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Logs & e

  1. Jan 11, 2004 #1
    Need a kickstart with this one:
    f(t)=t^(3/2)log(of 2)Sqrt(t+1)

    Integral of (3-x)7^[(3-x)^2] dx

    7^[(3-x)2] = e^[(3-x)2ln 7]
    u=(3-x)^2
    du/dx = -2(3-x)
    (3-x)dx = -1/2 du
    not even sure what so far is right..
     
  2. jcsd
  3. Jan 11, 2004 #2

    HallsofIvy

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    I'm not sure what the question is!

    "f(t)=t^(3/2)log(of 2)Sqrt(t+1)"

    Okay, what's the question???


    "Integral of (3-x)7^[(3-x)^2] dx

    7^[(3-x)2] = e^[(3-x)2ln 7]
    u=(3-x)^2
    du/dx = -2(3-x)
    (3-x)dx = -1/2 du
    not even sure what so far is right.."

    Seeing the exponent (3-x)2 and (3-x) multiplying the exponential, the first thing I would try is "let u= (3-x)2". Then du= -2(3-x)dx so the integral becomes

    -2 times Integral of 7udu.

    If you don't know the derivative and anti-derivative of 7u, remember that 7u= eu ln(7).
     
  4. Jan 11, 2004 #3
    Ah, sorry about the first one.

    f(t)=t^(3/2)log(of 2)Sqrt(t+1)
    I need to derive that.
     
  5. Jan 12, 2004 #4
    I feel u want derivative , if so then hint is

    take log on both sides and then differentiate
     
  6. Jan 12, 2004 #5

    HallsofIvy

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    I don't see any reason to take the logarithm. It's looks like a pretty direct application of the product rule and chain rule.

    f(t)=t3/2(log2[/sup](√(t+1))

    f'= (t3/2)'log2[/sup](√(t+1))+(t2)(log2[/sup](√(t+1))'

    (t3/2)'= (3/2)t1/2, of course.

    To differentiate log2(x) recall that log2(x)= ln(x)/ln(2) so (log2(x))'= 1/(xln(2)).
     
  7. Jan 12, 2004 #6
    There are many ways of doing a problem, though both are easy to use.

    yes it is a direct problem involving the product rule and chain rule
     
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