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MATHEMATICSClass 9th (KPK)
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UNIT # 3 LOGARITHM
Exercise # 3.1 As we know that SCIENTIFIC NOTATION: π π = π£π£π£π£ Scientific notation is a way of writing
numbers that are too big or too small to be easily written in decimal form.
Put the values π π = 1.86 Γ 105 Γ 8.64 Γ 104
π π = 1.86 Γ 8.64 Γ 105 Γ 104 π π = 16.0704 Γ 105+4 π π = 16.0704 Γ 109 π π = 1.60704 Γ 101 Γ 109 π π = 1.60704 Γ 1010
Representation The positive number "π₯π₯" is represented in
scientific notation as the product of two numbers where the first number "ππ" is a real number greater than 1 and less than 10 and the second is the integral power of "ππ" of 10.
Thus light travels 1.60704 Γ 101 Γ 109 miles in
a day
π₯π₯ = ππ Γ 10ππ Exercise # 3.1 Rules for Standard Notation to Scientific Notation Page # 80
(i) In a given number, place the decimal after first non-zero digit.
Q1: Write each number in scientific notation. (i) 405,000 (ii) If the decimal point is moved towards left, then the
power of β10β will be positive. Solution:
405,000 (iii) If the decimal is moved towards right, then the
power of β10β will be negative. In Scientific Form:
4.05 Γ 104 The numbers of digits through which the decimal
point has been moved will be the exponent.
(ii) 1,670,000 Rules for Standard Notation to Scientific Notation Solution:
(i) If the exponent of 10 is Positive, move the decimal towards Right.
1,670,000 In Scientific Form: (ii) If the exponent of 10 is Negative, move the
decimal toward Left. 1.67 Γ 106
(iii) Move the decimal point to the same number of
digits as the exponent of 10. (iii) 0.00000039
Solution: 0.00000039 Example # 7 Page # 80 In Scientific Form: How many miles does light travel in 1 day? The
speed of the light is 186,000 mi/ sec. write the answer in scientific notation.
3.9 Γ 10β7 (iv) 0.00092 Solution: Solution: ππππππππ = π£π£ = 1 ππππππ = 24 βππ
π£π£ = 24 Γ 60 Γ 60 π π πππ π = 86400 π£π£ = 8.64 Γ 104 π π πππ π ππππππππππ = π£π£ = 186000 ππππ/π π πππ π π£π£ = 1.86 Γ 105 ππππ/π π πππ π
0.00092 In Scientific Form: 9.2 Γ 10β4
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Ex # 3.1 Ex # 3.1 (v) 234,600,000,000 (iii) ππ.ππππΓ ππππππ
Solution: Solution: 234,600,000,000 2.07 Γ 107 In Scientific Form: In Standard Form: 2.346 Γ 1011 20700000
(vi) 8,904,000,000 (iv) ππ.ππππΓ ππππβππ Solution: Solution: 8,904,000,000 3.15 Γ 10β6 In Scientific Form: In Standard Form: 8.904 Γ 109 0.00000315 (vii) 0.00104 (v) ππ.ππππΓ ππππβππππ
Solution: Solution: 0.00104 6.27 Γ 10β10 In Scientific Form: In Standard Form: 1.04 Γ 10β3 0.000000000627 (viii) 0.00000000514 (vi) ππ.ππππΓ ππππβππ Solution: Solution: 0.00000000514 5.41 Γ 10β8 In Scientific Form: In Standard Form: 5.14 Γ 10β9 0.0000000541
(ix) ππ.ππππΓ ππππβππ (vii) ππ.ππππππΓ ππππβππ Solution: Solution: 0.05 Γ 10β3 7.632 Γ 10β4 In Scientific Form: In Standard Form: 5.0 Γ 10β2 Γ 10β3 0.0007632 5.0 Γ 10β2β3 5.0 Γ 10β5 (viii) ππ.ππ Γ ππππππ Solution: Q2: Write each number in standard notation. 9.4 Γ 105
(i) ππ.ππ Γ ππππβππ In Standard Form: Solution: 940000 8.3 Γ 10β5 In Standard Form: (ix) βππ.ππΓ ππππππ 0.000083 Solution:
(ii) ππ.ππ Γ ππππππ β2.6 Γ 109 Solution: In Standard Form: 4.1 Γ 106 β2600000000 In Standard Form: 410000
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Ex # 3.1 Ex # 3.2 Q3: How long does it take light to travel to Earth
from the sun? The sun is ππ.ππΓ ππππππmiles from Earth, and light travels ππ.ππππΓ ππππππ mi/s.
Page # 83 Q1: Write the following in logarithm form. (i) ππππ = ππππππ Solution: Solution: Given: 44 = 256 Distance between earth and sun = 9.3 Γ 107 miles In logarithm form Speed of light = 1.86 Γ 105 mi/s log4 256 = 4 As we have: π π = π£π£π£π£
π π π£π£
= π£π£
Or
π£π£ =π π π£π£
Put the values:
π£π£ =9.3 Γ 107
1.86 Γ 105
π£π£ = 5 Γ 107 Γ 10β5
π£π£ = 5 Γ 107β5 π£π£ = 5 Γ 102 π£π£ = 500 π π πππ π π£π£ = 480 sec + 20 π π πππ π π£π£ = 8 min 20 π π πππ π
(ii) ππβππ =ππππππ
Solution:
2β6 =1
64
In logarithm form
log21
64= β6
(iii) ππππππ = ππ Solution: 100 = 1 In logarithm form log10 1 = 0 (iv) π₯π₯
34 = ππ
Exercise # 3.2 Solution:
π₯π₯34 = ππ
In logarithm form
logπ₯π₯ ππ =34
Logarithm If πππ₯π₯ = ππ then the index π₯π₯ is called the
logarithm of ππ to the base ππ and written as logππ ππ = π₯π₯. We called logππ ππ = π₯π₯ like log of y to the base a equal to π₯π₯.
