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Math Help - Need help with exponential equation

  1. #1
    Newbie
    Joined
    Oct 2007
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    Need help with exponential equation

    Hello,

    I have a job interview tomorrow, where I will be given a maths test. It's been several years since I did maths last and I'm quite rusty. I'm having difficulty with the following problem in the practice paper:

    e ^ (2x)    +       2e^x        =        8

    My attempt went as follows:

    ln e ^ (2x) + ln 2 e ^ x = ln 8

    ln 2 e ^ (3x) = ln 8

    2 e ^ (3x) = 8

    e ^ (3x) = 4

    3x = ln 4

    x = (ln 4)/3

    However when I substitute this back in, it doesn't seem to work. Where am I going wrong?
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    My attempt went as follows:

    ln e ^ (2x) + ln 2 e ^ x = ln 8


    No, that is not correct.
    ln(a +b) is not ln(a) +ln(b).
    Rather, ln(a*b) = ln(a) +ln(b)

    By the way, your equation do not need logs yet.
    If you look closely, it is a quadratic equation in e^x

    e^(2x) +2e^x = 8
    Or, that is,
    (e^x)^2 +2e^x -8 = 0

    It is like having
    y^2 +2y -8 = 0
    if y = e^x

    So, factoring that,
    (y +4)(y-2) = 0

    y +4 = 0
    y = -4
    Or, back to the e^x for y,
    e^x = -4
    Take the ln of both sides,
    ln(e^x) = ln(-4)
    x*ln(e) = ln(-4)
    x = ln(-4)
    But there are no logs of negative numbers (no real number logarithmns for negative numbers), so reject e^x = -4

    y -2 = 0
    Or,
    e^x -2 = 0
    e^x = 2
    x = ln(2) = 0.693147181 -----------------answer.
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  3. #3
    Newbie
    Joined
    Oct 2007
    From
    Manchester
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    hello, here's an answer for you

    e^2x + 2e^x = 8

    e^2x + 2e^x -8 = 0 (treat it like a normal quadratic)

    (e^x +4)(e^x-2) = 0

    therefore
    e^x +4 = 0 or e^x -2 = 0

    e^x-2 = 0
    ln e^x = ln 2
    x = ln 2

    as the x in e^x +4 = 0 turns out to be ln(-4) which can't be, we ingore this answer meaning that it has to be the above answer.


    i think your problem was that when you loged the equation, you gave both e's a seprate ln when they should have been made into ln( e^2x + 2e^x) which would be impossible to split up into seprete lns.

    hope that helps, (good luck in the interview)
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  4. #4
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    Oct 2007
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    Thank you both; thats perfect

    I'm rustier than I thought
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