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Natural log question.

  1. Dec 9, 2006 #1
    I have the problem 3e^(x+2) = e^(-x)

    and I need to find X. It's been a while since i've looked at any Calculus 2 stuff so im not sure on what to do here.

    I take the natural log of both sides

    LN(3e^(x+2) =LN(e^(-x))= (x+2)ln3e = -xlne

    So (x+2)Ln3e= -x

    Where do I go from here?
     
  2. jcsd
  3. Dec 9, 2006 #2

    cristo

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    It's easier to collect the terms before you log both sides. Try collecting the exponentials on one side. Do you know, for example, how to simplify [tex] e^ae^b [/tex] ?
     
  4. Dec 9, 2006 #3
    the same was as you would any number x^3 * x^2 = X^5

    So e^(x+2) * e^(-x) = (x+2-x) or e^2, correct?
     
  5. Dec 9, 2006 #4

    cristo

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    yes, [tex] e^{x+2}e^{-x}=e^2 [/tex] However, when manipulating the above equation, you need to multiply both sides by [tex] e^{x} [/tex] in order to cancel the [tex] e^{-x} [/tex] on the RHS
     
  6. Dec 10, 2006 #5

    Mute

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    Also note that ln(3*e^a) is not a*ln(3e), so unless you just didn't put some brackets around (3e)^(x+2), your current expression on the left-hand-side of your last line is incorrect.
     
  7. Dec 10, 2006 #6

    HallsofIvy

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    You don't go anywhere. Your first step was wrong: ln(3ex+2)= ln 3+ (x+2) not (x+2)ln(3e).

    Your final equation is ln(3)+ x+2= -x. Would you know how to solve
    A+ x+ B= -x for constants A and B?
     
  8. Dec 11, 2006 #7

    dextercioby

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    Others pointed out flaws, i just wanna be sure that you can solve

    [tex] e^{-2x}=3e^{2} [/tex]

    Daniel.

    P.S. I get [itex] x=-\frac{1}{2}\ln 3 -1 + ik\pi \ , \ k\in\mathbb{Z} [/itex]
     
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