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Math Help - logarithmic question

  1. #1
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    logarithmic question

    having trouble with this one

    solve for x using logs

    2e^3x=4e^5x

    thanks!
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  2. #2
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    First, I would divide by 2 to produce:

    e^{3x} = 2e^{5x}

    Take the natural logarithm.

    \ln(e^{3x}) = \ln(2e^{5x})

    3x = \ln(2) + 5x

    Subtract 5x.

    -2x = \ln(2)

    Divide by -2.

    x = -\frac{\ln(2)}{2}
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  3. #3
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    Helolo, aldric!

    Another approach . . .


    Solve for x using logs: . 2e^{3x}\:=\:4e^{5x}

    We have: . 4e^{5x} \;=\;2e^{3x}


    Divide by 4e^{3x}\!:\;\;\frac{4e^{5x}}{4e^{3x}} \;=\;\frac{2e^{3x}}{4e^{3x}} \quad\Rightarrow\quad e^{2x} \;=\;\tfrac{1}{2}


    Take logs: . \ln\left(e^{2x}\right) \;=\;\ln\left(\tfrac{1}{2}\right)

    . . . . . . . 2x\cdot\underbrace{\ln(e)}_{\text{This is 1}} \;=\;\ln\left(2^{-1}\right)

    . . . . . . . . . . . . 2x \;=\;-\ln(2)

    . . . . . . . . . . . . x \;=\;-\tfrac{1}{2}\ln(2)

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  4. #4
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    i.e. x=ln \left(\frac{1}{\sqrt{2}}\right).
    At least, It has nice from. =)
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