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Thread: simple logs

  1. #1
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    simple logs

    Hello there,

    Im a little stuck on this:

    e^2x + e^x - 2 = 0

    it says hint use e^2x = ((e^x)^2)

    im meant to get 0 but i get 0.23...
    can you help?
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  2. #2
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    Quote Originally Posted by darksupernova View Post
    Hello there,

    Im a little stuck on this:

    e^2x + e^x - 2 = 0

    it says hint use e^2x = ((e^x)^2)

    im meant to get 0 but i get 0.23...
    can you help?
    $\displaystyle e^{2x} + e^x - 2 = 0$

    $\displaystyle (e^x)^2 + e^x - 2 = 0$.

    You have a quadratic equation in $\displaystyle e^x$. So let $\displaystyle X = e^x$ so that the equation looks like

    $\displaystyle X^2 + X - 2 = 0$

    $\displaystyle (X + 2)(X - 1) = 0$

    So $\displaystyle X = -2$ or $\displaystyle X = 1$.


    Therefore $\displaystyle e^x = -2$ or $\displaystyle e^x = 1$.

    But since $\displaystyle e^x > 0$ for all $\displaystyle x$, that means that only the second case is possible.


    So $\displaystyle e^x = 1$

    $\displaystyle x = \ln{1}$

    $\displaystyle x = 0$.
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  3. #3
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    i didnt spot that! Thanks very much!
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