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What is log(a+b)?

  1. Nov 12, 2006 #1
    What is log(a+b)? This is one of those questions that has been bothering me since the day learned about logs. log(a*b) = log(a)+log(b) but is there a symmetric relation like log(a+b) = gol(a)*gol(b) or something like that? Or is there even a made-up function that deals with log(a+b?
     
  2. jcsd
  3. Nov 12, 2006 #2
    well for starters theres no such thing as gol. Youre thinking of like FoG and G of F which doesnt apply in this case as log stands for logarithm. And no log(a+b) is just log(a+b)
     
  4. Nov 12, 2006 #3
    I know! I just made that up. Its the reverse of log. I was just trying to say that wouldn't it be nice that there is a function called gol which has the above property. So is there a special function that mathematicians have invented which has that property?
     
  5. Nov 12, 2006 #4
    There is an inverse for the logarithm which is the exponential, but there is no "reverse."
     
  6. Nov 12, 2006 #5
    g(a*b)=log(a*b) = log(a) + log(b)=g(a)+g(b);
    f(a+b)=e^(a+b) = e^a e^b=f(a)*f(b)
    coincidence? maybe

    g(f(a+b))=?
    f(g(a*b))=?
     
    Last edited: Nov 12, 2006
  7. Nov 13, 2006 #6

    CRGreathouse

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    g(f(a+b)) = log(exp(a+b)) = a+b
    f(g(a+b)) = exp(log(a+b)) = a+b for a+b>0
     
  8. Nov 13, 2006 #7

    CRGreathouse

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    There's nothing particularly nice. Sometimes it is convenient to use

    [tex]\log(a+b)=\log(a(b/a+1))=\log(a)+\log(b/a+1)[/tex]

    if a and b are of differing magnitudes. I used that identity when making a calculator that stored numbers in the form b^b^b^...b^X, where only the number of levels of exponentiation and the final exponent were tracked. (I'm torn on what base to use; 2, e, 10, native word size, or its square root.)
     
    Last edited by a moderator: May 15, 2009
  9. Nov 13, 2006 #8
    Oo thats clever, nice one :)
     
  10. Nov 14, 2006 #9

    dextercioby

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    Well, there's one

    [tex] \ln \left(a+b\right)=\int_{1}^{a+b} \frac{dx}{x} [/tex]

    and you can see it's invariant under the transformation [itex] a\rightarrow b \ , \ b\rightarrow a [/itex].

    Daniel.
     
  11. Apr 18, 2008 #10
    log(a+b) help

    First
    Log(a+b) is Log(a+b), but if you want to solve a regular problem like this :

    Log 2 = 0.3 Log 3 = 0.48 (aprox) and you can´t use your calculator then

    Log 5 ?

    You must think in Log (3 + 2) but there´s no way to solve this way.

    Log (10/2) = Log 5 then using Logarithm fundamentals:

    Log 10 - Log 2 = 1 - log 2 = 1 -0.3 = 0.7
     
  12. May 13, 2009 #11
    Re: log(a+b)=?

    Hi. I studied this question in my vocation and I concludes this not possible. Look:

    conditions:
    1) log(a+b)= log(a) $ log(b)
    2) ln(a+b)= ln(a) $ ln(b)
    3) a=b=1


    consequences

    1) log(1+1)= log (1) $ log (1)
    2) ln(1+1)= ln (1) $ ln (1)

    so

    1) log (2) = 0 $ 0
    2) ln(2) = 0 $ 0

    them

    log(2)=ln(2)

    but it not possible. So:

    There is no operation Mathematics simple that can solves that question.


    I am studying non-simple mathematical operations that can resolve this issue.
     
  13. May 13, 2009 #12
    Re: log(a+b)=?

    Assume such a function gol(x), did exist. Then

    gol(a)gol(b) = log(a+b).

    log(4) = log(2+2)
    gol(2)gol(2) = 2

    implies that gol(2) = sqrt(2) for logarithms in base 2.

    log(6) = log(3+3)
    gol(3)gol(3) = log(6).

    implies gol(3) = sqrt(log 6).

    Now log(5) = log(2+3)
    so gol(2)gol(3) = log(5).

    But gol(2) = sqrt(2) and gol(3) = sqrt(log 6).

    A simple calculation shows that sqrt(2)sqrt(log 6) is not equal to log(5).

    sqrt(2)sqrt(log 6) = log(5)?
    sqrt(2log6) = log(5)?
    sqrt(log36) = log(5)?
    log(36) = log(5)^2?
    5.1699250014423123629074778878956 = 5.3913500778272559669432034405889?
    no.

    So no such function can exist... it wouldn't be a function in the technical sense because it would require either 2 or 3 to map to different things depending on the situation.
     
  14. May 15, 2009 #13

    Mark44

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    Re: log(a+b)=?

    Make that

    [tex]\log(a+b)=\log(a(b/a+1))=\log(a) + log(b/a+1)[/tex]

    and I'll be a lot happier; i.e., that last multiplication should be an addition.
     
  15. May 15, 2009 #14

    CRGreathouse

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    Re: log(a+b)=?

    That's a bad typo; I had it right in my program. I'd edit the post, but the edit window expired about three years ago.
     
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