(v) ππβππ =ππππππ
Logarithm Form Exponential Form
logππ ππ = π₯π₯
πππ₯π₯ = ππ log8 64 = 2
82 = 64
Solution:
3β4 =1
81
In logarithm form
log31
81= β4
(vi) ππππ
ππππ = ππππ
Solution:
6423 = 16
In logarithm form
log64 16 =23
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Ex # 3.2 Ex # 3.2 Q2: Write the following in exponential form. Q3: Solve:
(i) π₯π₯π₯π₯π₯π₯ππ οΏ½πππππποΏ½ = βππ
Solution:
logππ οΏ½1ππ2οΏ½ = β1
In exponential form
ππβ1 =1ππ2
(i) π₯π₯π₯π₯π₯π₯βππ ππππππ = ππ Solution:
logβ5 125 = π₯π₯ In exponential form οΏ½β5οΏ½
π₯π₯= 125
125
x
= 5 Γ 5 Γ 5
25x
= 53 Now π₯π₯2
= 3 Multiply B.S by 2 2 Γ
π₯π₯2
= 2 Γ 3
π₯π₯ = 6
(ii) π₯π₯π₯π₯π₯π₯ππππππππππ
= βππ
Solution:
log21
128= β7
In exponential form
2β7 =1
128
(iii) π₯π₯π₯π₯π₯π₯ππ ππ = ππππ (ii) π₯π₯π₯π₯π₯π₯ππ ππ = βππ Solution: Solution:
log4 π₯π₯ = β3 In exponential form 4β3 = π₯π₯ Now 1
43= π₯π₯ 1
4 Γ 4 Γ 4= π₯π₯
164
= π₯π₯
Or
π₯π₯ =1
64
logππ 3 = 64 In exponential form ππ64 = 3 (iv) π₯π₯π₯π₯π₯π₯ππ ππ = ππ
Solution: logππ ππ = 1 In exponential form ππ1 = 1
(v) π₯π₯π₯π₯π₯π₯ππ ππ = ππ Solution: logππ 1 = 0 (iii) π₯π₯π₯π₯π₯π₯ππππ ππ = ππ In exponential form Solution: ππ0 = 1 log81 9 = π₯π₯ In exponential form
(vi) π₯π₯π₯π₯π₯π₯ππππππ
=βππππ
Solution:
log418
=β32
In exponential form
4β32 =
18
81π₯π₯ = 9 (92)π₯π₯ = 91
92π₯π₯ = 91 Now 2π₯π₯ = 1
Divide B.S by 2 2π₯π₯2
=12
2π₯π₯ =12
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Ex # 3.2 Ex # 3.2 (iv) π₯π₯π₯π₯π₯π₯ππ(ππππ+ ππ) = ππ (vii) π₯π₯π₯π₯π₯π₯ππ(ππ.ππππππ) = βππ Solution:
log3(5π₯π₯ + 1) = 2 In exponential form 32 = 5π₯π₯ + 1 9 = 5π₯π₯ + 1 Subtract 1 form B.S 9 β 1 = 5π₯π₯ + 1β 1 8 = 5π₯π₯ Divide B.S by 5 85
=5π₯π₯5
85
= π₯π₯
π₯π₯ =85
Solution: logπ₯π₯(0.001) = β3
In exponential form π₯π₯β3 = 0.001
π₯π₯β3 =1
1000
π₯π₯β3 =1
103
π₯π₯β3 = 10β3 So
π₯π₯ = 10
(viii) π₯π₯π₯π₯π₯π₯ππ
ππππππ
= βππ Solution:
logπ₯π₯1
64= β2
In exponential form
π₯π₯β2 =1
64
π₯π₯β2 =1
8 Γ 8
π₯π₯β2 =1
82
π₯π₯β2 = 8β2 So π₯π₯ = 8
(v) π₯π₯π₯π₯π₯π₯ππ ππ = ππ
Solution: log2 π₯π₯ = 7 In exponential form 27 = π₯π₯ Now 2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 2 = π₯π₯ 128 = π₯π₯ π₯π₯ = 128 (vi) π₯π₯π₯π₯π₯π₯ππ ππ.ππππ = ππ
Solution: logπ₯π₯ 0.25 = 2 In exponential form π₯π₯2 = 0.25
ππππ =ππππππππππ
Taking square root on B.S
οΏ½π₯π₯2 = οΏ½ 25100
π₯π₯ =5
10
π₯π₯ =12
(ix) π₯π₯π₯π₯π₯π₯βππ ππ = ππππ Solution:
logβ3 π₯π₯ = 16 In exponential form
οΏ½β3οΏ½16
= π₯π₯ 161
23
= π₯π₯
1623 = π₯π₯
38 = π₯π₯ 3 Γ 3 Γ 3 Γ 3 Γ 3 Γ 3 Γ 3 Γ 3 = π₯π₯ 6561 = π₯π₯ π₯π₯ = 6561
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Exercise # 3.3 Ex # 3.3 COMMON LOGARITHM Introduction Page # 86 The common logarithm was invented by a British
Mathematician Prof. Henry Briggs (1560-1631). Q1: Find the characteristics of the common
logarithm of each of the following numbers. Definition (i) 57 Logarithms having base 10 are called common
logarithms or Briggs logarithms. In Scientific form:
5.7 Γ 101 Note: Thus Characteristics = 1 The base of logarithm is not written because it is
considered to be 10.
(ii) 7.4 Logarithm of the number consists of two parts. In Scientific form: Characteristics and Mantissa 7.4 Γ 100 Example: 1.5377 Thus Characteristics = 0 Characteristics The digit before the decimal point or Integral part
is called characteristics (iii) 5.63
In Scientific form: Mantissa 5.63 Γ 100 The decimal fraction part is mantissa. Thus Characteristics = 0 In above example 1 is Characteristics and . 5377 is Mantissa. (iv) 56.3 In Scientific form: USE OF LOG TABLE TO FIND MANTISSA: 5.63 Γ 101 A logarithm table is divided into three parts. Thus Characteristics = 1
(i) The first part of the table is the extreme left column contains number from10 to 99.
(v) 982.5 (ii) The second part of the table consists of 10 columns
headed by 0, 1, 2, β¦. 9. The number under these columns are taken to find mantissa.
In Scientific form: 9.825 Γ 102 Thus Characteristics = 2 (iii) The third part consists of small columns known as
mean difference headed by 1, 2, 3, β¦ 9. These columns are added to the Mantissa found in second column.
(vi) 7824 In Scientific form: 7.824 Γ 103 Thus Characteristics = 3 To Find Mantissa Let we have an example: 763.5 (vii) 186000 Solution: In Scientific form:
(i) First ignore the decimal point. 1.86 Γ 105 (ii) Take first two digits e.g. 76 and proceed along
this row until we come to column headed by third digit 3 of the number which is 8825.
