Characteristics and mantissa of a logarithm
Consider the following.
We will take logs to the base 10 as we are dealing with common logarithms.
From the above cases, it is clear that the logarithm of a number increases as the number increases and decreases as the number decreases. It also follows that if a number lies between 0 and 10, its logarithm lies between 0 and 1. If the number lies between 10 and 100, the logarithm lies between 1 and 2, and so on.
So, the logarithm of a number is partly integral (that is, it is an integer) and partly fractional.
The integral part of a logarithm is called the characteristic and the fractional or decimal part of the logarithm is called the mantissa.
Rule for determining the characteristic of a common logarithm
Example 1:
Number |
Integral Digits
N |
Characteristic
n-1 |
1657 |
4 |
3 |
208 |
3 |
2 |
56.7 |
2 |
1 |
81 |
2 |
1 |
97.3614 |
2 |
1 |
8.4123 |
1 |
0 |
6 |
1 |
0 |
If N, the given number, is less than 1; that is, it is a fraction or decimal and contains the first significant figures after n zeros (for example, in .0000561 the significant figures are 561 after four zeros) immediately after the decimal point, then as stated earlier it follows that a decimal fraction with no zero immediately following the decimal point lies between 10-1 and 100 (for example, 0.34). A decimal fraction with one zero following immediately after the decimal point lies between 10-2 and 10-1 (for example, 0.0519), etc.
or log N = -(n+1) + a proper fraction
Thus, the characteristic of the logarithm of a decimal fraction N with n zeros following the decimal point is -(n+1).
Example 2
with the same significant figures.
For example, the mantissa for
1) 3125, 31.25, 3.125, 0.3125 and .0003125 have the same mantissa, although their characteristics are different.
Example 3
If the mantissa of 5813 is .7644
write the logs of
a. 5813000 b. 5.813 c. 0.0005813 d. 58.13
Given that the mantissa of log 5813 = .7644
- 581300
Number of integral digits = 6
Characteristic of log = n-1 = 6-1 = 5
log 581300 = 5 + .7644
= 5.7644
- 5.813
Number of integral digits = 1
Characteristic = n-1 = 1-1 = 0
log 5.813 = 0 + .7644
= 0.7644
- 0.0005813
Number of zeros after decimal point = 3
- 58.13
Number of integral digits = 2
Characteristic = n-1
= 2-1
= 1
log 58.13 = 1+ .7644
= 1.7644
Sometimes, we may obtain a logarithm that is wholly negative. In this case, we need to use arithmetical operations to write the logarithm so that the mantissa is positive.
Example 4
The result -2.17985 may be transformed by adding -1 to the integral part and +1 to the decimal part.
- 2.17895 = -2 -1 + (1 - .7985)
= -3 + (0.82015)
=
Example 5
-
Write down the characteristics of the logarithms of
- 3174
- 625.7
- 0.374
- 0.00135
- 1.26 * 10-10
- The Mantissa of log 37203 is .5705. Write down the logs of
- 37.203
- 3.7203
- 372030000
- .000037203
- .37203
- The logarithm of 7623 is 3.8821. Write down the numbers whose logarithms are
- Given log 2 = .3010, log 3 = .4771, log 7 = .8451. Find the number of digits or zeros in
- Given log 2 = .3010 and log 3 = .4771. Find the numerical value of x in the following equations.
Answers to problems
-
- Number = 3174
Number of integral digits = n = 4
Characteristic = n-1 = 4-1
= 3
-
Number = 625.7
Number of integral digits = n = 3
Characteristic = n-1
= 3-1
= 2
-
Number = 0.374
-
Number = 0.00135
- Number = 1.26 * 10-10
- Mantissa of log 37203 is .5705
- The logarithm of 7623 is 3.8821
- To write the number whose log is 0.8821
log 7623 = 3.8821
Let log x = 0.8821
Characteristic = 0
⇒ the number has a single digit
∴ x = 7.623
- To write the number whose log is 7.8821
log 7623 = 3.8821
Let log x = 7.8821
Characteristic = 7
⇒ number of integral digits = 7+1 = 8
∴ x = 76230000
- To write the number whose log is 1.8821
log 7623 = 3.8821
Let log x = 1.8821
Characteristic of x = 1
⇒ number of integral digits = 1+1 = 2
∴ x = 76.23
-
- Given log 2 = .3010, log 3 = .4771, log 7 = .8451. Find the number of digits in (42)42
Solution:
- Given log 2 = .3010, log 3 = .4771, and log 7 = .8451. Find the number of digits in (81/80)1000
Solution:
-
- Given log 2 = .3010 and log 3 = .4771. Find the numerical value of x in 3x+2 = 405
- Given log 2 = .3010, log 3 = .4771 to obtain a numerical value for x
5x-3 = 8
5x-3 = 23
Taking logs and using log am = m log a
(x-3) log 5 = 3 log 2
xlog 5 - 3 log 5 = 3 log 2
x log 5 = 3 log 2+ 3 log 5