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Homework Help: Trancendental Functions

  1. Apr 16, 2006 #1
    Im stuck on these probelms
    13ln[exp[(x^2+2)] - exp[ln(13x^2+26)]]

    exp[ln(13x^2+26) cancels out to 13x^2+26 but i dont see how that helps with the entire probelm

    x^(e^x) my friend suggested power rule...but that seems a litle too simple and would create a quite a mess

    thank you in advance
    Last edited: Apr 16, 2006
  2. jcsd
  3. Apr 16, 2006 #2
    For the simplification question, whatever's in the exp() or ln() when one is inside the other gets out because ln and exp are inverses... so you're left to simplify some algebraic expressions! (Some cancelling out happens)

    And for finding the derivative of x^(e^x): (assuming it's defined as a function f)
    f = x^(e^x)
    Take the natural log of both sides
    ln f = (e^x) ln x

    And use the chain rule and product rule and solve for f'. :)
    Last edited: Apr 16, 2006
  4. Apr 16, 2006 #3


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    I presume that if you know that, then you also know that exp[ln(13x^2+ 26)]= 13x^2+ 26 (exp and ln are "inverse" functions and "cancel" each other.

    And you don't see how reducing 13ln[exp[(x^2+2)] - exp[ln(13x^2+26)] to 13(x^2+ 2)- (13x^2+ 26) helps at all?? Believe, me that simplifies trivially!

    Use "logarithmic differentiation". If y= x^(e^x), then ln y= (e^x)x. On the left, (ln y)'= (1/y)y' and you should be able to differentiate (e^x)x using the product rule. Solve the resulting equation for y'.
  5. Apr 17, 2006 #4
    The way it's written reduces to 13ln(e^expression - 13*expression), where expression is x^2+2, so it doesn't reduce trivially. Unless it's written wrong in the OP.
  6. Apr 17, 2006 #5
    hmm how exactly do you get to the algebraic expression 13(x^2+ 2)- (13x^2+ 26) the function dave posted was the farthest i got..:confused:
  7. Apr 17, 2006 #6
    I think you're misreading the OP's post. Lets write this out in LaTex so there's no confusion. What it looks like is written is,

    [tex]13\ln (e^{x^2+2}) - e^{\ln (13x^2 + 26)} [/tex]

    If this is correct, than what Pseudo and Halls have said stands.

    To OP:
    To reduce this, note that e and ln are inverse functions, as Pseudo and Halls have already mentioned. So, for any positive real number a, you get the following

    [tex] \ln e^a = a [/tex]

    [tex] e^{\ln a} = a[/tex]

    [tex] \ln (e^{x^2 + 2}) = x^2 + 2 [/tex]
    Last edited: Apr 17, 2006
  8. Apr 17, 2006 #7
    the function is 13ln (exp[(x^2+2)] - exp[ln(13x^2+26)])

    NOT seperately as in (13ln[exp[(x^2+2)] ) - (exp[ln(13x^2+26)])

    sorry if i was unclear, writing in laTex is too tedious for me
    Last edited: Apr 17, 2006
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