Thus Characteristics = 5 viii. 0.71 (iii) Now take fourth digit i.e. 5 and proceed along
this row in mean difference column which is 5. In Scientific form:
7.1 Γ 10β1 (iv) Now add 8825 + 3 = 8828 Thus Characteristics = β1
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Ex # 3.3 Ex # 3.3 (v) π₯π₯π₯π₯π₯π₯ππ.ππππππππππ Solution: Q2: Find the following. log 0.00159
(i) π₯π₯π₯π₯π₯π₯ππππ.ππ In Scientific form: Solution: 1.59 Γ 10β3 log 87.2 Thus Characteristics = β3 In Scientific form: To find Mantissa, using Log Table: 8.72 Γ 101 So Mantissa = .2014 Thus Characteristics = 1 Hence log 0.00159 = 3. 2014 To find Mantissa, using Log Table: So Mantissa = .9405 (vi) π₯π₯π₯π₯π₯π₯ππ.ππππππππ Hence log 87.2 = 1.9405 Solution: log 0.0256 (ii) π₯π₯π₯π₯π₯π₯ππππππππππ In Scientific form:
Solution: 2.56 Γ 10β2 log 37300 Thus Characteristics = β2 In Scientific form: To find Mantissa, using Log Table: 3.73 Γ 104 So Mantissa = .4082 Thus Characteristics = 4 Hence log 0.0256 = 2. 4082 To find Mantissa, using Log Table: So Mantissa = .5717 (vii) π₯π₯π₯π₯π₯π₯ππ.ππππππ Hence log 37300 = 4.5717 Solution: log 6.753 (iii) π₯π₯π₯π₯π₯π₯ππππππ In Scientific form: Solution: 6.753 Γ 100 log 753 π π .ππ
8293 + 2 = 8295
Thus Characteristics = 0 To find Mantissa, using Log Table Mantissa = .8295 Hence log 6.753 = 0.8295
In Scientific form: 7.53 Γ 102 Thus Characteristics = 2 To find Mantissa, using Log Table: Q3: Find logarithms of the following numbers. So Mantissa = .8768 (i) ππππππππ Hence log 753 = 2.8768 Solution: 2476 (iv) π₯π₯π₯π₯π₯π₯ππ.ππππ πΏπΏπππ£π£ π₯π₯ = 2476
Solution: Taking log on B.S log 9.21 log π₯π₯ = log 2476 In Scientific form: In Scientific form: 9.21 Γ 100 2.476 Γ 103 Thus Characteristics = 0 Thus Characteristics = 3 To find Mantissa, using Log Table: To find Mantissa, using Log Table
π π .ππ
3927 + 11 = 3938
So Mantissa = .3927 + 11 Mantissa = .3938 Hence log 2476 = 3.3938
So Mantissa = .9643 Hence log 9.21 = 0.9643
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Ex # 3.3 Ex # 3.3 (ii) 2.4 (v) 0.783
Solution: Solution: 2.4 0.783 πΏπΏπππ£π£ π₯π₯ = 2.4 πΏπΏπππ£π£ π₯π₯ = 0.783 Taking log on B.S Taking log on B.S log π₯π₯ = log 2.4 log π₯π₯ = log 0.783 In Scientific form: In Scientific form: 2.4 Γ 100 7.83 Γ 10β1 Thus Characteristics = 0 Thus Characteristics = β1 To find Mantissa, using Log Table: To find Mantissa, using Log Table: So Mantissa = .3802 So Mantissa = .8938 Hence log 2.4 = 0.3802 Hence log 0.783 = 1. 8938 (iii) 92.5 (vi) 0.09566 Solution: Solution: 92.5 0.09566 πΏπΏπππ£π£ π₯π₯ = 92.5 πΏπΏπππ£π£ π₯π₯ = 0.09566 Taking log on B.S Taking log on B.S log π₯π₯ = log 92.5 log π₯π₯ = log 0.09566 In Scientific form: In Scientific form: 9.25 Γ 101 9.566 Γ 10β2 Thus Characteristics = 1 Thus Characteristics = β2 To find Mantissa, using Log Table: To find Mantissa, using Log Table: So Mantissa = .9661 π π .ππ
9805 + 3 = 9808
So Mantissa = . 9808 Hence log 0.09566 = 2 . 9808 Hence log 92.5 = 1.9661
(iv) 482.7
Solution: 482.7 (vii) 0.006753 πΏπΏπππ£π£ π₯π₯ = 482.7 Solution: Taking log on B.S 0.006753 log π₯π₯ = log 482.7 πΏπΏπππ£π£ π₯π₯ = 0.006753 In Scientific form: Taking log on B.S 4.827 Γ 102 log π₯π₯ = log 0.006753 Thus Characteristics = 2 In Scientific form: To find Mantissa, using Log Table: 6.753 Γ 10β3 π π .ππ
6830 + 6 = 6836
So Mantissa = .6836 Hence log 482.7 = 2.6836
Thus Characteristics = β3 To find Mantissa, using Log Table: So Mantissa = . 8295
π π .ππ
8293 + 2 = 8295
Hence log 0.006735 = 3 . 8295
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Ex # 3.3 Ex # 3.4 (viii) 700 Solution: Page # 88 700 Q1: Find anti-logarithm of the following numbers. πΏπΏπππ£π£ π₯π₯ = 700 (i) 1.2508 Taking log on B.S Solution: log π₯π₯ = log 700 1.2508 In Scientific form: πΏπΏπππ£π£ log π₯π₯ = 1.2508
Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 1.2508
π π .ππ
1778+3 = 1781
π₯π₯ = Antiβ log 1.2508 Characteristics = 1 Mantissa = .2508 So π₯π₯ = 1.781 Γ 101 π₯π₯ = 17.81
7.00 Γ 102 Thus Characteristics = 2 To find Mantissa, using Log Table: So Mantissa = . 8451 Hence log 700 = 2 . 8451 Exercise # 3.4 ANTI-LOGARITHM If log π₯π₯ = ππ then π₯π₯ is the anti-logarithm of y and
written as π₯π₯ = πππππ£π£ππ β log ππ
(ii) 0.8401 Explanation with Example: Solution: 2 . 3456 0.8401
(i) Here the digit before decimal point is Characteristics i.e. 2
πΏπΏπππ£π£ log π₯π₯ = 0.8401 Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 0.8401 π₯π₯ = Antiβ log 0.8401 π π .ππ
6918+2 = 6920
Characteristics = 0 Mantissa = .8401 So π₯π₯ = 6.920 Γ 100 π₯π₯ = 6.920
(ii) And Mantissa= . 3456
To find anti-log, we see Mantissa in Anti-log Table
(i) Take first two digits i.e. . 34 and proceed along this row until we come to column headed by third digit 5 of the number which is 2213.
