Application of Common Logarithm
Exercise 5.1
1. Write the characteristics of the following logarithms.
(i) log 6987 (ii) log 256987 (iii) log 0.000089 (iv) log 4.68
(v) log 0.7491 (vi) log 1 (vii) log 35.492 (viii) log 36
(ix) log 0.00305 (x) log 1.965
Solution: (i) log 6987
Characteristics=4 – 1
= 3
(ii) log 256987
Characteristics = 6 – 1
= 5
(iii) log 0.000089
Characteristics = - (4 + 1)
= - 5
(iv) log 4.68
Characteristics = 1 – 1
= 0
(v) log 0.7491
Characteristics= - (0+1)
= -1
(vi) log 1
Characteristics= 1 – 1
= 0
(vii) log 35.492
Characteristics= 2 – 1
= 0
(viii) log 36
Characteristics= 2 – 1
= 0
(ix) log 0.00305
Characteristics= - (2 + 1)
= -3
(x) log 1.965
Characteristics= 1 – 1
2. If log 2925 = 3.46612 then find the values of log 29.25, log 2.925, log 0.002925 and log 292500.
Solution: Given, log 2925 = 3.46612
Log 29.25 = log 2925 × 10 -2
= - 2 + log 2925
= -2 + 3.46612
= 1.46612
Log 2.925 = log 2925 × 10 -3
= -3 + 3.46612
= 0.46612
Log 0.002925 = log 2925 × 10-6
= -6 + 3.46612
= -3.46612
= 3̅.46612
Log 292500 = log 2925 × 102
= 2 + 3.46612
= 5.46612
(i) 3.1465 (ii) 1̅.8621 (iii) 4̅.6663
(iv ) -2.7917 (v) 2.5591
Solution: (i) Antilog 3.1465 = 0.14012 × 103+1
= 0.14012 × 104
= 1401.2
(ii) Antilog 1̅.8621 = 0.72795 × 10-1+1
= 0.72795
(iii) Antilog 4̅.6663 = 0.46377 × 10-4 +1
= 0.46377 × 10-3
= 0.00046377
(iv) Antilog -2.7917= Antilog ( - 2 – 0.7917)
= Antilog (– 3 + 1 – 0.7917)
= Antilog (- 3 + 0.2083)
= Antilog ( 3̅.2083)
=0.16155 × 10-3+1
= 0.16155 × 10-2
= 0.0016155
(v) Antilog 2.5591 = 0.36232 × 102+1
= 0.36232 × 103
=362.32
4. Given that log2 = 0.30103, log3 = 0.47712 and log5 = 0.69897
(i) Log 0.002 (ii) log 25 (iii) log 7.2
(iv ) log 300 (v) log 0.15
Solution: Given that log2 = 0.30103, log3 = 0.47712 and log5 = 0.69897
(i) Log 0.002 = log 2 × 10-3
= -3 + log 2
= -3 + 0.30103
= 3̅.30103
(ii) Log 25 = log 52
= 2 log 5
= 2 × 0.69897
= 1.39794
(iii) Log 7.2 = log 72 × 10-1
= -1 + log 72
= -1 + log (23 × 32)
= -1 + log 23 + log 32
= -1 + 3 log 2 + 2 log 3
= -1 + 3 × 0.30103 + 2 × 0.47712
= -1 + 0.90309 + 0.95424
= -1 + 1.85733
= -1 + 1 + 0.85733
= 0.85733
(iv) Log 300 = log 3 × 102
= 2 + log 3
= 2 + 0.47712
= 2.47712
(v) Log 0.15 = log 15 × 10-2
= -2 + log (3 × 5)
