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Math Help - Trying to figure this out?

  1. #1
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    Trying to figure this out?

    We were given 1 example to do math questions. But I don't get how I got the answer.

    With with Log and Ln, but in this one it's Ln.
    2.6=1.4e^(0.3t)
    I have to solve for t.
    So you have to divide both sides by 1.4 and add Ln as well.
    Ln(2.6/1.4) = Ln e^(0.3t)
    Now my example in my book does not make sense. I did not make a note next to it.
    What are the next examples to get to the answer of t=2.06.

    If someone would kindly help me out that would be appreciated.
    Thanks
    Joanne
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  2. #2
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    Quote Originally Posted by bradycat View Post
    We were given 1 example to do math questions. But I don't get how I got the answer.

    With with Log and Ln, but in this one it's Ln.
    2.6=1.4e^(0.3t)
    I have to solve for t.
    So you have to divide both sides by 1.4 and add Ln as well.
    Ln(2.6/1.4) = Ln e^(0.3t)
    Now my example in my book does not make sense. I did not make a note next to it.
    What are the next examples to get to the answer of t=2.06.

    If someone would kindly help me out that would be appreciated.
    Thanks
    Joanne

    You don't "add" log or ln to both sides: you apply it to both sides, and that only if you're sure they are positive, so:

    \displaystyle{\ln\left(\frac{2.6}{1.4}\right)=\ln(  e^{0.3t})=0.3t\Longrightarrow t=\frac{\ln\left(\frac{2.6}{1.4}\right)}{0.3}} , as

    \ln e^x=x since e^x\,,\ln x are inverse function to each other.

    Tonio
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  3. #3
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    The functions \ln x and e^x are inverse, i.e., \ln(e^x)=x for all x and e^{\ln x}=x for all x>0. Therefore, \ln(2.6/1.4) = \ln e^{0.3t}=0.3t, from which t can be found.
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  4. #4
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    You are on the right track,

    2.6 = 1.4 e^{0.3t}

    \frac{2.6}{1.4}= e^{0.3t}

    \ln \frac{2.6}{1.4}= 0.3t

    \displaystyle\frac{\ln \frac{2.6}{1.4}}{0.3}= t
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  5. #5
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    OK I get that now. But what happens to the e on the 2nd line?? Is that the e that is applied to the left side as Ln??? Do I have that correct?

    Thank you all for the help to clear this up.
    Joanne

    OK I called a friend of mine and the Ln is e with a base of e, so the answer is 1. So it's not needed in a way, correct?
    Last edited by bradycat; December 4th 2010 at 11:47 AM.
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  6. #6
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    They are the inverse of each other.

    I.e x+3-3 = x-3+3 = x

    \displaystye \ln e^x = e^{\ln x} = x

    and also,

    \displaystye \log_{a} a^x = a^{\log_{a} x} = x
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