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Integration - Find the x coordinate and the area under the curve

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/photo/my-images/703/dfdfu.jpg

    2. Relevant equations

    To find the x coordinate

    1. Make both equations equal, expose e and take logs. I'm not sure how to do this and I've tried but keep getting the wrong answer.

    2. To find the area, subract the higher curve from the lower curve and integrate. I can do this. I only need help finding the x coordinate.

    3. The attempt at a solution

    1. y=e^(x-3)/2 and y =e^(2x-7)

    e^(x-3)/2 =e^(2x-7)

    Now take logs of both sides. How?
     
  2. jcsd
  3. Dec 4, 2011 #2
    [tex] \ln (e^{f(x)})=f(x) [/tex] But watch out for the fraction over 2. First, move it over to the other side. Then be careful, as [itex] \ln (2e^{2x-7}) \neq 2e^{2x-7} [/itex]
     
  4. Dec 4, 2011 #3

    Thanks for your reply. I get:

    1/2e^(x-3)=e^(2x-7)
    e^(x-3)=2(e^(2x-7))
    lne^(x-3)=ln(e^(2x-7))
    (x-3)/2=2x-7
    x=11/3

    'Express your answer in the form a-lnb.How?
     
  5. Dec 4, 2011 #4
    Whoops! your 2 disappeared for a line, then it came back. You can't do that. You still need to figure out what [itex] \ln (2e^{2x-7}) [/itex] is. Hint: [itex] \ln (a) + \ln (b)=\ln (ab) [/itex]
     
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