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Math Help - Is this correct

  1. #1
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    Is this correct

    is it?

    i just want to make sure.

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by subzero06 View Post
    is it?

    i just want to make sure.

    Nope, not right. \ln ({\color{red}-} 0.73) should send up a whole bunch of red flags, by the way
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  3. #3
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by subzero06 View Post
    is it?

    i just want to make sure.
    Jhevon gives good advice: when possible, make sure that your answer makes sense within the context of the problem.

    \ln e^{-0.73t}=-0.73t.
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  4. #4
    MHF Contributor
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    Talking

    Might it help to move the exponential to get a positive power...?

    . . . . . P\, +\, 56.5Pe^{-0.73t}\, =\, 230

    . . . . . P\, +\, \frac{56.5P}{e^{0.73t}}\, =\, 230

    . . . . . e^{0.73t}P\, +\, 56.5P\, =\, 230 e^{0.73t}

    . . . . . 56.5P\, =\, 230 e^{0.73t}\, -\, e^{0.73t}P

    . . . . . 56.5P\, =\, \left(230\, -\, P\right) e^{0.73t}

    . . . . . \frac{56.5P}{230\, -\, P}\, =\, e^{0.73t}

    . . . . . \ln{\left(\frac{56.5P}{230\, -\, P}\right)}\, =\, 0.73t

    ...and so forth.



    P.S. To review why "ln(e^(some power))" simplifies to just the (some power), try here.
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