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Thread: logs problem

  1. Dec 21st 2009, 02:36 PM #1
    math321
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    logs problem

    $\displaystyle 20000=Pe^{0.073*12}$



    solve for p
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  2. Dec 21st 2009, 03:07 PM #2
    skeeter
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    Quote Originally Posted by math321 View Post
    $\displaystyle 20000=Pe^{0.073*12}$



    solve for p
    $\displaystyle P = \frac{20000}{e^{0.073(12)}}$


    you should have asked this in your previous thread since it belongs to the same question ...

    http://www.mathhelpforum.com/math-he...-interest.html
    Last edited by skeeter; Dec 21st 2009 at 03:18 PM.
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