LOGARITHMS

for
CLASS-IX
2023-24 onwards

BY

PRASHANT A. NAIK

(DCT’S VASANTRAO DEMPO HSS OF ARTS,SCIENCE &

COMMERCE,CUJIRA-ST.CRUZ GOA)

Acknowledgements

I would like to thank Chairman of Goa Board Shri Bhagirath Shetye

sir for his valuable suggestions, Ms. Sybil Fernandes (Convener-
BOS), Prashant Dessai and other BOS members of Goa Board in
Mathematics for their gracious assistance and support.

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 LOGARITHMS

 Concept of logarithm

 Finding logarithm of a number

 Antilogarithm

 Exercises

 Common Logarithms and antilogarithms tables

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 LOGARITHMS

John Napier, Scottish mathematician, physicist and astronomer originated

the concept of logarithms as a mathematical device to aid in calculations.

LAWS OF INDICES

For any real numbers 𝑎, 𝑏, 𝑥 𝑎𝑛𝑑 𝑦 (𝑎 , 𝑏 > 0, ≠ 1)the following laws are valid

1. 𝑎 𝑥 × 𝑎 𝑦 = 𝑎 𝑥+𝑦
𝑎𝑥
2. = 𝑎 𝑥−𝑦
𝑎𝑦

3. (𝑎 𝑥 )𝑦 = 𝑎 𝑥𝑦
4. (𝑎 × 𝑏)𝑥 = 𝑎 𝑥 × 𝑏 𝑥
𝑎 𝑥 𝑎𝑥
5. (𝑏) = 𝑏𝑥

6. 𝑎0 = 1
1
7. 𝑎−𝑥 = 𝑎𝑥
𝑥
𝑦 𝑦 𝑥
8. 𝑎 𝑦 = √𝑎 𝑥 = ( √𝑎)

Concept of Logarithm

Logarithm is usually written as ‘log’. Let us understand the concept of logarithm.

1. We know that 23 = 8 (exponential form).

We write this statement as log 2 8 = 3 (logarithmic form) and read it as “logarithm of 8 to the
base 2 is equal to 3”.

2. 10−3 = 0.001(exponential form).

We write as log10 0.001 = 3 (logarithmic form) and read it as “logarithm of 0.001 to the base 10
is equal to bar 3”.
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 3. 34 = 81 (exponential form).
We write as log 3 81 = 4 (logarithmic form) and read it as “logarithm of 81 to the base 3 is
equal to 4”.

In the above examples one must note that 2,10,3 are called bases of indices 3,-3 and 4
respectively.

The indices of bases are called logarithms.

Now observe and study the following examples

1. 102 = 100 is written as log10 100 = 2

This implies that 2 is the index of base 10 to get the number 100

2. 103 = 1000 is written as log10 1000 = 3

⟹ 3 is the index of base 10 to get the number 1000

3. 10−2 = 0.01 is written as log10 0.01 = 2

⟹ -2 is the index of base 10 to get the number 0.01

4. 32 = 9 is written as log 3 9 = 2
⟹ 2 is the index of base 3 to get the number 9

5. 53 = 125 is written as log 5 125 = 3

⟹ 3 is the index of base 5 to get the number 125

From the above examples one can understand the concept of logarithm

Definition of logarithm

The logarithm of any number to a given base is the value of the index to which the base must be
raised to get the given number

If bm = N, then m is said to be the logarithm of the number N and is written as log 𝑏 𝑁 = 𝑚 where
b and N are positive real numbers and (𝑏 ≠ 1) ; m is a real number

NOTE
1. We know that 𝑏1 = 𝑏
∴ log b b = 1
⟹ Logarithm of any number to the same base is always 1

2. We know that 𝑏 0 = 1
∴ log b 1 = 0

Also
We know that 𝑎0 = 1
∴ log a 1 = 0
⟹ Logarithm of 1 to any base is always zero

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 3. We know that 23 = 8 (𝟖 > 𝟎, 𝒊𝒆 𝟖 𝒊𝒔 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆)
log 2 𝟖 = 3

Also 2−3 = 0.125 (𝟎. 𝟏𝟐𝟓 > 𝟎, 𝒊𝒆 𝟎. 𝟏𝟐𝟓 𝒊𝒔 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆)

log 2 𝟎. 𝟏𝟐𝟓 = 3

log 2 (−8) = 𝐝𝐨𝐞𝐬 𝐧𝐨𝐭 𝐞𝐱𝐢𝐬𝐭 ie we will not get −8 if we raise the base 2 to any index

⟹ Logarithm of negative number does not exist

Common logarithms and use of logarithm tables

Logarithms with base 10 are called common logarithms

Henry Briggs, English mathematician known for changing the original logarithms invented by
John Napier into common (base 10) logarithms, which are sometimes known as Briggsian
logarithms in his honor.

