
#1
Nov1206, 08:13 PM

P: 460

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 madeup function that deals with log(a+b?




#2
Nov1206, 08:37 PM

P: 479

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)




#3
Nov1206, 09:30 PM

P: 460

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?




#4
Nov1206, 10:12 PM

P: 483

What is log(a+b)?
There is an inverse for the logarithm which is the exponential, but there is no "reverse."




#5
Nov1206, 11:49 PM

P: 487

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))=? 



#6
Nov1306, 01:33 AM

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P: 3,680

f(g(a+b)) = exp(log(a+b)) = a+b for a+b>0 



#7
Nov1306, 01:34 AM

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[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.) 



#8
Nov1306, 11:07 PM

P: 479

Oo thats clever, nice one :)




#9
Nov1406, 03:20 AM

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P: 11,866

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. 



#10
Apr1808, 03:00 PM

P: 1

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 canīt use your calculator then Log 5 ? You must think in Log (3 + 2) but thereīs 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 



#11
May1309, 12:18 PM

P: 1

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 nonsimple mathematical operations that can resolve this issue. 



#12
May1309, 02:16 PM

P: 490

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. 



#13
May1509, 12:52 AM

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P: 21,069

[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. 



#14
May1509, 01:35 AM

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