= -2 + log 3 + log 5
= -2 + 0.47712 + 0.69897
= -2 + 1.17609
= -2 + 1 + 0.17609
= -1 + 0.17609
= 1̅.17609
5. Find the number of digits of the following numbers.
(i) 220 (ii) 225 (iii) 317 (iv) 515
(v ) 620 (vi) 713 (vii) 2200 × 310
(viii) 312 × 28 (ix) 2100 × 618
Solution: (i) let x = 220
Log x = log 220
= 20 log 2
= 20 × 0.30103
= 6.0206
Here, characteristics = 6
No. of digit in the number = 6 + 1= 7
(ii) Let x = 225
Log x = log 225
= 25 log 2
= 25 × 0.30103
= 7.52575
Characteristics = 7
Hence, the number of digit = 7 + 1 = 8
(iii) Let x = 317
Log x = log 317
= 17 log 3
= 17 × 0.47712
= 8.11104
Characteristics = 8
Hence, the number of digit = 8 + 1 = 9
(iv) Let x = 515
Log x = log 515
= 15 log 5
= 15 × 0.69897
= 10.48455
Characteristics = 10
Hence, number of digit = 10 + 1 = 11
(v) Let x = 620
Log x = log 620
= 20 log 6
= 20 [log 2 + log 3]
= 20 [0.30103 + 0.47712]
= 20 × 0.77815
= 15.563
Characteristics = 15
Hence, number of digit = 15 + 1 = 16
(vi) Let x = 713
Log x = log 713
= 13 log 7
= 13 × 0.8451
= 10.9863
Characteristics = 10
Hence, number of digit = 10 + 1 = 11
(vii) Let x = 2200 × 310
Log x = log (2200 × 310)
= log 2200 + log 310
= 200 log 2 + 10 log 3
= 200 × 0.30103 + 10 × 0.47712
= 60.206 + 4.7712
= 64.9772
Characteristics = 64
Hence, number of digit = 64 + 1 = 65
(viii) Let x = 312 × 28
Log x = log (312 × 28)
= log 312 + log 28
= 12 log 3 + 8 log 2
= 12 × 0.47712 + 8 × 0.30103
= 5.72544 + 2.40824
= 8.13368
Characteristics = 8
Hence, number of digit = 8 + 1 = 9
(ix) Let x = 2100
Log x = log 2100
= 100 log 2
= 100 × 0.30103
= 30.103
Characteristics = 30
Hence, number of digit = 30 + 1 = 31
(x) Let x = 618
Log x = log 618
= 18 log 6
= 18 log (2 × 3)
= 18 [log 2 + log 3]
= 18 [0.30103 + 0.47712]
= 18 × 0.77815
= 14.0067
Characteristics = 14
Hence, total digit number = 14 + 1= 15
(i) 2-64 (ii) (1/2)1000 (iii) 3-6 (iv) (1/3)100
(v ) (0.0016)20 (vi) 3-7 (vii) (16.8)-12 (viii) 7-4
(ix) 3-21 (x) 2-10
Solution: (i) Let x = 2-64
Log x = log 2-64
= -64 log 2
= - 64 × 0.30103
= -19.26592
= -20 + 1 – 0.26592
= -20 + 0.73408
= 2̅0̅.73408
Characteristics = 2̅0̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 20 – 1 = 19
(ii) Let x = (1/2)1000
Log x = 1000 log (1/2)
= 1000 [log 1 – log 2]
= 1000 [0 – 0.30103]
= 1000 × - 0.30103
= - 301.03
= - 301 – 0.03
= - 302 + 1 – 0.03
= - 302 + 0.97
= 3̅0̅̅2̅.97
Characteristics = 3̅0̅̅2̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal = 302 – 1 = 301
(iii) Let x = 3-6
Log x = log 3-6
= -6 log 3
= -6 × 0.47712
= - 2.86272
= - 2 – 0.86272
= -3 + 1 – 0.86272
= -3 + 0.13728
= 3̅.13728
Characteristics = 3̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 3 – 1 = 2
(iv) Let x = (1/3)100
Log x = log (1/3)100
= 100 log (1/3)
= 100 log 3-1
= -100 log 3
= -100 × 0.47712
= - 47.712
= -48 + 1 – 0.712
= -48 + 0.288
= 4̅8̅.288
Characteristics = 4̅8̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 48 – 1 = 47
(v) Let x = (0.0016)20
Log x = log (0.0016)20
= 20 log (0.0016)
= 20 [log 16 × 10-4)
= 20 [log 24 + log 10-4]
= 20 [4 log 2 + -4 log 10]
= 20 [4× 0.30103 -4]
= 20 [1.20412 – 4]
= 20 × - 2.79588
