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Math Help - Quick Question about Logarithms and Exponents

  1. #1
    Junior Member
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    Quick Question about Logarithms and Exponents

    I have an assignment question looks like this

    2e^-^2^x) = 5

    I figured, first time around, that I could just apply natural logarithm to both, and get
    -4x = ln(5)

    and then x = ln(5)/-4

    This does not result in the correct answer.

    Why? I can get the correct answer (by dividing both sides by two initially)

    If I have 2e^(-2x) and apply the natural logarithm to it, what does it become?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Quick Question about Logarithms and Exponents

    Divide through by 2 first, then convert from exponential to logarithmic form.

    If you take the natural log of both sides, you have:

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

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

    -2x=\ln(5)-\ln(2)=\ln\left(\frac{5}{2} \right)

    Same as if you had just divided through by 2 initially.
    Last edited by MarkFL; May 1st 2013 at 10:56 PM.
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