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Properties of logs question

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    the question is this:
    show[​IMG]
    is 2[tex]\sqrt{3}[/tex]

    whereby e is natural logarithm base

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 3, 2008 #2

    lurflurf

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    Re: Logarithm

    What have you tried?
    Remember
    x^y=e^[x log(y)]
     
  4. Dec 3, 2008 #3

    Integral

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    Re: Logarithm

    What do you know about the properties of logs?
     
  5. Dec 3, 2008 #4
    Re: Logarithm

    i suppose everything.. i just cant show tat it is 2sqrt3
     
  6. Dec 3, 2008 #5

    Integral

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    Re: Logarithm

    Do you have a text?

    What are some of the properties of logs. They should be highlighted in boxes. RTFM
     
  7. Dec 3, 2008 #6
  8. Dec 3, 2008 #7

    Integral

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    Re: Logarithm

    look at the product rule:

    Log(M*N) = logM + LogN

    Now consider Log (M 2) = log (M*M) = log M + log M = 2log M

    This is justification for the rule that is not shown on that page.

    Log (MN) = N log M

    Now apply that to the exponents of your problem.
     
  9. Dec 3, 2008 #8
    Re: Logarithm

    but it is just the exponent constant we cant apply the logarithm rules unless to the power.
     
  10. Dec 3, 2008 #9

    Integral

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    Re: Logarithm

    Repeat in english please
     
  11. Dec 3, 2008 #10
    Re: Logarithm

    oh.. let me rephrase
    the equation is just the the sums of 2 exponent constant thus we cant apply the any logarithm rules except to the power of the constant
     
  12. Dec 3, 2008 #11

    Integral

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    Re: Logarithm

    You can apply the rules of logs to logs where ever they appear. Start by appling the rules to the logs which appear as exponents in your proplem.


    [tex] e^{xlny} = e^{lny^x} [/tex]
     
  13. Dec 3, 2008 #12
    Re: Logarithm

    yup i applied already
    3e^(ln (1/sqrt3) ) + e^(ln sqrt3)
     
  14. Dec 3, 2008 #13

    HallsofIvy

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    Re: Logarithm

    You need: a ln(b)= ln(ba) and eln(x)= x.
     
  15. Dec 3, 2008 #14

    Integral

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    Re: Logarithm

    now you are getting some where.

    you do know that :

    [tex] e^{lnx} = x [/tex]
     
  16. Dec 3, 2008 #15
    Re: Logarithm

    cool it works.
    i have never seen this law before!
    eln(x)= x

    thank you HALLSOFIVY and INTEGRAL .:smile:
    mind explaining this law eln(x)= x ?
     
  17. Dec 3, 2008 #16

    Integral

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    Re: Logarithm

    On the page YOU linked read the POWER RULE of logs.

    then understand that ln = loge
     
  18. Dec 3, 2008 #17
    Re: Logarithm

    ya.. i know about it.
    how about this?
    e^ln(x)= x
     
  19. Dec 3, 2008 #18
    Re: Logarithm

    The natural logarithm is the 'inverse' of the e-exponential. If ex is f(x), and ln(x)=g(x), then you can rewrite the equation eln(x) to f(g(x)), but g(x)=f-1(x) (it's inverse), so f(f-1(x))=x. Or qualitatively, applying a function (the exponential) to it's inverse (the natural logarithm), or vice versa, returns it's argument ('input'), they 'undo' each other.

    I'm sure there is a more rigorous and mathematically correct proof out there though.
     
  20. Dec 3, 2008 #19
    Re: Logarithm

    i do agree with your proving however until that inverse portion what do you really do to inverse it? i know it is somehow inverse of its law but how do you really come about to do it?
     
  21. Dec 3, 2008 #20
    Re: Logarithm

    btw you see, ln x means x=e^? and e^x means e x e x e x e x ...
    ^
    l
    x factors
    i dont really see the relationship
     
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