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Math Help - inverse function

  1. #1
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    inverse function

    Hi i'm having problem finding inverse of this exponential equation.

    f(x)=4-2^(-x+1)

    f(y)=4-2^(-y+1)

    solve for y:

    x=4-2^-y+1

    x= 4- (1/2^(-y+1))

    What do i do next?
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Anemori View Post
    Hi i'm having problem finding inverse of this exponential equation.

    f(x)=4-2^(-x+1)

    f(y)=4-2^(-y+1)

    solve for y:

    x=4-2^-y+1

    x= 4- (1/2^(-y+1))

    What do i do next?
    x=4-2^{-y+1}

    \Rightarrow 4-x=2^{-y+1}

    \Rightarrow \log{(4-x)}=\log{2^{-y+1}}

    \Rightarrow \log{(4-x)}=(-y+1)\log{2}

    \Rightarrow \frac{\log{(4-x)}}{\log 2}=-y+1

    \Rightarrow y=1-\frac{\log{(4-x)}}{\log 2}
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Anemori View Post
    Hi i'm having problem finding inverse of this exponential equation.

    f(x)=4-2^(-x+1)

    f(y)=4-2^(-y+1)

    solve for y:

    x=4-2^-y+1

    x= 4- (1/2^(-y+1))

    What do i do next?
    An equivalent solution to the one posted above:

    x = 4 - 2^{1 - y}

    4 - x = 2^{1 - y}

    log_2(4 - x) = 1 - y

    y = 1 - log_2(4 - x)

    f^{-1}(x) = 1 - log_2(4 - x)

    If you think about it you should see that in general, log_b(b^u) = u, for suitable choice of b.
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