How do you properly add logarithms with negative characteristics by hand?

AI Thread Summary
The discussion revolves around the proper addition of logarithms with negative characteristics using log tables. It highlights the confusion between two notations: the bar notation (e.g., log(0.0637) = \bar{2}.8041) and the standard negative decimal notation (e.g., -1.1959). The key point is that the bar notation indicates only the number under the bar is negative, while the entire value in the standard notation is negative. The correct addition of logarithms requires careful attention to these notations to avoid errors in calculation. Understanding these differences is crucial for accurate logarithmic calculations by hand.
gerid21
Messages
2
Reaction score
0
Homework Statement
Calculate ##890\times12.34\times0.0637## using logarithms of base 10
Relevant Equations
Log tables, calculator
From the log tables:
##log(890) = 2.9494, \space
log(12.34)=1.0913, \space
log(0.0637)=\bar{2}.8041##

I calculate by hand:
##\begin{array}{r}
&2.9494\\
+&1.0913\\&\bar{2}.8041\\\hline &2.8448
\end{array}##

Thus:
##log^{-1}(2.8448) \approx 699.6 \space##

Which is the correct answer.

Now I know ##log(0.0637)=\bar{2}.8041=-2+0.8041=-1.1959##

Yet if I did

##
\begin{array}{r}
&2.9494\\
+&1.0913\\&-1.1959\\\hline &3.2366
\end{array}##

##log^{-1}(3.2366)\approx1724\space## This is obviously wrong.

But if I were to do the problem solely with a calculator, I have to plug in ##log(0.0637)= -1.1959\space## and not ##-2.8041## to get the correct answer. I'm confused at what's going on. I know it has something to do with the negative characteristic of the logarithm. But then if I'm adding it by hand, why doesn't ##-1.1959## work?
 
Last edited:
Physics news on Phys.org
gerid21 said:
##
\begin{array}{r}
&2.9494\\
+&1.0913\\&-1.1959\\\hline &3.2366
\end{array}##
When I add these numbers, I get 2.8448.
 
DrClaude said:
When I add these numbers, I get 2.8448.

So essentially $$\bar{2}.8041$$ means only the number under the bar is negative and the rest is positive whereas $$-1.1959$$ means the entire number before and after the decimal is negative?
 
gerid21 said:
So essentially $$\bar{2}.8041$$ means only the number under the bar is negative and the rest is positive whereas $$-1.1959$$ means the entire number before and after the decimal is negative?
The latter is the standard mathematical notation. I'm not that familiar with log tables, but it seems that the former is the common notation used in those tables:
https://www.wikiwand.com/en/Common_logarithm
 
Log tables, how quaint!

gerid21 said:
So essentially $$\bar{2}.8041$$ means only the number under the bar is negative and the rest is positive whereas $$-1.1959$$ means the entire number before and after the decimal is negative?

Yes, and probably you knew and then forgot, because log tables only go from 1 to 9.999 which far from includes all numbers, so you would have to have known that to look up and use. Maybe this not so drilled into students these days, and that's no loss.

(The two parts are called 'mantissa' and 'exponent' by the way, to jog your memory or if you want to look them up.)
 
  • Like
Likes Delta2
DrClaude said:
I'm not that familiar with log tables, but it seems that the former is the common notation used in those tables:
Yes - otherwise they would be twice as long!
 
pbuk said:
Yes - otherwise they would be twice as long!
They could be millions of times as long and it wouldn't be enough!
 
  • Like
Likes SammyS
gerid21 said:
Now I know ##log(0.0637)=\bar{2}.8041=-2+0.8041=-1.1959##

Yet if I did

##
\begin{array}{r}
&2.9494\\
+&1.0913\\&-1.1959\\\hline &3.2366
\end{array}##
Strange you understand that ##\bar{2}.8041=-2+0.8041## and correctly add it up in that form, but don't realize that the more common ##-1.1959 = -1 -0.1959## and add that up wrongly.

But if you did it properly, it works
##
\begin{array}{r}
&2.9494\\
+&1.0913\\\hline &+4.0407\\&-1.1959\\\hline &+2.8448
\end{array}##

I wonder for which method you would do it right when you subtract to do a division?
 
Back
Top