(ii) Now take fourth digit i.e. 6 and proceed along this row which is 3.
(iii) Now add 2213 + 3 = 2216 (iii) 2.540 So to find anti-log, write it in Scientific form like Solution: πππππ£π£ππ β log 2.3456 = 2 . 2216 Γ 10ππβππππ 2.540 πππππ£π£ππ β log 2.3456 = 2 . 216 Γ 102 πΏπΏπππ£π£ log π₯π₯ = 2.540
Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 2.540 π₯π₯ = Antiβ log 2.540 Characteristics = 2 Mantissa = .540 So π₯π₯ = 3.467 Γ 102 π₯π₯ = 346.7
πππππ£π£ππ β log 2.3456 = 221 . 6
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Ex # 3.4 Ex # 3.4 (iv) ππ.ππππππππ Q2: Find the values of ππ from the following
equations: Solution: 2. 2508 (i) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ πΏπΏπππ£π£ log π₯π₯ = 2. 2508
Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 2. 2508 π₯π₯ = Antiβ log 2. 2508
π π .ππ
1778+3 = 1781
Characteristics = β2 Mantissa = .2508 So π₯π₯ = 1.781 Γ 10β2 π₯π₯ = 0.01781
Solution: log π₯π₯ = 1. 8401 Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S
Antiβ log (log π₯π₯) = Antiβ log 1. 8401 π₯π₯ = Antiβ log1. 8401
π π .ππ
6918 + 2= 6920
Characteristics = β1 Mantissa = .8401 So π₯π₯ = 6.920 Γ 10β1 π₯π₯ = 0.6920
(v) ππ.ππππππππ Solution: (ii) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ 1. 5463 Solution: πΏπΏπππ£π£ log π₯π₯ = 1. 5463
Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 1. 5463 π₯π₯ = Antiβ log 1. 5463
π π .ππ
3516+2 = 3518
Characteristics = β1 Mantissa = .5463 So π₯π₯ = 3.518 Γ 10β1 π₯π₯ = 0.3518
log π₯π₯ = 2.1931 Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S
Antiβ log (log π₯π₯) = Antiβ log 2.1931 π₯π₯ = Antiβ log 2.1931
π π .ππ
1560 + 0= 1560
Characteristics = 2 Mantissa = .1931 So π₯π₯ = 1.560 Γ 102 π₯π₯ = 156.0
(vi) 3.5526 (iii) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ
Solution: Solution: 3.5526 log π₯π₯ = 4.5911 πΏπΏπππ£π£ log π₯π₯ = 3.5526
Taking anti-log on B.S π΄π΄πππ£π£ππ β log(log x) = Antiβ log 3.5526 π₯π₯ = Antiβ log 3.5526
π π .ππ
3565+5 = 3570
Characteristics = 3 Mantissa = .5526 So π₯π₯ = 3.570 Γ 103 π₯π₯ = 3570
Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S Antiβ log (log π₯π₯) = Antiβ log 4.5911 π₯π₯ = Antiβ log 4.5911
π π .ππ
3899 + 1= 3900
Characteristics = 4 Mantissa = .5911 So π₯π₯ = 3.900 Γ 104 π₯π₯ = 39000.0
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Ex # 3.4 Ex # 3.5 (i) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ LAWS OF LOGARITHM
Solution: (i) logππ ππππ = logππ ππ + logππ ππ log π₯π₯ = 3. 0253 or logππππ = logππ + log ππ Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S
Antiβ log (log π₯π₯) = Antiβ log 3. 0253 π₯π₯ = Antiβ log 3. 0253
π π .ππ
1059 + 1= 1060
Characteristics = β3 Mantissa = .0253 So π₯π₯ = 1.060 Γ 10β3 π₯π₯ = 0.001060
Example: log 2 Γ 3 = log 2 + log 3 (ii) logππ
ππππ
= logππ ππ β logππ ππ or log
ππππ
= logππβ log ππ Example: log
35
= log 3 β log 5 (ii) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ
log 6 β log 3 = log63
= log 2 Solution: log π₯π₯ = 1.8716 (iii) logππ ππππ = ππ logππ ππ Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S
Antiβ log (log π₯π₯) = Antiβ log 1.8716 π₯π₯ = Antiβ log 1.8716
π π .ππ
7430 + 10= 7440
Characteristics = 1 Mantissa = .8716 So π₯π₯ = 7.440 Γ 101 π₯π₯ = 74.40
or logππππ = ππ logππ Example: log 23 = 3 log 2 logππ ππ logππ ππ = logππ ππ log2 3 log3 5 = log3 5 logππ ππ =
logππ ππlogππ ππ
Example:
(iv)
log7 ππlog7 π£π£
= logπ‘π‘ ππ (iii) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ Solution: Note: log π₯π₯ = 2. 8370 (i) logππ ππ = 1 Taking ππππππππ β π₯π₯π₯π₯π₯π₯ on B.S
Antiβ log (log π₯π₯) = Antiβ log 2. 8370 π₯π₯ = Antiβ log 2. 8370 Characteristics = β2 Mantissa = .8370 So π₯π₯ = 6.871 Γ 10β2 π₯π₯ = 0.06781
(ii) log10 10 = 1 (iii) log 10 = 1 (iv) log10 1 = 0 (v) log 1 = 0 (vi)
logππ ππ =logππ ππlogππ ππ
This is called Change of Base Law
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Ex # 3.5 Ex # 3.5 Proof of Laws of Logarithm one by one (iii) π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ
(i) π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ Proof: Proof: πΏπΏπππ£π£ logππ ππ = π₯π₯
In Exponential form: πππ₯π₯ = ππ Or ππ = πππ₯π₯ Taking power βnβ on B.S ππππ = (πππ₯π₯)ππ
ππππ = πππππ₯π₯ Taking loga on B.S
