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Math Help - Inverse Exponential Function?

  1. #1
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    Inverse Exponential Function?

    Ive been having a bit of trouble with this one.

    Find the Inverse.

    g(x)= 3^x + 1



    I think that the answer is this but Im not sure

    g^-1(x)=\log_{x}3 + 1

    Or would the answer be

    g^-1(x)=\log_{x}3 - 1

    Thanks for any and all help given.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Legal Eagle View Post
    Ive been having a bit of trouble with this one.

    Find the Inverse.

    g(x)= 3^x + 1



    I think that the answer is this but Im not sure

    g^-1(x)=\log_{x}3 + 1

    Or would the answer be

    g^-1(x)=\log_{x}3 - 1

    Thanks for any and all help given.
    neither is correct

    y = 3^x + 1

    for inverse, switch x and y and solve for y

    \Rightarrow x = 3^y + 1

    \Rightarrow x - 1 = 3^y

    \Rightarrow \log_3 (x - 1) = \log_3 3^y

    \Rightarrow \log_3 (x - 1) = y

    thus, g^{-1}(x) = \log_3 (x - 1)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Legal Eagle View Post
    Ive been having a bit of trouble with this one.

    Find the Inverse.

    g(x)= 3^x + 1



    I think that the answer is this but Im not sure

    g^-1(x)=\log_{x}3 + 1

    Or would the answer be

    g^-1(x)=\log_{x}3 - 1

    Thanks for any and all help given.
    Now you should be able to see that that is not the inverse. Using the function you defined is

    g\left(g^{-1}(x)\right)=x?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Now you should be able to see that that is not the inverse. Using the function you defined is

    g\left(g^{-1}(x)\right)=x?
    indeed! thank you for that input. however, i should stress, it is important to check that g^{-1}(g(x)) = x as well
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