= - 55.9176
= - 55 – 0.9176
= -56 + 1 – 0.9176
= -56 + 0.0824
= 5̅6̅.0824
Characteristics = 5̅6̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 56 – 1 = 55
(vi) Let x = 3-7
Log x = log 3-7
= -7 log 3
= -7 × 0.47712
= - 3.33984
= -3 – 0.33984
= -4 + 1 – 0.33984
= - 4 + 0.66016
= 4̅.66016
Characteristics = 4̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 4 – 1 = 3
(vii) Let x = (16.8)-12
Log x = log (16.8)-12
= -12 log 16.8
= -12 [log (168 × 10-1)]
= - 12 [log 10-1 + log (23×3×7)]
= -12 [-1 + 3 log 2 + log 3 + log 7]
= -12 [-1 + 3× 0.30103 + 0.47712 + 0.84510]
= -12 [-1 + 0.90309 + 0.47712 + 0.84510]
= -12 × 1.22531
= - 14.70372
= -14 – 0.70372
= -15 + 1 – 0.70372
= -15 + 0.29628
= 1̅5̅.29628
Characteristics = 1̅5̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 15 – 1 = 14
(viii) Let x =7-4
Log x = log 7-4
= -4 log 7
= -4 × 0.84510
= -3.3804
= - 3 – 0.3804
= -4 + 1 – 0.3804
= -4 + 0.6196
= 4̅.6196
Characteristics = 4̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 4 – 1 = 3
(ix) Let x = 3-21
Log x = log 3-21
= -21 log 3
= -21 × 0.47712
= -10.01952
= -10 – 0.01952
= -11 + 1 – 0.01952
= -11 + 0.98048
= 1̅̅1̅.98048
Characteristics = 1̅1̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 11 – 1 = 10
(x) Let x = 2-10
Log x = log 2-10
= -10 log 2
= -10 × 0.30103
= -3.0103
= - 3 – 0.0103
= -4 + 1 – 0.0103
= -4 + 0.9897
= 4̅.9897
Characteristics = 4̅
Hence, the number of zeroes between the decimal point and the first significant digit after decimal
= 4 – 1 = 3
(i) (7.92)5 (ii) 
(iii) (0.264)1/5
(iv) 
(v ) 
Solution: (i) let x = (7.92)5
Log x = log (7.92)5
Log x = 5 log 7.92
Log x = 5 ( 0 + 0.89873)
Log x = 4.49365
x = Antilog (4.49365)
= 31162.05
(i) let x =
Log x = log
Log x = 
log (0.00000165)
Log x = [ log (1.65 × 10-7]
Log x = [ log 1.65 + log 10-7]
Log x = [-7 + 1.21748]
Log x = × -5.78252
Log x = - 1.9275
Log x = -1 – 0.9275
Log x = -2 + 1 – 0.9275
Log x = -2 + 0.0725
Log x = 2̅.0725
X = Antilog (2̅.0725)
X = 0.01181
(ii) Let x = 
Log x = log
Log x = 
log (0.264)
Log x = (log 2.64 × 10-1)
Log x = ( -1 + 0.42160)
Log x = × -0.57840
Log x = - 0.11568
Log x = -1 + 1 – 0.11568
Log x = -1 + 0.88432
Log x = 1̅.88432
x =Antilog (1̅.88432)
x = 0.76616
(iii) let x =
Log x = log
Log x = log 
Log x = [ log (625)4 + log (0.32)8 – log 52- log (0.3125)3- log (0.00432)2]
Log x = [ 4 log 625 + 8 log (0.32) – 2 log 5 – 3 log (0.3125) – 2 log (0.00432)]
Log x = [ 4 × 2.7959 + 8 × (1̅.5051) – 2 × 0.6990 – 3(1̅.4949) – 2(3̅.6355)]
Log x = [11.1836 + 8 (-0.4949) – 1.398 – 3(-0.5051) – 2 (-2.3645)]
Log x = [11.1836 – 3.9592 – 1.398 + 1.5153 + 4.729]
Log x = × 12.0707
Log x = 2.4142
x = Antilog 2.4142
x = 259.5
(iv) let x =
log x = log
log x = 
log 
log x = [log 125 + log 147 – log 21 – log 32]
log x = [ log 53 + log (3×72) – log ( 3×7) – log 25]
log x = [ 3 log 5 + log 3 + log 72 – log 3 – log 7 – 5 log 2]
log x = [ 3 log 5 + log 7 – 5 log 2]
log x = [ 3 × 0.6989 + 0.8451 – 5 × 0.30103]
log x = [ 2.0967 + 0.8451 – 1.50515]
log x = × 1.4367
log x = 0.9578
Therefore, x = Antilog (0.9578)
= 9.076226
8. Show that 
= 12.843 (upto three decimal places).