logππ ππππ = logππ πππππ₯π₯ logππ ππππ = πππ₯π₯ logππ ππ logππ ππππ = πππ₯π₯(1) β΄ logππ ππ = 1 logππ ππππ = πππ₯π₯ logππ ππππ = ππ logππ ππ
πΏπΏπππ£π£ logππ ππ = π₯π₯ ππππππ logππ ππ = ππ Write them in Exponential form: πππ₯π₯ = ππ ππππππ πππ¦π¦ = ππ Now multiply these: πππ₯π₯ Γ πππ¦π¦ = ππππ Or ππππ = πππ₯π₯ Γ πππ¦π¦ ππππ = πππ₯π₯+π¦π¦ Taking loga on B.S
logππ ππππ = logππ πππ₯π₯+π¦π¦ logππ ππππ = (π₯π₯ + ππ) logππ ππ logππ ππππ = (π₯π₯ + ππ)(1) β΄ logππ ππ = 1 logππ ππππ = π₯π₯ + ππ logππ ππππ = logππ ππ + logππ ππ
(ii) π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ (iv) π₯π₯π₯π₯π₯π₯ππππ π₯π₯π₯π₯π₯π₯ππ ππ = π₯π₯π₯π₯π₯π₯ππ ππ Proof:
Proof: πΏπΏπππ£π£ logππ ππ = π₯π₯ ππππππ logππ ππ = ππ Write them in Exponential form: πππ₯π₯ = ππ ππππππ πππ¦π¦ = ππ Now multiply these: π΄π΄π π πππ₯π₯π¦π¦ = (πππ₯π₯)π¦π¦ π΅π΅π΅π΅π£π£ (πππ₯π₯)π¦π¦ = ππ ππππ πππ₯π₯π¦π¦ = (ππ)π¦π¦ = ππ ππππ πππ₯π₯π¦π¦ = ππ Taking loga on B.S
logππ πππ₯π₯π¦π¦ = logππ ππ (π₯π₯ππ) logππ ππ = logππ ππ π₯π₯ππ(1) = logππ ππ β΄ logππ ππ = 1 Now logππ ππ logππ ππ = logππ ππ
πΏπΏπππ£π£ logππ ππ = π₯π₯ ππππππ logππ ππ = ππ Write them in Exponential form: πππ₯π₯ = ππ ππππππ πππ¦π¦ = ππ Now Divide these: πππ₯π₯
πππ¦π¦=ππππ
Or ππππ
=πππ₯π₯
πππ¦π¦
ππππ
= πππ₯π₯βπ¦π¦
Taking loga on B.S
logππππππ
= logππ πππ₯π₯βπ¦π¦
logππππππ
= (π₯π₯ β ππ) logππ ππ
logππππππ
= (π₯π₯ β ππ)(1) β΄ logππ ππ = 1
logππππππ
= π₯π₯ β ππ
π»π»πππππ π ππ logππππππ
= logππ ππ β logππ ππ
Example # 14 page # 90
β1 + log ππ
Solution:
= β1 + log ππ
= β log 10 + log ππ
= log 10β1 + log ππ
= log1
10+ log ππ
= log 0.1 + log ππ
= log 0.1ππ
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Ex # 3.5 Ex # 3.5
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ Page # 91 π₯π₯ =
32
log10 10
π₯π₯ =32
(1) β΄ π₯π₯π₯π₯π₯π₯ππ ππ = ππ
π₯π₯ =32
Q1: Use logarithm properties to simplify the expression.
(i) π₯π₯π₯π₯π₯π₯ππ βππ Solution: log7 β7 πΏπΏπππ£π£ π₯π₯ = log7 β7
π₯π₯ = log7(7)12
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ
π₯π₯ =12
log7 7
π₯π₯ =12
(1) β΄ π₯π₯π₯π₯π₯π₯ππ ππ = ππ
π₯π₯ =12
(iv) π₯π₯π₯π₯π₯π₯ππ ππ+ π₯π₯π₯π₯π₯π₯ππ ππππ Solution: log9 3 + log9 27 πΏπΏπππ£π£ π₯π₯ = log9 3 + log9 27
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ + π₯π₯π₯π₯π₯π₯ππ ππ π₯π₯ = log9 3 Γ 27 π₯π₯ = log9 81 π₯π₯ = log9 92 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ π₯π₯ = 2 log9 9 π₯π₯ = 2(1) β΄ π₯π₯π₯π₯π₯π₯ππ ππ = ππ π₯π₯ = 2
(ii) π₯π₯π₯π₯π₯π₯ππππππ
Solution: log8
12
log812
πΏπΏπππ£π£ log812
= π₯π₯
In exponential form:
8π₯π₯ =12
(23)π₯π₯ = 2β1 23π₯π₯ = 2β1 Now 3π₯π₯ = β1 Divide B.S by 3, we get
π₯π₯ =β13
(v) π₯π₯π₯π₯π₯π₯ππ
(ππ.ππππππππ)βππ Solution: log
1(0.0035)β4
πΏπΏπππ£π£ π₯π₯ = log1
(0.0035)β4
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
π₯π₯ = log 1β log(0.0035)β4 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππ = ππ and π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ Thus π₯π₯ = 0 β (β4) log 0.0035
πΉπΉ.πΎπΎ
3.5 Γ 10β3
Here πΆπΆβ = β3 And ππ = .5441 So π₯π₯ = 4(β3 + .5441) π₯π₯ = 4(β2.4559) π₯π₯ = β9.8236
(iii) π₯π₯π₯π₯π₯π₯ππππ βππππππππ Solution: log10 β1000
πΏπΏπππ£π£ π₯π₯ = log10(103)12
π₯π₯ = log10(10)32
Trick
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Ex # 3.5 (vi) π₯π₯π₯π₯π₯π₯ππππ (iii) π₯π₯π₯π₯π₯π₯ππ β ππ
Solution: Solution: log 45 log 5 β 1
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππ = ππ log 5 β 1 = log 5 β log 10
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
log 5 β 1 = log5
10
log 5 β 1 = log 0.5
πΏπΏπππ£π£ π₯π₯ = log 45 π₯π₯ = log 3 Γ 3 Γ 5 π₯π₯ = log 32 Γ 5 π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ ππππππ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ π₯π₯ = 2 log 3 + log 5 π₯π₯ = 2 log 3.00 + log 5.00 π₯π₯ = 2(0 + .4771) + (0 + .6990) π₯π₯ = 2(0.4771) + (0.6990) π₯π₯ = 0.9542 + 0.6990 π₯π₯ = 1.6532
(iv)
πππππ₯π₯π₯π₯π₯π₯ππ β ππ π₯π₯π₯π₯π₯π₯ππππ+ πππ₯π₯π₯π₯π₯π₯ππ
Solution: 1
2log π₯π₯ β 2 log 3ππ + 3log π§π§
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ
= log π₯π₯12 β log(3ππ)2 + log π§π§3
= logβπ₯π₯ β log 9ππ2 + log π§π§3
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
π¨π¨πππ¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ 12
log π₯π₯ β 2 log 3ππ + 3log π§π§ = logβπ₯π₯π§π§3
9ππ2
Q2: Express each of the following as a single
logarithm.
