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  1. MHB Master
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    #1
    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...

    Last edited by karush; February 18th, 2020 at 16:39.

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    #2
    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? $f(x) = e^{2x}$ ???

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    #3 Thread Author


    well graphing it looks like its (c)

    so how???

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    #4
    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 ...

  6. MHB Master
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    #5 Thread Author


    ok looks like your suggestion of $y=x^{2x}$ is correct

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    #6
    Quote Originally Posted by karush View Post
    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!

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    #7 Thread Author
    post #2 looks like a suggestion to me!

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    #8
    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!

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    #9
    Quote Originally Posted by HallsofIvy View Post
    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.

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    #10
    But i don't see anything in the first post that is connected with $ \displaystyle x^{2x}$.

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