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Integrate ln

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Find area of region bounded by curve with equation y=e^2x , the x-axis and the lines x=-ln3 and x=-ln2.

    2. Relevant equations
    log. law + integration

    3. The attempt at a solution
    well here is how i started this:
    y=e^2x after integration
    =>(1/2e^-(ln6)-(1/2^-(ln4) )
    =>(1/2e^1/6) - (1/2e^1/4)
    =>(1/2e^1/6)/ (1/2e^1/4)
    =>(1/6(lne)1/2) / (1/4(lne)1/2)
    (1/6X1/2) / (1/4X1/2)
    => is this correct?
  2. jcsd
  3. Dec 9, 2009 #2


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    Homework Helper

    e^(-ln6) is not equal to e^1/6; it's equal to e^ln(1/6). What happens to any number if you take its natural log and raise e to the result?
  4. Dec 9, 2009 #3
    y=lnx => x=e^y ?
    Last edited: Dec 9, 2009
  5. Dec 9, 2009 #4
    still confused... but i thought just as -ln3 = 3^-1 = 1/3 ,same would happen to this but guess thats not the case ?
  6. Dec 9, 2009 #5


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    Homework Helper

    It is the case, but you wrote:


    then, in the next step,


    That would mean -(ln6) is equal to 1/6, which it isn't. Instead, -(ln6) is equal to ln(1/6).
  7. Dec 9, 2009 #6
    ah k, now i get it yea.. 1/6 would only be equal to 6^-1 , erm.. sorry for being such a nuisance but now what to do with the ln(1/6)? , i cant ln both sides .. =/
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