(i) ππ π₯π₯π₯π₯π₯π₯ππ β ππ π₯π₯π₯π₯π₯π₯ππ
Solution: 3 log 2 β 4 log 3
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ 3 log 2 β 4 log 3 = log 22 β log 34 3 log 2 β 4 log 3 = log 8 β log 81
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
3 log 2 β 4 log 3 = log8
81
(ii) ππ π₯π₯π₯π₯π₯π₯ππ + ππ π₯π₯π₯π₯π₯π₯ππ β ππ
Solution: Q3: Find the value of β²ππβ² from the following equations. 2 log 3 + 4 log 2 β 3
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ 2 log 3 + 4 log 2 β 3 = log 32 + log 24 β 3(1) π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππ = ππ So 2 log 3 + 4 log 2β 3 = log 9 + log 16 β 3(log 10) π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ 2 log 3 + 4 log 2 β 3 = log 9 Γ 16 β log 103 2 log 3 + 4 log 2 β 3 = log 9 Γ 16 β log 1000
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
2 log 3 + 4 log 2 β 3 = log144
1000
2 log 3 + 4 log 2 β 3 = log 0.144
(i) π₯π₯π₯π₯π₯π₯ππ ππ+ π₯π₯π₯π₯π₯π₯ππ ππ = π₯π₯π₯π₯π₯π₯ππ ππ Solution: log2 6 + log2 7 = log2 ππ π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ log2 6 Γ 7 = log2 ππ log2 42 = log2 ππ Thus ππ = 42
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(ii) π₯π₯π₯π₯π₯π₯βππ ππ = π₯π₯π₯π₯π₯π₯βππ ππ + π₯π₯π₯π₯π₯π₯βππ ππ β π₯π₯π₯π₯π₯π₯βππ ππ Q4: Find π₯π₯π₯π₯π₯π₯ππ ππ . π₯π₯π₯π₯π₯π₯ππ ππ . π₯π₯π₯π₯π₯π₯ππ ππ . π₯π₯π₯π₯π₯π₯ππ ππ . π₯π₯π₯π₯π₯π₯ππ ππ . π₯π₯π₯π₯π₯π₯ππ ππ
Solution: Solution: logβ3 ππ = logβ3 5 + logβ3 8 β logβ3 2
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
logβ3 ππ = logβ35 Γ 8
2
logβ3 ππ = logβ3402
logβ3 ππ = logβ3 20 Thus ππ = 20
πΏπΏπππ£π£ π₯π₯ = log2 3 . log3 4 . log4 5 . log5 6 . log6 7 . log7 8 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ So π₯π₯ = log2 4 . log4 5 . log5 6 . log6 7 . log7 8 π₯π₯ = log2 5 . log5 6 . log6 7 . log7 8 π₯π₯ = log2 6 . log6 7 . log7 8 π₯π₯ = log2 7 . log7 8 π₯π₯ = log2 8 π₯π₯ = log2 23 π₯π₯ = 3 log2 2 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππ ππ = ππ
(iii) π₯π₯π₯π₯π₯π₯ππ πππ₯π₯π₯π₯π₯π₯ππ ππ
= π₯π₯π₯π₯π₯π₯ππ ππ π₯π₯ = 3(1) π₯π₯ = 3
Solution: log7 ππ
log7 π£π£= logππ ππ
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππ ππ =π₯π₯π₯π₯π₯π₯ππ πππ₯π₯π₯π₯π₯π₯ππππ
logt ππ = logππ ππ Thus ππ = π£π£
Ex # 3.6 Page # 93 Q1: Simplify ππ.ππππΓ ππππ.ππ with the help of logarithm. Solution: (i) 3.81 Γ 43.4
log 3.81 πΆπΆβ = 0 ππ = .5809
log 43.4 πΆπΆβ = 1 ππ = .6375
Let π₯π₯ = 3.81 Γ 43.4 Taking log on B.S
log π₯π₯ = log 3.81 Γ 43.4
As logππππ = logππ + log ππ
log π₯π₯ = log 3.81 + log 43.4
log π₯π₯ = (0 + .5809) + (1 + .6375)
log π₯π₯ = 0.5809 + 1.6375
log π₯π₯ = 2.2184
Taking anti β log on B.S
Antiβ log (log π₯π₯) = Antiβ log 2.2184
π₯π₯ = Antiβ log 2.2184
1652 + 2= 1654
Here Characteristics = 2 Mantissa = .2184 So π₯π₯ = 1.654 Γ 102 π₯π₯ = 16.54
(iv) π₯π₯π₯π₯π₯π₯ππ ππππ β π₯π₯π₯π₯π₯π₯ππ ππ = π₯π₯π₯π₯π₯π₯ππ ππ
Solution: log6 25 β log6 5 = log6 ππ
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
log6255
= log6 ππ
log6 5 = log6 ππ Thus ππ = 5
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Ex # 3.6 (ii) ππππ.ππππΓ ππ.ππππππππππΓ ππ.ππππππππ
Solution: 73.42 Γ 0.00462 Γ 0.5143 Let π₯π₯ = 73.42 Γ 0.00462 Γ 0.5143 Taking log on B.S
log π₯π₯ = 73.42 Γ 0.00462 Γ 0.5143 As logππππ = logππ + log ππ log π₯π₯ = log 73.42 + log 0.00462 + log 0.5143 log π₯π₯ = (1 + .8658) + (β3 + .6646) + (β1 + .7113) log π₯π₯ = 1.8658 + (β2.3354) + (β0.2887) log π₯π₯ = 1.8658 β 2.3354 β 0.2887 log π₯π₯ = β0.7583 Add and Subtract β1 log π₯π₯ = β1 + 1β 0.7583 log π₯π₯ = β1 + .2417 log π₯π₯ = 1 . 2417 Taking antiβ log on B. S antiβ log (log π₯π₯) = antiβ log 1 . 2417 π₯π₯ = anti β log 1 . 2417 Here Characteristics = β1 Mantissa = .2417 So π₯π₯ = 1.745 Γ 10β1 π₯π₯ = 0.1745
log 73.42 πΆπΆβ = 1 8657 + 1 ππ = .8658
log 0.00462 πΆπΆβ = β3 ππ = .6646 log 0.5143 πΆπΆβ = β1
7110 + 3 ππ = .7113
1742 + 3
= 1745
(iii) ππππππ.ππ Γ ππ.ππππππππ
ππππ.ππππ
Solution: 784.6 Γ 0.0431
28.23
πΏπΏπππ£π£ π₯π₯ =784.6 Γ 0.0431
28.23
Taking log on B.S
log π₯π₯ = log784.6 Γ 0.0431
28.23
