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Math Help - Trouble finding inverse of y = 10^x / (10^x + 1)

  1. #1
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    Trouble finding inverse of y = 10^x / (10^x + 1)

    This is surely where a teacher is handy to have. Luckily, I also have this forum!

    I'm obviously missing something, but I can't figure out the inverse of:

    y = \frac{10^x}{10^x + 1}

    I can isolate one x variable, but can't figure out how to isolate it entirely.

    I've tried a bunch of ways, but the main one is:

    Take the log base 10 of both sides:
    log_{10}(y) = log_{10}(\frac{10^x}{10^x + 1})

    Split off the division into subtraction:
    log_{10}(y) = log_{10}(10^x) - log_{10}(10^x + 1)

    Then I can isolate one x:
    log_{10}(y) = x - log_{10}(10^x + 1)

    I just can't seem to see the way forward. Please, someone help put me out of my misery. Would be much appreciated!
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  2. #2
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    e^(i*pi)'s Avatar
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    10^xy + y - 10^x = 0

    10^xy-10^x = -y

    10^x(y-1) = -y

    10^x = \dfrac{-y}{y-1}

    x = \log \left(\dfrac{-y}{y-1}\right) = \log(-y) - \log(y-1)


    f^{-1}(x) = \log \left(\dfrac{-x}{x-1}\right) = \log(-x) - \log(x-1)


    It would appear that no real number satisfies the domain
    Last edited by e^(i*pi); December 22nd 2010 at 02:17 PM. Reason: log(a/b) = log(a) - log(b)
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Firstly express:

    10^x=\dfrac{y}{1-y}

    ...

    Fernando Revilla

    Edited: I didn't see the previous post.
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  4. #4
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    How about

    \displaystyle\ y=\frac{10^x}{10^x+1}=\frac{10^{-x}}{10^{-x}}\;\frac{10^x}{10^x+1}=\frac{1}{1+10^{-x}}\Rightarrow\ 1+10^{-x}=\frac{1}{y}\Rightarrow\ 10^{-x}=\frac{1-y}{y}
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  5. #5
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    Thanks to you all! I see my mistake. I tried to go too quickly to logs, but I needed to manipulate it quite a bit before I even get there. I love that I got two different ways to arrive at the solution. Going through Archie Meade's solution to the end, I get the textbook answer of y = log_{10}(\frac{x}{1-x}).

    And e^(i*pi)'s solution reduces to the same thing if I just do the somewhat obvious y = log_{10}(\frac{-x}{x -1}) = log_{10}(\frac{-x}{-(-x + 1)}) = log_{10}(\frac{x}{1 - x}).

    Thanks again. And on to the next problem I go.
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  6. #6
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    \displaystyle \frac{10^x}{10^x + 1} = \frac{10^x + 1 - 1}{10^x + 1} = \frac{10^x + 1}{10^x + 1} - \frac{1}{10^x + 1} = 1 - \frac{1}{10^x + 1}.

    So if \displaystyle y = 1 - \frac{1}{10^x + 1}, to find the inverse, the \displaystyle x,y values swap, giving

    \displaystyle x = 1 - \frac{1}{10^y + 1}

    \displaystyle \frac{1}{10^y+1} = 1 - x

    \displaystyle 10^y + 1 = \frac{1}{1-x}

    \displaystyle 10^y = \frac{1}{1-x} - 1

    \displaystyle 10^y = \frac{1}{1-x}- \frac{1-x}{1-x}

    \displaystyle 10^y = \frac{x}{1-x}

    \displaystyle y = \log_{10}{\left(\frac{x}{1-x}\right)}

    \displaystyle y = \log_{10}{(x)} - \log_{10}{(1-x)}
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  7. #7
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    Nice, a third way to solve it. You guys are great.
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  8. #8
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    e^(i*pi)'s Avatar
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    Don't forget the domain x>0 and 1-x >0 \implies x < 1

    Thus the domain is 0< x< 1

    I'm not sure if your question asks for it
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