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Math Help - Helping with long inverse equation

  1. #1
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    Helping with long inverse equation

    Suppose i had this equation , which i do , well i have one similar.
    I keep getting stuck on finding the inverse becuase of so many things in it , i am not sure about how where to take the ln of the euqaiotn to get rid of the e.


    f(x)= 40+30(2-e^{(-0.2x)})

    Could some one walk me through it?
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  2. #2
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    Quote Originally Posted by el123 View Post
    Suppose i had this equation , which i do , well i have one similar.
    I keep getting stuck on finding the inverse becuase of so many things in it , i am not sure about how where to take the ln of the euqaiotn to get rid of the e.


    f(x)= 40+30(2-e^{(-0.2x)})

    Could some one walk me through it?
    y = 40+30(2 - e^{-0.2x})

    y = 40 + 60 - 30e^{-0.2x}

    y = 100 - 30e^{-0.2x}

    x = 100 - 30e^{-0.2y}

    100-x = 30e^{-0.2y}

    \frac{100-x}{30} = e^{-0.2y}

    \ln\left(\frac{100-x}{30}\right) = -0.2y

    -5\ln\left(\frac{100-x}{30}\right) = y

    y = \ln\left(\frac{30}{100-x}\right)^5
    Last edited by skeeter; August 9th 2009 at 05:34 AM. Reason: fixed sign error.
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  3. #3
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    Why did you switch the sign around when you exchanged y and x?
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  4. #4
    Super Member 11rdc11's Avatar
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    Thats what you have to do to find an inverse. Switch y and x. To check to see if you found the inverse plug in your new equation into the original and it should equal y=x


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  5. #5
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    Quote Originally Posted by el123 View Post
    Why did you switch the sign around when you exchanged y and x?
    my mistake ... I'll edit it.
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  6. #6
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    Thanks again skeeter!
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