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππ
= π₯π₯π₯π₯π₯π₯ππ β π₯π₯π₯π₯π₯π₯ππ
log π₯π₯ = log 784.6 Γ 0.0431 β log 28.23
As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ log π₯π₯ = log 784.6 + log 0.0431 β log 28.23
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Ex # 3.6 log 784.6 πΆπΆβ = 2 8943 + 3 ππ = .8946
log 0.0431 πΆπΆβ = β2 ππ = .6345 log 28.23 πΆπΆβ = 1 4502 + 5 ππ = .4507
log π₯π₯ = (2 + .8946) + (β2 + .6345) + (1 + .4507)
log π₯π₯ = 2.8946 + (β1.3655) + (1.4507)
log π₯π₯ = 2.8946 β 1.3655 β 1.4507
log π₯π₯ = 0 . 0784
Taking antiβ log on B. S
antiβ log (log π₯π₯) = antiβ log 0 . 0784
π₯π₯ = anti β log 0 . 0784 Here Characteristics = 0 Mantissa = . 0784 So π₯π₯ = 1.198 Γ 100 π₯π₯ = 1.198
1197 + 1
= 1198
(iv) ππ.ππππππππΓ ππππππ.ππππ.ππππππππππΓ ππππππππ
Solution:
0.4932 Γ 653.70.07213 Γ 8456
πΏπΏπππ£π£ π₯π₯ =0.4932 Γ 653.70.07213 Γ 8456
Taking log on B.S
log π₯π₯ = log0.4932 Γ 653.70.07213 Γ 8456
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππ
= π₯π₯π₯π₯π₯π₯ππ β π₯π₯π₯π₯π₯π₯ππ
log π₯π₯ = log(0.4932 Γ 653.7) β log(0.07213 Γ 8456)
As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ
log π₯π₯ = log 0.4932 + log 653.7 β (log 0.07213 + log 8456)
log π₯π₯ = log 0.4932 + log 653.7 β log 0.07213 β log 8456
log π₯π₯ = (β1 + .6930) + (2 + .8154) β (β2 + .8581) β (3 + .9271)
log π₯π₯ = (β1 + .6930) + (2 + .8154) β (β2 + .8581) β (3 + .9271)
log π₯π₯ = (β0.3070) + (2.8154) β (β1.1419) β (3.9271)
log π₯π₯ = β0.3070 + 2.8154 + 1.1419 β 3.9271
log π₯π₯ = β0.2768
log 0.4932
πΆπΆβ = β1 6928 + 2 ππ =. 6930 log 653.7 πΆπΆβ = 2 8149 + 5 ππ = . 8154 log 0.07213 πΆπΆβ = β2 8579 + 2 ππ = . 8581 log 8456 πΆπΆβ = 3 9269 + 3 ππ = . 9272
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Ex # 3.6 Add and Subtract β1
log π₯π₯ = β1 + 1β 0.2768 log π₯π₯ = β1 + .7232 log π₯π₯ = 1 . 7232 Taking antiβ log on B. S antiβ log (log π₯π₯) = antiβ log 1 .7232 π₯π₯ = anti β log 1 . .7232 Here Characteristics = β1 Mantissa = .7232 So π₯π₯ = 5.286 Γ 10β1 π₯π₯ = 0.5286
5284 + 2
= 5286
(v) (ππππ.ππππ)ππβππππππ.ππ
βππ.ππππππππππ
Solution: (78.41)3β142.3
β0.15624
πΏπΏπππ£π£ π₯π₯ =(78.41)3β142.3
β0.15624
Taking log on B.S
log π₯π₯ = log(78.41)3β142.3
β0.15624
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππ
= π₯π₯π₯π₯π₯π₯ππ β π₯π₯π₯π₯π₯π₯ππ
log π₯π₯ = log(78.41)3β142.3β log β0.15624
As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ
log π₯π₯ = log(78.41)3 + logβ142.3 β log β0.15624
log π₯π₯ = log(78.41)3 + log(142.3)12 β log(0.1562)
14
log π₯π₯ = 3 log 78.41 +12
log 142.3β14
log 0.1562
log π₯π₯ = 3 log(78.41) +12
log(142.3) β14
log(0.1562)
log π₯π₯ = 3 (1 + .8944) +12
( 2 + .1532) β14
(β1 + .1937)
log 78.41 πΆπΆβ = 1 8943 + 1 ππ =. 8944 log 142.3 πΆπΆβ = 2 1523 + 9 ππ = . 1523 log 0.1562 πΆπΆβ = β1 1931 + 6 ππ = . 1937
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log π₯π₯ = 3 (1.8944) +12
( 2.1532) β14
(β0.8063)
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Ex # 3.6 Review Ex # 3 log π₯π₯ = 5.6832 + 1.0766 + 0.2016
log π₯π₯ = 6.9614 Taking antiβ log on B. S antiβ log (log π₯π₯) = antiβ log 6.9614 π₯π₯ = anti β log 6.9614 Here
9141 + 8= 9149
Characteristics = 6 Mantissa = .9614 So π₯π₯ = 9.149 Γ 106 π₯π₯ = 9149000
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ log β723 =
13
log 72
ππππππ β723 =13
(ππππππ2 Γ 2 Γ 2 Γ 3 Γ 3)
ππππππ β723 =13
(ππππππ23 Γ 32)
As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ
ππππππ β723 =13
(ππππππ23 + ππππππ 32)
ππππππ β723 =13
(3 ππππππ2 + 2 ππππππ3)
ππππππ β723 =13
[3(0.3010) + 2(0.4771)]
ππππππ β723 =13
[0.9030 + 0.9542]
ππππππ β723 =13
[1.8572]
ππππππ β723 = 0.6191
Q2: Find the following if π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ,
π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ, π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ, π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ
(i) π₯π₯π₯π₯π₯π₯ππππππ
Solution:
log 105 log 105 = log 3 Γ 5 Γ 7 (iv) π₯π₯π₯π₯π₯π₯ππ.ππ As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ Solution: log 105 = log 3 + log 5 + log 7 log 2.4
