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Thread: Two More Natural Log Problems

  1. October 19th 2010, 06:36 PM #1
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    Two More Natural Log Problems

    Help to get me in the right direction will be much appreciated.
    Two More Natural Log Problems-math2.jpg
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  2. October 19th 2010, 06:57 PM #2
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    You are correct up to

    e^6 = 4 - x

    e^6 - 4 = -x

    x = 4 - e^6.


    For the second

    \displaystyle{e^{\frac{x}{4}} = 2^{5 - 9x}}

    \displaystyle{\frac{x}{4} = \ln{(2^{5 - 9x})}}

    \displaystyle{\frac{x}{4} = (5 - 9x)\ln{2}}

    \displaystyle{x = 4(5 - 9x)\ln{2}}.
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