Welcome to our community

Be a part of something great, join today!

[SOLVED] 6.1.202 AP Calculus Inverse of e^x

karush

Well-known member
Jan 31, 2012
2,678
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
Mar 1, 2012
655
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
Jan 31, 2012
2,678


well graphing it looks like its (c)

so how???
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
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 ...
 

Attachments

karush

Well-known member
Jan 31, 2012
2,678

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
ok looks like your suggestion of $y=x^{2x}$ is correct:cool:
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
Jan 31, 2012
2,678
post #2 looks like a suggestion to me!
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, but post 2 suggested that the original problem might be to find the inverse function of [tex]f(x)= e^{2x}[/tex], not of [tex]f(x)= x^{2x}[/tex] as you say in post 5!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Yes, but post 2 suggested that the original problem might be to find the inverse function of [tex]f(x)= e^{2x}[/tex], not of [tex]f(x)= x^{2x}[/tex] 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
Jan 29, 2012
1,151
But i don't see anything in the first post that is connected with [tex]x^{2x}[/tex].
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
But i don't see anything in the first post that is connected with [tex]x^{2x}[/tex].
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.