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F'(x) of log x

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data
    f'(x) of log10 x


    2. Relevant equations
    It's quite a nightmare to write on here if your not particularly fast with the code thing!

    3. The attempt at a solution
    I eventually reached 1/x * log10e
    However, the answer should be 1/(x ln 10).
    Is there a stage from what I obtained that leads to the answer?
    If not, then there's something incorrect 'further up' the calculations and I'll go back and have a look... Just would appreciate some input as to whether or not I've arrived at a correct stage.
    Thanks.
     
  2. jcsd
  3. Jan 2, 2009 #2

    jgens

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    Gold Member

    Your answer is correct. The discrepancies between the two answers can be easily resolved with the change of base property of logarithms.

    Ex: (log_a (b)) = 1/(log_b (a)).
     
  4. Jan 2, 2009 #3

    HallsofIvy

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    For any numbers, a, b, x,
    [tex]log_b(x)= \frac{log_a(x)}{log_a(b)}[/tex]

    In particular,
    [tex]log_{10}(x)= \frac{ln(x)}{ln(10)}[/itex]

    Therefore
    [tex]\frac{dlog_{10}(x)}{dx}= \frac{d ln(x)}{dx}\frac{1}{ln(10)}[/tex]
    [tex]= \frac{1}{x}\frac{1}{ln(10)}= \frac{1}{x ln(10)}[/tex]

    BUT, by that same initial formula,
    [tex]log_{10}(e)= \frac{ln(e)}{ln(10}= \frac{1}{ln(10)}[/tex]

    so your answer is also perfectly correct!

    In general
    [tex]log_a(b)= \frac{1}{log_b(a)}[/tex]
     
    Last edited: Jan 5, 2009
  5. Jan 2, 2009 #4
    Thanks!
    I arrived at that just now. Thought I'd log in to update my post and ask to see if the method was correct, but it's been confirmed! Thanks for the quick responses (and the detail)!
    :smile: Speak soon (hopefully not too soon, otherwise it means I'm stuck).
     
  6. Jan 2, 2009 #5

    HallsofIvy

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    Speak what soon? You said that you have already confirmed that what you had was correct and two people have already agreed!
     
  7. Jan 5, 2009 #6
    No! I was just being polite!:smile:
    You also arrived at the answer through a far quicker method than I used...
     
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