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Math Help - Inverse of Exponential Function

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    Inverse of Exponential Function

    Find the inverse of



    I know the answer, I'm just not sure how to get to it.
    Last edited by mr fantastic; December 7th 2011 at 02:32 AM. Reason: Copied title into main body of post.
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    Re: Inverse of Exponential Function

    Quote Originally Posted by Remriel View Post


    I know the answer, I'm just not sure how to get to it.
    For the inverse function, the x and y values swap, so you'll get

    \displaystyle \begin{align*} x &= \frac{2^y}{1 + 2^y} \\ x&= \frac{1 + 2^y - 1}{1 + 2^y} \\ x &= \frac{1 + 2^y}{1 + 2^y } - \frac{1}{1 + 2^y} \\ x &= 1 - \frac{1}{1 + 2^y} \\ \frac{1}{1 + 2^y} &= 1 - x \\ 1 + 2^y &= \frac{1}{1 - x} \\ 2^y &= \frac{1}{1 - x} - 1 \\ 2^y &= \frac{1}{1 - x} - \frac{1 - x}{1 - x} \\ 2^y &= \frac{1 - (1 - x)}{1 - x} \\ 2^y &= \frac{x}{1 - x} \\ \ln{\left(2^y\right)} &= \ln{\left(\frac{x}{1 - x}\right)} \\ y\ln{2} &= \ln{x} - \ln{(1 - x)} \\ y &= \frac{\ln{x} - \ln{(1 - x)}}{\ln{2}} \end{align*}
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    Re: Inverse of Exponential Function

    How did you get -1 in the numerator in the second equation?

    Also, your answer is different from the book which says: .
    Last edited by Remriel; November 28th 2011 at 04:54 PM.
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    Re: Inverse of Exponential Function

    Quote Originally Posted by Remriel View Post
    How did you get -1 in the numerator in the second equation?
    Added a cleverly disguised 0, in this case, 1 - 1...
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    Re: Inverse of Exponential Function

    Quote Originally Posted by Remriel View Post

    Also, your answer is different from the book which says: .
    It doesn't matter which base of the logarithm you use (just choose the most convenient).
    So try the same thing like Prove it did but take 2 as the base of the logarithm.
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    Re: Inverse of Exponential Function

    Hello, Remriel!

    \text{Find the inverse function: }\:f(x) \:=\:\frac{2^x}{1+2^x}

    I know the answer . . . .
    I'm always impressed by that!

    We are given: . y \;=\;\frac{2^x}{1 + 2^x}


    Switch variables: . x \;=\;\frac{2^y}{1+2^y}


    \text{Solve for }y\!:\;\;x(1+2^y) \;=\;2^y

    m . . . . . . . . x + x\!\cdot\!2^y \;=\;2^y

    . . . . . . . . . 2^y - x\!\cdot\!2^y \;=\;x

    . . . . . . . . . 2^y(1-x) \;=\;x

    . . . . . . . . . . . . . . 2^y \;=\;\frac{x}{1-x}

    Take logs, base 2: . \log_2(2^y) \;=\;\log_2\!\left(\frac{x}{1-x}\right)

    . . . . . . . . . . . . . y\!\cdot\!\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\log_2\!\left(\frac{x}{1-x}\right)


    Therefore: . y \;=\;\log_2(x) - \log_2(1-x)

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    Re: Inverse of Exponential Function

    Quote Originally Posted by Remriel View Post
    How did you get -1 in the numerator in the second equation?

    Also, your answer is different from the book which says: .
    We can manipulate Prove It's answer to get your one using the log laws.

    By the subtraction rule: y = \dfrac{\ln(x) - \ln(1-x)}{\ln(2)} = \dfrac{\ln \left(\frac{x}{1-x}\right)}{\ln(2)}

    By the change of base rule: \log_2(y) = \dfrac{\ln(y)}{\ln(2)} of which the RHS is the safe as the expression above.

    Putting in the expression for y: \log_2 \left(\dfrac{x}{1-x}\right)

    Thus the two answers are equivalent.

    Also note that for the inverse x \neq 0, 1 - what does that tell you about the original function's range (admittedly you don't need to include this bit for your question)
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    Re: Inverse of Exponential Function

    I understand now. Thanks everyone.
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