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Thread: Exponential Equation help!

  1. #1
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    Exponential Equation help!

    How would I go about doing problems such as

    e^x+1=1/e

    and

    e^x+4=1/e^2x

    Thanks!
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  2. #2
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    Quote Originally Posted by The Masked Trumpet View Post

    e^x+1=1/e
    $\displaystyle e^x+1=\frac{1}{e}$

    $\displaystyle e^x=\frac{1}{e}-1$

    $\displaystyle x=\ln\left(\frac{1}{e}-1\right)$

    Quote Originally Posted by The Masked Trumpet View Post
    How
    e^x+4=1/e^2x
    $\displaystyle e^x+4=\frac{1}{e^{2x}}$

    $\displaystyle e^{2x}(e^x+4)=1$

    $\displaystyle e^{2x}(e^x+4)=1$

    $\displaystyle e^{3x}+4e^{2x}=1$

    Now substituting

    $\displaystyle a = e^x$

    $\displaystyle a^{3}+4a^{2}-1=0$

    You now have to solve this cubic.
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  3. #3
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    K, I think I understand those.

    How would I do one like this,

    e^(x+1)=1/e
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  4. #4
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    You should consider this

    $\displaystyle e^x = y \Rightarrow x = \log_e(y)$

    or

    $\displaystyle e^x = y \Rightarrow x = \ln(y)$

    Therefore using the second transformation

    $\displaystyle e^{x+1}=\frac{1}{e} $

    $\displaystyle x+1=\ln\left(\frac{1}{e}\right) $

    $\displaystyle x=\ln\left(\frac{1}{e}\right)-1 $
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