Note: If base is not mentioned then base is always taken as 10

Logarithm with base e is called natural logarithm where e is a Napier constant(Euler’s number)
which is an irrational number lying between 2 and 3 (𝑒 = 2.718281828459045 … ).We usually take
the value of 𝑒 = 2.718.
Change of base law is used to convert natural logarithm to common logarithm and vice versa

Logarithm of a number

The logarithm of a number consists of two parts

1. Characteristic (integral part of the logarithm)

2. Mantissa (decimal part of the logarithm)

For a positive real number 𝑥

𝑙𝑜𝑔 𝑥 = 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑜𝑓 𝑥 + 𝑚𝑎𝑛𝑡𝑖𝑠𝑠𝑎 𝑜𝑓 𝑥
In log 675.8 = 2.8298,the integral part is 2 and decimal part is 0.8298

⟹ 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 = 2 and 𝑚𝑎𝑛𝑡𝑖𝑠𝑠𝑎 = 0.8298

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 Method of finding characteristic by inspection

1. Positive numbers greater than 1

2. Positive numbers less than 1

1. Characteristic of a number greater than 1

Rule : One less than the number of digits on the left of a decimal point of a given number
Let the number of digits before decimal point are m. Then the characteristic of that number is
m-1

2. Characteristic of a number less than 1

Rule: One more than the number of zeros immediately after the decimal point in the decimal part of
a number with bar on the top.
Let the number of zeros immediately after the decimal point be n.Then the characteristic of that
number is −(𝒏 + 𝟏) 𝒂𝒏𝒅 𝒊𝒔 𝒅𝒆𝒏𝒐𝒕𝒆𝒅 𝒂𝒔 𝒏 + 𝟏 .Bar is used in the characteristic to indicate only
the number below the bar is negative.

Examples:

Sr.No Number Characteristic

1. 478 2
2. 5.4 0
3. 98 1
4. 2915 3
5. 32.57 1
6. 0.5791 𝟏
7. 0.05008 𝟐
8. 0.007184 𝟑
9. 0.00051 𝟒
10. 0.00007 𝟓

Note: Characteristic is generally found by writing the given number in scientific form. The power of
10 indicate the characteristic.

We find the mantissa of a logarithm of a number from the logarithm table

Note:-Logarithm table is given at the end of the chapter

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A logarithm table consists of three parts

1. First column contains two digit numbers from 10 to 99

2. Next there are 10 columns with headings 0,1,2,3,4,5,6,7,8 and 9

3. Next there are 9 columns with headings 1,2,3,4,5,6,7,8 and 9 called Mean Differences

Method to find the mantissa of a given number

1. Remove the decimal point and express the given number in 4 digits if not.

2. Consider first two digits and note down the row starting with this two digits.

3. Next consider third digit and note down the number in this row under the third digit from 10

columns

4. Next consider the fourth digit and note down the number in this row under the fourth digit

from 9 columns (mean differences).

5. Add the numbers noted in step 3 and step 4

6. Mantissa is number obtained in step 5 with decimal point before the number

Examples:

Sr.No. Number Number for finding Four digit number for Mantissa
mantissa finding mantissa
1. 7 7 7000 0.8451
2. 2.8 28 2800 0.4472
3. 18.9 189 1890 0.2765
4. 576.1 5761 5761 0.7605
5. 718 718 7180 0.8561
6. 647.579 647579 6476 0.8113
7. 4.3825 43825 4383 0.6418
8. 5967.5 59675 5968 0.7758
9. 0.021356 21356 2136 0.3296
10. 0.00985 985 9850 0.9934

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Solved Examples

1. Find the logarithm of 473.5

LOGARITHM TABLE

Mean Differences
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 5 6 7 7 8
47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1 2 3 4 5 5 6 7 8
48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 1 2 3 4 4 5 6 7 8

For finding Mantissa

In 47th row add the number under column 3 with the number under column 5 from Mean difference

6749 + 5 = 6754

Characteristic = 2
Mantissa = 0.6754
𝐥𝐨𝐠 𝟒𝟕𝟑. 𝟓 = 𝟐. 𝟔𝟕𝟓𝟒

2. Find the logarithm of 0.0013415

LOGARITHM TABLE

Mean Differences
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 34
12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 31
13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29

