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    (Original post by desertwarrior)
    don't you need to make x the subject for iteration... i thought making it equal to 0 was to determine whether a root was present if there was a sign change between 2 numbers between which you suspect the root lies between????
    This is not iteration it is the Remainder Theorem.
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    Ah, solved it guys.

    e^{2x}=10x

    \frac{1}{10}=\frac{x}{e^{2x}}

    \frac{1}{10}={{x}e^{-2x}}

    \frac{-2}{10}={{-2x}e^{-2x}}

    \frac{-1}{5}={{-2x}e^{-2x}}

    Wn(\frac{-1}{5})={-2x}

    (\frac{-1}{2})Wn(\frac{-1}{5})={x} for n = 1,2

    Where Wn is the Lambert W Function
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    (Original post by desertwarrior)
    Ah, solved it guys.

    e^{2x}=10x

    \frac{1}{10}=\frac{x}{e^{2x}}

    \frac{1}{10}={{x}e^{-2x}}

    \frac{-2}{10}={{-2x}e^{-2x}}

    \frac{-1}{5}={{-2x}e^{-2x}}

    Wn(\frac{-1}{5})={-2x}

    (\frac{-1}{2})Wn(\frac{-1}{5})={x} for n = 1,2

    Where Wn is the Lambert W Function
    The Lambert W Function will never ever ever be used in C3. Its not even used in further maths, its very much degree level We had a nice discussion about it in the cambridge applicants thread.
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    I'm going to read medicine at Cambridge this year. Maybe I should have taken mathematics...
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    (Original post by desertwarrior)
    Ah thanks Mr M!!! That's what i've been working on, but I get one solution converging to 1.2713.........
    From the shape of the y=e^2x and y=10x graphs I would assume that there surely should be 2 solutions.. but iteration can only find one from my understanding..

    am I forgetting something,
    P.s. repped you thanks
    desert, I gave you two iterative formulas giving two distinct solutions. Did you even try them?
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    (Original post by Mr M)
    desert, I gave you two iterative formulas giving two distinct solutions. Did you even try them?
    ah yeah sorry Mr M! Didn't read that properly.. got the actually solutions through the formula now though to get them exact as specified.. I wonder whether my teacher will have heard of it :P! thanks anyway
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    (Original post by desertwarrior)
    I'm going to read medicine at Cambridge this year. Maybe I should have taken mathematics...
    What happened to engineering/oxford?
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    (Original post by desertwarrior)
    ah yeah sorry Mr M! Didn't read that properly.. got the actually solutions through the formula now though to get them exact as specified.. I wonder whether my teacher will have heard of it :P! thanks anyway
    no problem
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    (Original post by refref)
    What happened to engineering/oxford?
    That was just my friend who solved the equation for me.
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    (Original post by desertwarrior)
    That was just my friend who solved the equation for me.
    Then tell him he is a typical engineer for using maths that he does not understand
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    (Original post by refref)
    Then tell him he is a typical engineer for using maths that he does not understand
    he has a maths degree from harvard....
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    Plot it:
    y=10x
    y=e^(2x)
    using your graphing calculator solve for solutions,
    Or, use wolframalpha:
    http://www.wolframalpha.com/input/?i=graph+y%3D10x+and+y%3De^(2x)
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    (Original post by desertwarrior)
    he has a maths degree from harvard....
    I don't think Harvard award maths degrees.

    (also, for the original question, a solution to the equation e^(kx) = cx can be written down as an infinite series for all c and for suitible k)
 
 
 
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