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Math Help - Solving log/exponential problem

  1. #1
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    Solving log/exponential problem

    So I have a question about this problem:

    5e^2-x = 7e^x

    i know you need to apply Ln to both sides, but first can you bring the 5 up to the exponent? or does this not apply...
    meaning can you go

    e^10-5x = e^7x
    10-5x = 7x
    10= 12x
    x = 5/6?
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  2. #2
    MHF Contributor harish21's Avatar
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    5e^{2-x} \neq e^{10-2x}

    start the problem by taking log on both sides.
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  3. #3
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    e^(i*pi)'s Avatar
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    I don't understand your first line at all? The 5 is not distributed into the exponent. 5e^t = e^t + e^t + e^t + e^t + e^t \neq e^{5t}

    Instead take the natural log of both sides, you can use the addition and power laws to simplify

    \ln 5 + \ln(e^{2-x}) = \ln(7) + \ln(e^x) \implies \ln 5 + 2-x = \ln 7 + x

    Solve the linear equation
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  4. #4
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    So it would be:

    ln5 +(2-x) = ln7 +x
    ln(5/7) = 2x-2
    x = ln(5/7) +2 /2
    x = 0.832
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  5. #5
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    e^(i*pi)'s Avatar
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    2x = \left(\ln\dfrac{5}{7} + 2\right)

    x = \dfrac{1}{2} \left(\left(\ln \dfrac{5}{7}\right) +2\right) = \dfrac{1}{2} \left(\ln \dfrac{5}{7}\right) + 1
    Last edited by e^(i*pi); February 17th 2011 at 04:22 PM. Reason: latex
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