What is log(a+b)?


by Swapnil
Tags: loga
Swapnil
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#1
Nov12-06, 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 made-up function that deals with log(a+b?
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FunkyDwarf
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#2
Nov12-06, 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)
Swapnil
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#3
Nov12-06, 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?

z-component
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#4
Nov12-06, 10:12 PM
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What is log(a+b)?


There is an inverse for the logarithm which is the exponential, but there is no "reverse."
leon1127
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#5
Nov12-06, 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))=?
CRGreathouse
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#6
Nov13-06, 01:33 AM
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Quote Quote by leon1127
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))=?
g(f(a+b)) = log(exp(a+b)) = a+b
f(g(a+b)) = exp(log(a+b)) = a+b for a+b>0
CRGreathouse
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#7
Nov13-06, 01:34 AM
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Quote Quote by Swapnil
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?
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.)
FunkyDwarf
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#8
Nov13-06, 11:07 PM
P: 479
Oo thats clever, nice one :)
dextercioby
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#9
Nov14-06, 03:20 AM
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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.
Simple
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#10
Apr18-08, 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
clayton1975mg
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#11
May13-09, 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 non-simple mathematical operations that can resolve this issue.
AUMathTutor
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#12
May13-09, 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.
Mark44
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#13
May15-09, 12:52 AM
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Quote Quote by CRGreathouse View Post
There's nothing particularly nice. Sometimes it is convenient to use

[tex]\log(a+b)=\log(a(b/a+1))=\log(a)\cdot\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.)
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.
CRGreathouse
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#14
May15-09, 01:35 AM
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Quote Quote by Mark44 View Post
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.
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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