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Math Help - [SOLVED] Formula for inverse function

  1. #1
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    [SOLVED] Formula for inverse function

    Let f(x)=20/(1+3e^-x). Find a formula for f^-1(x).

    So I tried to do it this way:

    y=f(x)=20/(1+3e^-x).

    x=20/(1+3e^-y)

    20/x=1+3e^-y

    log ((20/x)-1)/log(3e)

    Does this look right, or am I going about this in the wrong way?
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  2. #2
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    Hello, CalcQueen!

    You were okay up to the last step(s) (which you didn't show).


    Let f(x)\,=\,\frac{20}{1+3e^{-x}} . Find f^{-1}(x)

    So I tried to do it this way:

    . . y\:=\:\frac{20}{1+3e^{-x}}

    . . x\:=\:\frac{20}{1+3e^{-y}}

    . . 1+3e^{-y} \:=\:\frac{20}{x} . . . . good!

    . . \frac{\log\left (\frac{20}{x}-1\right)}{\log(3e)} . ?

    You had: . 1 + 3e^{-y} \:=\:\frac{20}{x}

    Then: . 3e^{-y} \:=\:\frac{20}{x} - 1 \quad\Rightarrow\quad 3e^{-y}\:=\:\frac{20-x}{x} \quad\Rightarrow\quad e^{-y} \:=\:\frac{20-x}{3x}

    Take logs: . \ln\left(e^{-y}\right) \:=\:\ln\left(\frac{20-x}{3x}\right) \quad\Rightarrow\quad (-y)\ln(e) \:=\:\ln\left(\frac{20-x}{3x}\right)


    Since \ln(e) = 1, we have: . -y \:=\:\ln\left(\frac{20-x}{3x}\right)

    Finally: . y \:=\:-\ln\left(\frac{20-x}{3x}\right) \quad\Rightarrow\quad y \:=\:\ln\left(\frac{20-x}{3x}\right)^{-1} \quad\Rightarrow\quad\boxed{ y \:=\:\ln\left(\frac{3x}{20-x}\right)}

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