log 2.4 = log2410
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
log 2.4 = log 24 β log 10
log 2.4 = log 2 Γ 2 Γ 2 Γ 3 β log 10
log 2.4 = log 23 Γ 3β log 10
As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ
log 2.4 = log 23 + log 3 β log 10
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ log 2.4 = 3 log 2 + log 3 β log 10
log 2.4 = 3(0.3010) + 0.4771 β log 10
log 2.4 = 0.9030 + 0.4771 β 1 β΄ π₯π₯π₯π₯π₯π₯ππππ = ππ
log 2.4 = 1.3801 β 1
log 2.4 = 0.3801
log 105 = 0.4771 + 0.6990 + 0.8451 log 105 = 2.0211
(ii) π₯π₯π₯π₯π₯π₯ππππππ log 108 log 108 = log 2 Γ 2 Γ 3 Γ 3 Γ 3 log 108 = log 22 Γ 33 As π₯π₯π₯π₯π₯π₯ππππ = π₯π₯π₯π₯π₯π₯ππ + π₯π₯π₯π₯π₯π₯ππ log 108 = log 22 + log 33 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ log 108 = 2 log 2 + 3 log 3 log 108 = 2(0.3010) + 3(0.4771) log 108 = 0.6020 + 1.4313 log 108 = 2.0333 (iii) π₯π₯π₯π₯π₯π₯ βππππππ
Solution: log β723
log β723 = log(72)13
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Ex # 3.6 Review Ex # 3 (v) π₯π₯π₯π₯π₯π₯ππ.ππππππππ Q5: Write in exponential form: π₯π₯π₯π₯π₯π₯ππ ππ = ππ
Solution: log5 1 = 0 log 0.0081
log 0.0081 = log81
10000
log 0.0081 = log34
104
log 0.0081 = log οΏ½3
10οΏ½4
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ log 0.0081 = 4 log
310
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
log 0.0081 = 4(log 3β log 10) log 0.0081 = 4(0.4771 β 1) β΄ log 10 = 1 log 0.0081 = 4(β0.5229)
log 0.0081 = β2.0916
In exponential form: 50 = 1 Q6: Solve for ππ: π₯π₯π₯π₯π₯π₯ππ ππππ = ππ log4 16 = π₯π₯ In exponential form: 4π₯π₯ = 16 4π₯π₯ = 42 So π₯π₯ = 2
Q7: Find the characteristic of the common logarithm 0.0083.
0.0083 In scientific notation: 8.3 Γ 10β3 So Characteristics β3 Q8: Find π₯π₯π₯π₯π₯π₯ππππ.ππ REVIEW EXERCISE # 3
log 12.4 In Scientific form: Page # 95 1.24 Γ 101 Q2: Write 9473.2 in scientific notation. Thus Characteristics = 1 9473.2 To find Mantissa, using Log Table: In scientific notation: Mantissa = . 0934 9.4732 Γ 103 Hence log 12.4 = 0. 0934 Q3: Write ππ.ππ Γ ππππππ in standard notation. Q9: Find the value of β²ππβ², 5.4 Γ 106 logβ5 3ππ = logβ5 9 + logβ5 2β logβ5 3 In standard form: Solution: 5400000 logβ5 3ππ = logβ5 9 + logβ5 2β logβ5 3
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππ ππ
π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ
= π₯π₯π₯π₯π₯π₯ππππ β π₯π₯π₯π₯π₯π₯ππ ππ
logβ5 3ππ = logβ59 Γ 2
3
logβ5 3ππ = logβ5 3 Γ 2 logβ5 3ππ = logβ5 6 Thus 3ππ = 6
ππ =62
ππ = 3
Q4: Write in logarithm form: ππβππ = ππ
ππππ
3β3 =1
27
In logarithm form:
log31
27= β3
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Q10 (ππππ.ππππ)ππ(ππ.ππππππππππ)ππ(ππ.ππππππππ)
(ππππππ.ππ)(ππ.ππππππ)ππ
Solution:
(63.28)3(0.00843)2(0.4623)(412.3)(2.184)5
πΏπΏπππ£π£ π₯π₯ =(63.28)3(0.00843)2(0.4623)
(412.3)(2.184)5
Taking log on B.S
log π₯π₯ = log(63.28)3(0.00843)2(0.4623)
(412.3)(2.184)5
log 63.28 πΆπΆβ = 1 8007 + 5 ππ =. 8012
log 0.00843 πΆπΆβ = β3 ππ = . 9258 log 0.4623 πΆπΆβ = β1 6646 + 3 ππ = . 6649 log 412.3 πΆπΆβ = 2 6149 + 3 ππ = . 6152 log 2.184 πΆπΆβ = 0 3385 + 8 ππ = . 3393
π΄π΄π π logππππ
= logππβ log ππ
log π₯π₯ = log((63.28)3(0.00843)2(0.4623)) β log((412.3)(2.184)5)
As logππππ = logππ + log ππ
log π₯π₯ = log(63.28)3 + log(0.00843)2 + log 0.4623β (log 412.3 + log(2.184)5) log π₯π₯ = 3 log 63.28 + 2 log 0.00843 + log 0.4623β (log 412.3 + 5 log 2.184) log π₯π₯ = 3 log 63.28 + 2 log 0.00843 + log 0.4623β log 412.3β 5 log 2.184
log π₯π₯ = 3(1 + .8012) + 2(β3 + .9258) + (β1 + .6649)β (2 + .6152) β 5(0 + .3393)
log π₯π₯ = 3(1.8012) + 2(β2. 0742) + (β0.3351) β (2.6152) β 5(0.3393)
log π₯π₯ = 5.4036 β 4.1484 β 0.3351 β 2.6152 β 1.6965
log π₯π₯ = β3.3916
Add and Subtract β4 log π₯π₯ = β4 + 4β 3.3916 log π₯π₯ = β4 + .6084 log π₯π₯ = 4 . 6084 Taking antiβ log on B. S antiβ log (log π₯π₯) = antiβ log 4 . 6084 π₯π₯ = anti β log 4 . 6084 Here Characteristics = β4 4055 + 4
= 4059 Mantissa = . 6084 So π₯π₯ = 4.059 Γ 10β4 π₯π₯ = 0.000405
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