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Math Help - [SOLVED] solving an exponential equation

  1. #1
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    [SOLVED] solving an exponential equation

    I'm finding this difficult to solve: 2x+e^x=0

    I tried putting 2x on the other side, then taking logs to base e to get:

    x\ln (e)=\ln(-2x)\Rightarrow x=\ln(-2) + \ln(x)\Rightarrow x - \ln(x)=\ln(-2)

    I was trying to isolate x, but there's no such thing as \ln(-2).
    And even if there was I wouldn't know how to isolate x from x-\ln(x).
    So I guess that isn't the right approach.

    Then I tried:
    e^x=-2x \Rightarrow \frac{e^x}{-x}=2 \Rightarrow  \ln(\frac{e^x}{-x})=\ln(2)
    \Rightarrow \ln(e^x)-\ln(-x)=\ln(2)\Rightarrow x-\ln(-x)=\ln(2)

    But I still can't isolate x and don't know if that's really any closer to solving it.
    I'm a bit stumped.

    Got any tips? Cheers.
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  2. #2
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    Quote Originally Posted by tleave2000 View Post
    I'm finding this difficult to solve: 2x+e^x=0

    I tried putting 2x on the other side, then taking logs to base e to get:

    x\ln (e)=\ln(-2x)\Rightarrow x=\ln(-2) + \ln(x)\Rightarrow x - \ln(x)=\ln(-2)

    I was trying to isolate x, but there's no such thing as \ln(-2).
    And even if there was I wouldn't know how to isolate x from x-\ln(x).
    So I guess that isn't the right approach.

    Then I tried:
    e^x=-2x \Rightarrow \frac{e^x}{-x}=2 \Rightarrow \ln(\frac{e^x}{-x})=\ln(2)
    \Rightarrow \ln(e^x)-\ln(-x)=\ln(2)\Rightarrow x-\ln(-x)=\ln(2)

    But I still can't isolate x and don't know if that's really any closer to solving it.
    I'm a bit stumped.

    Got any tips? Cheers.
    This equation can not be solved explicitally for x.

    Consider the function

    f(x)=2x+e^x

    f(0)=2(0)+e^{0}=1 and

    f(\frac{-1}{2}) =2(\frac{-1}{2})+e^{-\frac{1}{2}}=\frac{-\sqrt{e}+1}{\sqrt{e}}<0

    So it has a root in \left[ \frac{-1}{2},0\right]

    This could be approximated numerically by using Newtons method.

    I hope this helps
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  3. #3
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    Yeah, thanks a lot, that's interesting. I'm reading about Newton's method and equations that can't be solved exactly now.

    Is the log of a negative number anything like the square root of a negative number?
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