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

  • #1
10
1
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:
  • Like
Likes Battlemage!

Answers and Replies

  • #2
10
1
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
22,089
3,286
So you basically approximate the arithmetic mean by the geometric mean?
 
  • Like
Likes fresh_42
  • #4
10
1
So you basically approximate the arithmetic mean by the geometric mean?
I don't even know what a geometric mean would be, I only know the basics of logarithms... I'm 15 years old
 
  • #5
22,089
3,286
Google it.
 
  • #6
34,156
5,773
Log(a+b) = Log(2) + Log(SQRT)(a*b)
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.
 
  • Like
Likes Battlemage!
  • #7
10
1
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.
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
34,156
5,773
Sorry, I'm new to this forum, didn't even know we could use these "uncommon outside the mathematic language" characters such as √ here
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
294
44
This is really cool. Thanks.
 

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

Replies
0
Views
1K
Replies
13
Views
3K
Replies
6
Views
1K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
6K
Replies
2
Views
746
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
Top