# Log calculation

## Homework Statement

:I had to find 102 *9.8/10^-4 from log[/B]

## Homework Equations

So i decided to find it in this way
log of 102+log of 9.8 -log of 10^-4
and then whatever answer i get ,take antilog of it.
[/B]

## The Attempt at a Solution

log of 102+log of 9.8 -log of 10^-4 came out to 6.9998[/B]
Antilog of 6.9998=9995000
I am not getting where I am wrong.

Bystander
Homework Helper
Gold Member
You need to carry one more significant figure in the actual logs through the calculation.

You need to carry one more significant figure in the actual logs through the calculation.
But in log table these are the only significant figures given.How can I know other significant figure.Calculator is not allowed in our class test,only log tables are given.

Bystander
Homework Helper
Gold Member
You can approximate one more by a linear interpolation between the values for the two values that bracket the number you need.

linear interpolation

Bystander
Homework Helper
Gold Member
Just ran it on the calculator, and through the log table --- you didn't go wrong --- whoever wrote the exercise cheated and used a calculator rather than the log table. Forget what I said about interpolations.

• gracy
One more question.I have to solve this with help of log table.98 *10^-4/3.14*0.25*12.5
So I did the same thing.
log of 98+(-)4*log of 10 -log of 3.14 +log of 0.25+log of 12.5
=1.9912+(-)4*1 -[0.4969+(-0.6020)+1.0969]
= - 2.0088 -[0.9918]
=- 3.0006
Now I had to find antilog of whatever the answer comes to get tha answer of 98 *10^-4/3.14*0.25*12.5
But how can I find antilog of a negative number?

Mark44
Mentor
One more question.I have to solve this with help of log table.98 *10^-4/3.14*0.25*12.5
So I did the same thing.
log of 98+(-)4*log of 10 -log of 3.14 +log of 0.25+log of 12.5
=1.9912+(-)4*1 -[0.4969+(-0.6020)+1.0969]
= - 2.0088 -[0.9918]
=- 3.0006
Now I had to find antilog of whatever the answer comes to get tha answer of 98 *10^-4/3.14*0.25*12.5
But how can I find antilog of a negative number?
-3.0006 = -4 + .9994
Use your table to find the antilog of .9994, and then adjust it by a suitable factor of 10.

• gracy
Use your table to find the antilog of .9994, and then adjust it by a suitable factor of 10.
And what to do with -4 ?It will be converted into 10^-4.right?

Last edited:
Mark44
Mentor
And what to do with -4 ?It will be converted into 10^-4.right?
Yes.

How did you know to do this.-3.0006 = -4 + .9994.
I mean how can I do such type of steps in other questions?

SteamKing
Staff Emeritus
Homework Helper
And what to do with -4 ?It will be converted into 10^-4.right?

Look at this progression:

log (10) = 1
log (100) = 2
log (1000) = 3
log (0.10) = -1
log (0.01) = -2
log (0.001) = -3

The fractional part of the common log is called the mantissa. This is what you find in the log tables. The whole number part of the log is called the characteristic. It gives you the power of 10 for the antilog, after you have looked up the mantissa in the log table:

http://en.wikipedia.org/wiki/Common_logarithm

• gracy
Thanks steamking and Mark44.

The fractional part of the common log is called the mantissa. This is what you find in the log tables. The whole number part of the log is called the characteristic. It gives you the power of 10 for the antilog, after you have looked up the mantissa in the log table:
In my example 0.994 is mantissa.And -4 is characteristic.Right?

And we want fractional part only to find antilog that's why we broke -3.0006 into -4 + .9994,not -5 +1.994 or -6+2.994 or so on.....
Right?

Mark44
Mentor
And we want fractional part only to find antilog that's why we broke -3.0006 into -4 + .9994,not -5 +1.994 or -6+2.994 or so on.....
Right?
Right.

Also, we want the mantissa to be between 0 and 1 (0 ≤ mantissa < 1.0). -3.0006 is the same as -3 + (-.0006), and the table doesn't have negative mantissas, so that was the reason for writing -3.0006 as -4 + .9994. IOW, to get a positive mantissa.

Chestermiller
Mentor
How did you know to do this.-3.0006 = -4 + .9994.
I mean how can I do such type of steps in other questions?
This is how they taught us to do it when I took intermediate algebra in high school. (Thank you Mrs. Sheerin).

Chet