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Math Help - [SOLVED] finding x, using the natural logarithm

  1. #1
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    Question [SOLVED] finding x, using the natural logarithm

    I am having trouble with this question. I keep getting X to cancel out somehow.

    e^(ax) = Ce^(bx), a cannot = b, Find X

    Please help me!
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  2. #2
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    Quote Originally Posted by dee18 View Post
    I am having trouble with this question. I keep getting X to cancel out somehow.

    e^(ax) = Ce^(bx), a cannot = b, Find X

    Please help me!
    Take the natural log of both sides.

     \ln(e^{ax}) = \ln(Ce^{bx})

    Using log rules...

     ax\ln(e) = bx\ln(Ce)

    More rules...

    ax(1) = bx[\ln(C)+\ln(e)]

    Finally ax=bx(\ln(C)+1)

    Can you solve it now?
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  3. #3
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    e^{ax} = Ce^{bx}

    \Rightarrow C = \frac{e^{ax}}{e^{bx}}

    \Rightarrow C = e^{(a-b)x}

    Now recall that:

    \ln{A^x} = x\ln{A}

    Solve for x now.
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  4. #4
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    Sigh, Chop Suey never fails to do the same problem faster and more elegantly...
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  5. #5
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    Thanks guys! I figured it out
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