# Is there a relationship between log(a+b) and other logarithmic functions?

• Swapnil
In summary, there is no special function invented by mathematicians that has the property of log(a+b) = gol(a)*gol(b). However, there are some identities and operations that involve logarithms which can be useful in certain situations.
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?

well for starters there's no such thing as gol. Youre thinking of like FoG and G of F which doesn't apply in this case as log stands for logarithm. And no log(a+b) is just log(a+b)

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?

There is an inverse for the logarithm which is the exponential, but there is no "reverse."

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))=?

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leon1127 said:
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

Swapnil said:
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

$$\log(a+b)=\log(a(b/a+1))=\log(a)+\log(b/a+1)$$

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.)

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Oo that's clever, nice one :)

Well, there's one

$$\ln \left(a+b\right)=\int_{1}^{a+b} \frac{dx}{x}$$

and you can see it's invariant under the transformation $a\rightarrow b \ , \ b\rightarrow a$.

Daniel.

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

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.

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.

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

$$\log(a+b)=\log(a(b/a+1))=\log(a)\cdot\log(b/a+1)$$

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

$$\log(a+b)=\log(a(b/a+1))=\log(a) + log(b/a+1)$$

and I'll be a lot happier; i.e., that last multiplication should be an addition.

Mark44 said:
Make that

$$\log(a+b)=\log(a(b/a+1))=\log(a) + log(b/a+1)$$

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.

## 1. What is the definition of log(a+b)?

The logarithm of a sum, log(a+b), is equal to the sum of the logarithms of each individual term, log(a) + log(b). This is known as the log rule for addition.

## 2. How is log(a+b) related to exponents?

The logarithm of a sum, log(a+b), is equivalent to the exponent that the base (commonly denoted as 'b') must be raised to in order to get the sum (a+b). In other words, log(a+b) is the inverse function of b^(log(a)+log(b)).

## 3. What happens to log(a+b) when a and/or b are equal to 0?

If either a or b is equal to 0, then log(a+b) is undefined because the logarithm function is not defined for non-positive numbers. In other words, the sum of two or more non-positive numbers does not have a logarithm.

## 4. Can log(a+b) be simplified further?

In most cases, log(a+b) cannot be simplified further, as it is already expressed as the sum of two logarithms. However, if a or b is a power of the base (b), then log(a+b) can be written as a single logarithm using the logarithm power rule.

## 5. How is log(a+b) used in real life?

The logarithm of a sum, log(a+b), is used in various fields of science and mathematics, such as finance, engineering, and statistics. It is often used to simplify complex calculations involving large numbers or exponential growth, and can also be used to model real-life phenomena such as population growth and radioactive decay.

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