For finding Mantissa

In 13th row add the number under column 4 with the number under column 2 from Mean difference

1271 + 6 = 1277

Characteristic = 3
Mantissa = 0.1277
𝐥𝐨𝐠 𝟎. 𝟎𝟎𝟏𝟑𝟒𝟏𝟓 = 𝟑. 𝟏𝟐𝟕𝟕

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 3. Find the logarithm of 8.57

LOGARITHM TABLE

Mean Differences
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 2 3 3 4 4 5
86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 1 2 2 3 3 4 4 5
87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0 1 1 2 2 3 3 4 4

For finding Mantissa

In 85th row take the number under column 7 only as there is no 0th column in the Mean Difference.

9330

Characteristic = 0
Mantissa = 0.9330
𝐥𝐨𝐠 𝟖. 𝟓𝟕 = 𝟎. 𝟗𝟑𝟑𝟎

Antilogarithm
The number corresponding to a given logarithm is called its antilogarithm
If log 𝑥 = 𝑦 then antilog 𝑦 = 𝑥
Examples
1. If log 25 = 1.3979 then antilog 1.3979 = 25
2. If log 0.005382 = 3. 7310 then antilog 3. 7310 = 0.005382

Note: We can find the antilog of positive mantissa only. If negative mantissa is given then we have
to convert negative mantissa into positive mantissa.

Examples of Positive and negative mantissa.

Sr.No. Number with positive mantissa Number with negative mantissa

1. 1.5728 = 1 + 0.5728 −3.2756 = −3 − 0.2756
2. 0.5136 = 0.5136 −12.6178 = −12 − 0.6178
3. 12.7419 = 12 + 0.7419 −34.7048 = −34 − 0.7048
4. 3. 7310 = 3 + 0.7310 −0.7856 = −0.7856
5. 14. 6742 = 14 + 0.6742 −1.2347 = −1 − 0.2347

Method to convert negative Mantissa into positive Mantissa

1. Add and subtract 1 to negative mantissa

2. Subtract negative mantissa from +1
3. Add -1 to integral part and express it using bar
4. Then write characteristic and positive mantissa with decimal point in between
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Examples:
1. antilog(−2.5123) = antilog(−2 − 0.5123) = antilog(−2 − 1 + 1 − 0.5123) = antilog(3. 4877)
2. antilog(−5.6251) = antilog(−5 − 0.6251) = antilog(−5 − 1 + 1 − 0.6251) = antilog(6. 3749)
3. antilog(−0.1764) = antilog(−1 + 1 − 0.1764) = antilog(1. 8236)

To find the antilog of a number we refer to a table with heading ANTILOGARITHMS

Finding antilog of positive mantissa by method of inspection

1. Consider only the decimal part as Antilog is to be found only in the decimal part without
considering the characteristic.
2. Find the antilog from the Antilogarithm table ( same method as used before for finding
logarithm)
3. If the characteristic of the logarithm is m (where 𝑚 ≥ 0) then insert the decimal point after
(m+1)th digit from the left.
4. If the characteristic of the logarithm is m (where 𝑚 < 0 𝑖𝑒 𝑚 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒) then insert the
decimal point so that non-zero number is at mth place.
5. Generally if the characteristic is large then express the antilog in scientific form where
characteristic is in power of 10

Note: Any antilog can be actually written in scientific form where characteristic is in power
of 10

Examples:

1. Find the antilog(2.0953)

ANTLOGARITHM TABLE

Mean Differences
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
.09 1230 1233 1236 1239 1242 1245 1247 1250 1253 1256 0 1 1 1 1 2 2 2 3
.10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 0 1 1 1 1 2 2 2 3
.11 1288 1291 1294 1297 1300 1303 1306 1309 1312 1315 0 1 1 1 1 2 2 2 3

In .09th row add the number under column 4 with the number under column 2 from Mean difference

1245 + 1 = 1246

antilog(2.0953) = 124.6

2. Find the antilog(15.3124)

antilog(15.3124) = 2.053 × 1015

3. Find antilog(0.573)
antilog(0.573) = 3.741

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 4. Find antilog(2. 95068)
antilog(2. 95068) = 0.08928

5. Find antilog(19. 63)

antilog(19. 63) = 4.266 × 10−19

6. Find antilog(−2.3721)
antilog(−2.3721) = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔(3. 6279) = 0.004245

7. Find antilog(−7.5463)
antilog(−7.5463) = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔(8. 4537) = 0.00000002843 = 2.843 × 10−8

