
MHB Master
#1
February 18th, 2020,
16:29
If $f^{1}(x)$ is the inverse of $f(x)=e^x$, then $f^{1}(x)=$
$a. \ln\dfrac{2}{x}$
$b. \ln \dfrac{x}{2}$
$c. \dfrac{1}{2}\ln x$
$d. \sqrt{\ln x}$
$e. \ln(2x)$
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...
Last edited by karush; February 18th, 2020 at 16:39.

February 18th, 2020 16:29
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MHB Journeyman
#2
February 18th, 2020,
17:57
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? $f(x) = e^{2x}$ ???

MHB Master
#3
February 18th, 2020,
20:45
Thread Author
well graphing it looks like its (c)
so how???

MHB Journeyman
#4
February 18th, 2020,
23:25
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 ...

MHB Master
#5
February 19th, 2020,
20:00
Thread Author
ok looks like your suggestion of $y=x^{2x}$ is correct

#6
February 20th, 2020,
12:01
Originally Posted by
karush
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!

MHB Master
#7
February 20th, 2020,
17:30
Thread Author
post #2 looks like a suggestion to me!

#8
March 5th, 2020,
08:25
Yes, but post 2 suggested that the original problem might be to find the inverse function of $ \displaystyle f(x)= e^{2x}$, not of $ \displaystyle f(x)= x^{2x}$ as you say in post 5!

MHB Seeker
#9
March 5th, 2020,
08:55
Originally Posted by
HallsofIvy
Yes, but post 2 suggested that the original problem might be to find the inverse function of $ \displaystyle f(x)= e^{2x}$, not of $ \displaystyle 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.

#10
March 5th, 2020,
12:56
But i don't see anything in the first post that is connected with $ \displaystyle x^{2x}$.