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Log function e^(ln12/2)?

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

    The full problem is y=e^(x/2) and y=e^(-x) revolved around the x-axis, from [ln6,ln12].
    The part I'm struggling with is evaluating 2e^(ln12/2) and 2e^(ln6/2).
    I am just reviewing for my Calculus 1 final.

    2. Relevant equations



    3. The attempt at a solution

    [tex]2e^{(ln12/2)}[/tex]

    [tex]2 e^{(1/2)(ln12)}[/tex]

    [tex]2 (e^{1/2}), e^{ln12}=12, and 2(12)e^{1/2}???[/tex]
     
  2. jcsd
  3. Dec 15, 2011 #2

    rock.freak667

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    Well you can easily show that e^(lnA) = A. So you can use that result as is.
     
  4. Dec 15, 2011 #3
    The problem is [tex]e^{(\frac{ln12}{2})}[/tex]
    not
    [tex]e^{ln\frac{12}{2}}[/tex], that would be e^(ln6)=6
     
  5. Dec 15, 2011 #4
    I have a full equation that looks like this:
    [tex]\pi((2e^{\frac{ln12}{2}}+e^{-12})-(2e^{\frac{ln6}{2}}+e^{-6}))[/tex]
     
  6. Dec 15, 2011 #5

    SammyS

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    I doubt that you have stated the full problem ?

    What are you finding? Volume, Area, ... ?

    What is being revolved around the x-axis ?
     
  7. Dec 15, 2011 #6
    Sorry, maybe I have made a mistake in my prior work. (this may need to be moved to the calculus section)
    The problem states: Use the washer method to find the volume of the solid when the region bounded by y=e^(x/2) and y=e^(-x), x=ln6, x=ln12, when the region is revolved around the x-axis.

    So I started with:
    [tex]\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx[/tex]

    [tex]\pi(2e^{x/2}+e^{-x})|_{ln6}^{ln12}[/tex]

    [tex]\pi((2e^{\frac{ln12}{2}}+e^{-ln12})-(2e^{\frac{ln6}{2}}+e^{-ln6}))[/tex]
     
    Last edited: Dec 15, 2011
  8. Dec 16, 2011 #7

    Mentallic

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    [tex]a^{\frac{1}{2}}=\sqrt{a}[/tex]

    and

    [tex]\left(a^b\right)^c=a^{bc}[/tex]

    using these two rules for indices, you should be able to answer your question.
     
  9. Dec 16, 2011 #8
    Thanks, so...
    [tex]2e^{\frac{ln12}{2}}=2\sqrt{e^{ln12}}=2\sqrt{12}[/tex]
    [tex]=4\sqrt{3}[/tex] and
    [tex]2\sqrt{e^{ln6}}=2\sqrt{6}[/tex]
    giving me:
    [tex]\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))[/tex]
    [tex]\pi(\frac{1}{12}+4\sqrt{3}-2\sqrt{6})[/tex]
    But that was wrong, and I am unsure how to get the correct answer.
    How does that become [itex]\frac{575\pi}{96}[/itex]?
     
  10. Dec 16, 2011 #9
    No, this isn't true.

    [tex]-\frac{1}{12},-\frac{1}{6}[/tex] here is mistake.
    Hm, I checked this in one program, and it isn't true also.
     
    Last edited: Dec 16, 2011
  11. Dec 16, 2011 #10

    SammyS

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    Using the washer method to find the volume of the solid when the region bounded by y=f(x) and y=g(x), x=a, x=b, is revolved around the x-axis, (Assumes f(x)>g(x) on the interval [a,b].) results in the following integral:
    [itex]\displaystyle \int_a^b\pi\left((f(x))^2-(g(x))^2\right)\,dx[/itex]​

    You failed to square the two functions.
     
  12. Dec 16, 2011 #11
    Man do I feel like an idiot. Thanks for pointing that out.
     
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