# I I found a *useful* method to calculate log(a+b), check it out

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1. Feb 28, 2016

### guifb99

So, I found this method, I don't think I was the first to, though, but I don't see any post related to this anywhere on the internet, so maybe there's a slim chance I was the first? Anyway, it doesn't really matter. The method does not give the precise result, only approximations, but I find it really useful, since when I have to use logarithms, I only work with approximations. Here it is:

Log(a+b) = Log(2) + Log(SQRT)(a*b)

It works with any base you choose, as long as it is the base for Log(a+b), Log(2) and Log(SQRT)(a*b) at the same time

See the "equals to" symbol as an "is approximate to" symbol

Also, "(SQRT)" means "square root of", quite obvious

And since Log(SQRT)(a*b) is (Log(a*b))/2, then

Log(a+b) = Log(2) + (Log(a*b))/2

And since (Log(a*b))/2 = (Log(a) + Log(b))/2, then

Log(a+b) = Log(2) + (Log(a) + Log(b))/2

In base 10, Log(2) is approximately 0,3, so

Log(a+b) = 0,3 + Log(SQRT)(a*b) in base 10

***IMPORTANT***

The bigger (a+b) is, the closer the results will be to the real values

If a>b

The smaller (a-b) is, the closer the results will be to the value

If b>a

The smaller (b-a) is, the closer the results will be to the value.

Example:
If you use this to find Log(11) and you want to use only natural numbers, you should use "6" and "5" for "a" and "b", it will not work if you use, instead, "10" and "1" for "a" and "b"

Also, "a" and "b" can be any REAL number to make this work, I don't know if this works with complex numbers, but I'm guessing it doesn't.

Anyway, I hope this helps you guys, any questions just ask me :)

Last edited: Feb 28, 2016
2. Feb 28, 2016

### guifb99

I'm brazilian so we use commas in cases like Log(2) = 0,3

But for americans, you can see the 0,3 as a 0.3 ;)

3. Feb 28, 2016

### micromass

Staff Emeritus
So you basically approximate the arithmetic mean by the geometric mean?

4. Feb 28, 2016

### guifb99

I don't even know what a geometric mean would be, I only know the basics of logarithms... I'm 15 years old

5. Feb 28, 2016

### micromass

Staff Emeritus
Google it.

6. Feb 28, 2016

### Staff: Mentor

Your notation in the last term on the right is confusing. Most people would normally write a square root of a*b as SQRT(a*b), not make it look like (SQRT) "times" (a * b).

The reason behind your IMPORTANT notes is this:

Suppose $\frac{a + b} 2 = \sqrt{ab}$
This equation is equivalent to the one I quoted above, assuming that a + b > 0 and ab > 0 for the logs to be defined.
Square both sides: $\frac 1 4 (a^2 + 2ab + b^2) = ab$
Multiply both sides by 4: $a^2 + 2ab + b^2 = 4ab$
Add -4ab to both sides: $a^2 - 2ab + b^2 = 0$, or $(a - b) = 0$

With the assumptions I listed above, all of the steps are reversible, so if a = b, then $\frac{a + b} 2$ is exactly equal to $\sqrt{ab}$, and the logs of both expressions would be equal as well.

7. Feb 28, 2016

### guifb99

Sorry, I'm new to this forum, didn't even know we could use these "uncommon outside the mathematic language" characters such as √ here, but thinking about it now, it makes sense that we would have these as this forum is basically about physics and mathematics. Anyway... thanks for explaining that, now I know why (a–b) must be close to zero in order to make this approximation works

Writing it again...

a>b ⇔ Lim(a–b)→0Log(a+b) ≈ Log(2) + Log√(a⋅b)

8. Feb 28, 2016

### Staff: Mentor

See https://www.physicsforums.com/help/latexhelp/.

I used \frac{a + b} {2} to make the fraction and \sqrt{ab} to make the square root. Put a pair of # characters at the beginning of the expression, and another pair at the end, and your browser will turn them into nicely formatted math.

9. Mar 21, 2016

### Battlemage!

This is really cool. Thanks.

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