How do you add or subtract bar numbers in logarithms?

[creativeassault]

I have a small problem with logarithms. We have to solve physics and chemistry problems using only logs. And I don't know how to do the following -
1.2341 + bar2.4412
Well actually my question is how do you add or subtract bar numbers ? (It's the numbers with '-' on top)

Rohit.

 

[HallsofIvy]

First of all, this is not a "logarithm" problem. In the years "B.C." (before calculators), people used logarithms to convert multiplication and powers into addition but just added normally.

The real problem is what you are calling "bar" numbers. What are bar numbers? One symbolism I know is to use a bar to indicate that a part of the decimal repeats infinitely. If that is the case, what part of the number is the bar over?

 

[creativeassault]

Hmm...
It's the charecteristic, 2 only. It means negative not repeating infinitely.

 

[HallsofIvy]

Hmm...
It's the charecteristic, 2 only. It means negative not repeating infinitely.

What? I know a number of definitions of "characteristic", none of which seem to fit here.

If the "bar" is simply a negative sign then 1.2341 -2.4412= -1.2071. Surely that is not what you meant!

 

[creativeassault]

I actually found something on the net ... http://wwwchem.uwimona.edu.jm:1104/courses/pH/phlogs.html#2

Umm and it solved my problem ... I guess I should'nt have asked a question before looking for more information ... my bad :(.

 

[HallsofIvy]

Oh, now I understand- the numbers you gave WERE the logarithms, written in a rather peculiar engineering notation. (My god, are people still doing this? I haven't seen that notation in years) The point of the notation is to avoid subtracting from 1.0000.

The notation "bar" 2.4412 for a logarithm means -2+ 0.4412 so adding to 1.2341 is the same as (1- 2)+ (.4412+ .2341)= (-1)+ .6753 which would be, in this notation, "bar" 1.6753.

1.2341 is the common logarithm of 17.1435. "bar"2.4412 (or -1.5588 in normal notation) is the common logarithm of 0.02762. One would get those values by, for the second, looking up 2.762 in a table of logarithms and then appending the "bar" 2 (or -2) since 0.02762= 2.762x 10^-2. (Of course, I got them using a calculator!)

The "sum" of 1.2341 and "bar"2.4412 is the logarithm of the product (17.1435)(0.02762). To finish one would look up the "anti-logarithm" of .6753 in the same table (but used in reverse) which is 4.7348. The "bar"1 means this is multiplied by 10^-1= 0.1 so we get 0.4735 (rounded to 4 places as the original numbers were).

Of course a calculator very quickly gives .4735.

This method and notation is positively antique!

 

[creativeassault]

Thanks HallsofIvy. Well it's still used in Indian institutions I guess ... (12 th grade) :P