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

In summary: I would do it the same way, but I think it would be helpful to know that there is a difference between the two forms.
  • #1
gerid21
2
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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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.)
 
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  • #6
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!
 
  • #7
pbuk said:
Yes - otherwise they would be twice as long!
They could be millions of times as long and it wouldn't be enough!
 
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  • #8
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?
 

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

1. How do you calculate logarithms by hand?

To calculate a logarithm by hand, you can use the formula logb(x) = y, where b is the base, x is the number, and y is the exponent. You can also use a logarithm table or a scientific calculator to find the answer.

2. What is the purpose of using logarithms in calculations?

Logarithms are used to simplify complex mathematical calculations, especially those involving large numbers. They also help in solving exponential equations and finding the unknown exponent or base.

3. How do you convert a logarithm to exponential form?

To convert a logarithm to exponential form, you can use the identity logb(x) = y is equivalent to by = x. This means that the base of the logarithm becomes the base of the exponential expression, and the exponent becomes the value of the expression.

4. Can you calculate logarithms with negative numbers?

No, logarithms can only be calculated for positive numbers. This is because the logarithm function is not defined for negative numbers.

5. What is the difference between natural logarithms and common logarithms?

Natural logarithms, denoted as ln(x), have a base of e, which is a mathematical constant approximately equal to 2.71828. Common logarithms, denoted as log(x), have a base of 10. This means that natural logarithms are useful in solving problems involving exponential growth and decay, while common logarithms are useful in problems involving orders of magnitude.

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