Solution: let x =
log x = log
log x = log (349388)
log x = (5.5434)
= 1.10868
Therefore, x = Antilog (1.10868)
= 12.843
9. Find the position of 1st significant digit on the right side of the decimal point of the number 
Solution: let x =
log x = log
= log 1 – log (16.8)12
= 0 – 12 × log 16.8
= -12 × 1.2253
= - 14.7036
= -15 + 1 – 0.7036
= -15 + 0.2964
= 1̅5̅.2964
Characteristics= 1̅5̅
Therefore, required place = 15
10. Applying logarithm, find at what rate per cent will ₹ 160,000 amount to ₹1,91,844 in 2 years at compound interest?
Solution: Here, Principal, P= ₹ 160,000
Amount, A = ₹ 1, 91, 844
Years, n = 2
Rate percent, r =?
We know that-
A = P 
1,91,844 = 160000 
=
=
log = log
log 47961 – log 40000 = 2 log 
4.68089 – 4.60206 = 2 log
0.07883 = 2 log
log = 
× 0.07883
log = 0.039415
= Antilog (0.039415)
1 + 
= 1.0950
= 1.0950 – 1
= 0.095
r = 9.5
Therefore, rate of interest = 9.5 %
11. With the help of logarithm find the number of years in which ₹216 will become ₹ 625 at 16% compound interest.
Solution: Given, Principal P = 216
Amount, A = 625
Rate of interest, r = 16
Years = ?
We know that,
A = P
625 = 216 
=
log = log
log 625 – log 216 = n [ log 116 – log 100]
log 54 – log (23 ×33) = n[ 2.0645 – log 102]
4 log 5 – log 23 – log 33 = n [2.0645 - 2×log10]
4 × 0.69897 –3 log 2 – 3 log 3= n [2.0645 - 2]
2.7959 –3×0.30103 - 3×0.47712= 0.0645 n
2.7959 – 0.90309 – 1.43136 = 0.0645 n
0.46145 = 0.0645 n
n = 7.148
Therefore, required year = 7.148
12. What will be the interest of ₹ 100000 in 3 years at 8% compound interest.
Solution: Here, P= 100000
n = 3
r= 8
Compound interest = ?
For Compound interest,
A = P 
= 100000 
= 100000 
A = 100000 
log A = log 105 + 3[log 27 – log 25]
log A = 5 log 10 + 3 log 33 – 3 log 52
log A = 5 log 5 + 5 log 2 + 9 log 3 – 6 log 5
log A = 5 × 0.69897 + 5 × 0.30103 + 9× 0.47712 - 6× 0.69897
log A = 3.49485 + 1.50515 + 4.29408 – 4.19382
log A = 5.10026
A = Antilog 5.10026
= 125971.20
Required CI = 125971.20 – 100000
= ₹ 25971.20
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