8. Find antilog(−0.5127)
antilog(−0.5127) = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔(1. 4873) = 0.3071

Some more examples

Sr.No. Number Antilogarithm

1. 0.2538 1.794
2. 1.2538 17.94
3. 2.2538 179.4
4. 3.2538 1794
5. 4.2538 17940
6. 5.2538 179400
7. 12.2538 1.794 × 1012
8. 1. 2538 0.1794
9. 2. 2538 0.01794
10. 3. 2538 0.001794
11. 4. 2538 0.0001794
12. 5. 2538 0.00001794
13. 31. 2538 1.794 × 10−31
14. −3.7462 0.0001794
15. −1.7462 0.01794
16. −11.7462 1.794 × 10−12

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 Exercise-1

Express the following in the logarithmic form

a) 53 = 125
3
b) √343 = 7
c) 10−5 = 0.00001
d) 80 = 1
1
e) (81)2 = 9

Exercise-2
Express the following in the exponential form
a) log 8 64 = 2
b) log15 15 = 1
c) log 6 216 = 3
d) log10 0.0001 = 4
e) log 3 1 = 0

Exercise-3
Find the value of the following by using the definition of logarithm
a) log 2 32
b) log 5 1
c) log11 1331
d) log √2 4
e) log10 0.01

Exercise-4

Write down the characteristic for the following numbers

a) 58
b) 174.5
c) 0.000072
d) 1.597 × 10−12
e) 2.713 × 109
Exercise-5
Insert the decimal point in the proper position so that the characteristic of

a) 7618 is 1
b) 9527 is 3
c) 5612 is 2
d) 2154 is 0
e) 4024 is 3
f) 8217 is 2

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 Exercise-6
Find the mantissa of the following using logarithm table

a) 2.134
b) 0.00517
c) 8
d) 27
e) 317.25
f) 6.7215
g) 576682
Exercise-7
Find the logarithm of the following numbers using logarithm table

a) 2.714
b) 179
c) 2
d) 0.7538
e) 0.0001254
f) 10000
Exercise-8
Find the antilogarithm of the following using antilogarithm table

a) 2.1463
b) 0.728
c) 3. 5176
d) 27.63
e) 12. 4714
f) −0.2738
g) −2.7812
h) −13.1364

Exercise-9

1. If 𝐥𝐨𝐠(𝟎. 𝟎𝟓𝟕𝟑𝟖) = 𝟐. 𝟕𝟓𝟖𝟖 then find the values of

a) log(5.738)
b) log(573.8)
c) log(0.0005738)
d) log(57.38)
e) antilog(2. 7588)
f) antilog(3.7588)
g) antilog(2.7588)

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 Exercise-10

If 𝐚𝐧𝐭𝐢𝐥𝐨𝐠(𝟐. 𝟔𝟓𝟏𝟐) = 𝟒𝟒𝟕. 𝟗 then find the values of

a) antilog(2. 6512)
b) antilog(0.6512)
c) antilog(3.6512)
d) antilog(13.6512)
e) antilog(21. 6512)
f) log(44.79)
g) log(0.0004479)

ANSWERS

Exercise-1 Exercise-2

a) log 5 125 = 3 a) 82 = 64
1
b) log 343 7 = 3 b) 151 = 15
c) 63 = 216
c) log10 0.00001 = 5 d) 10−4 = 0.0001
d) log 8 1 = 0
1 e) 30 = 1
e) log 81 9 = 2

Exercise-3 Exercise-4

a) 5 a) 1
b) 0 b) 2
c) 3 c) 5
d) 4 d) 12
e) 2 or −2 e) 9

Exercise-5 Exercise-6

a) 76.18 a) 0.3292
b) 9527.0 b) 0.7135
c) 0.05612 c) 0.9031
d) 2.154 d) 0.4314
e) 0.004024 e) 0.5015
f) 821.7 f) 0.8275
g) 0.7609

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 Exercise-7 Exercise-8

a) 0.4336 a) 140.1
b) 2.2529 b) 5.346
c) 0.3010 c) 0.003294
d) 1.8773 d) 4.266 × 1027
e) 4.0983 e) 2.961 × 10−12
f) 4.0000 f) 0.5323
g) 0.001655
h) 7.305 × 10−14

Exercise-9 Exercise-10

a) 0.7588 a) 0.04479
b) 2.7588 b) 4.479
c) 4. 7588 c) 4479
d) 1.7588 d) 4.479 × 1013
e) 0.05738 e) 4.479 × 10−21
f) 5738 f) 1.6512
g) 573.8 g) 4.6512

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