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Math Help - e^x+lnx=7

  1. #1
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    e^x+lnx=7

    Could someone help please.

    I don't know how to solve this

    e^x+lnx=7

    I know the answer is approx 1.8536 (Wolfram alpha)

    and I know what the graph y=e^x+lnx-7 looks like (Desmos)

    I have played with it but I have not had any success.

    Thank you.
    Last edited by Melody2; January 16th 2014 at 02:59 PM.
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  2. #2
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    Re: e^x+lnx=7

    Quote Originally Posted by Melody2 View Post
    e^x+lnx=7
    I know the answer is approx 1.8536 (Wolfram alpha)
    and I know what the graph y=e^x+lnx-7
    There is no simple solution.
    But you need to understand Newton Method.
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  3. #3
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    Re: e^x+lnx=7

    Thanks Plato,
    I think that I do understand Newton's method.
    I'll try it that way and see how I go.
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  4. #4
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    Re: e^x+lnx=7

    Newton's method, also known as the Newton-Raphson method, is probably the first method that would spring to mind. For this example though a simple fixed point iteration will suffice.

    Rewrite the equation as

    x=\ln(7-\ln x),

    from which form the iterative equation

    x_{n+1}=\ln(7-\ln(x_{n})).

    A simple sketch graph shows that the root is somewhere around x = 2, so taking this to be our starting value x_{0}, successive substitution gets us the sequence

    1.84164, 1.85463, 1.85353, 1.85362, 1.8536237 etc.
    Last edited by BobP; January 17th 2014 at 01:20 AM.
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