f^{-1}(x)=\dfrac{1}{2}\ln{x}$, then $f^{-1}(e) = \dfrac{1}{2} \ne 1$

have another look ...

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#### HallsofIvy

##### Well-known member
MHB Math Helper
ok looks like your suggestion of $y=x^{2x}$ is correct
And that was not what he suggested! Please be more careful what you are writing or you are just wasting our time!

#### karush

##### Well-known member
post #2 looks like a suggestion to me!

#### HallsofIvy

##### Well-known member
MHB Math Helper
Yes, but post 2 suggested that the original problem might be to find the inverse function of $$f(x)= e^{2x}$$, not of $$f(x)= x^{2x}$$ as you say in post 5!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, but post 2 suggested that the original problem might be to find the inverse function of $$f(x)= e^{2x}$$, not of $$f(x)= x^{2x}$$ as you say in post 5!
I inspected the pdf. It looks to me that the typo is in the original problem.
That is, I think the writers of the pdf made the mistake.
We can only guess about what it should have been.

#### HallsofIvy

##### Well-known member
MHB Math Helper
But i don't see anything in the first post that is connected with $$x^{2x}$$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
But i don't see anything in the first post that is connected with $$x^{2x}$$.
Ah yes. That's true. That was a typo when referring to a suggested possible typo about a typo in the opening post that was actually a presumed typo in the original pdf.