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  1. #1
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    Log

    Hi guys. I'm stuck on this one.

    2*e^(-x) = 3*e^(0,1x)

    Any help would be appreciated.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Alvy View Post
    Hi guys. I'm stuck on this one.

    2*e^(-x) = 3*e^(0,1x)

    Any help would be appreciated.
    Hi Alvy,

    Do this:

    \ln 2e^{-x}=\ln 3e^{.1x}

    \ln 2 + (-x)=\ln 3 + .1x

    -1.1x=\ln 3 - \ln 2

    x=\frac{\ln 3 - \ln 2}{-1.1}
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  3. #3
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    Hello, Alvy!

    Another approach . . .


    Solve for x\!:\;\;2e^{-x} \:=\: 3e^{0.1x}

    We have: . 3e^{0.1x} \;=\;2e^{-x}

    Multiply by \frac{e^x}{3}\!:\;\;e^{1.1x} \;=\;\frac{2}{3}

    \text{Take logs: }\;\ln\left(e^{1.1x}\right) \;=\;\ln\left(\frac{2}{3}\right) \quad\Rightarrow\quad 1.1x\underbrace{\ln(e)}_{\text{This is 1}} \;=\;\ln\left(\frac{2}{3}\right)


    Therefore: . 1.1x \;=\;\ln\left(\frac{2}{3}\right) \quad\Rightarrow\quad x \;=\;\frac{1}{1.1}\ln\left(\frac{2}{3}\right)

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