unit # 3 logarithm
TRANSCRIPT
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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
log2 1
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 =3
4
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
log3 1
81= β4
(vi) ππππππππ = ππππ
Solution:
6423 = 16
In logarithm form
log64 16 =2
3
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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 οΏ½ πππππποΏ½ = βππ
Solution:
logππ οΏ½ 1ππ2οΏ½ = β1
In exponential form ππβ1 =1ππ2
(i) π₯π₯π₯π₯π₯π₯βππ ππππππ = ππ
Solution:
logβ5 125 = π₯π₯
In exponential form οΏ½β5οΏ½π₯π₯ = 125
1
25
x
= 5 Γ 5 Γ 5
25x
= 53
Now π₯π₯2
= 3
Multiply B.S by 2
2 Γπ₯π₯2
= 2 Γ 3 π₯π₯ = 6
(ii) π₯π₯π₯π₯π₯π₯ππ ππππππππ = βππ
Solution:
log2 1
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₯
1
64= π₯π₯
Or π₯π₯ =1
64
logππ 3 = 64
In exponential form
ππ64 = 3
(iv) π₯π₯π₯π₯π₯π₯ππππ = 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₯ππ ππππ =βππππ
Solution:
log4 1
8=β3
2
In exponential form
4β32 =
1
8
81π₯π₯ = 9
(92)π₯π₯ = 91
92π₯π₯ = 91
Now 2π₯π₯ = 1
Divide B.S by 2 2π₯π₯2
=1
2
2π₯π₯ =1
2
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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
8
5=
5π₯π₯5
8
5= π₯π₯ π₯π₯ =
8
5
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 25
100
π₯π₯ =5
10 π₯π₯ =
1
2
(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₯
16
23 = 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) logππππππ = logππππ + 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) logππππππ = logππππβ logππ ππ
or log
ππππ = logππβ log ππ
Example:
log
3
5= log 3 β log 5
(ii) π₯π₯π₯π₯π₯π₯ππ = ππ.ππππππππ log 6 β log 3 = log
6
3= log 2
Solution:
log π₯π₯ = 1.8716 (iii) logππππππ = ππ logππππ
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
logππππ logππ ππ = logππ ππ
log2 3 log3 5 = log3 5
logππ ππ =
logππ ππlogππππ
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 ππlogππππ
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₯ππππ Proof:
Proof: πΏπΏπππ£π£ logππππ = 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
logππππππ = logππ πππππ₯π₯
logππππππ = πππ₯π₯ logππ ππ
logππππππ = πππ₯π₯(1) β΄ logππ ππ = 1
logππππππ = πππ₯π₯
logππππππ = ππ logππππ
πΏπΏπππ£π£ logππππ = 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
logππππππ = logππ πππ₯π₯+π¦π¦
logππππππ = (π₯π₯ + ππ) logππ ππ
logππππππ = (π₯π₯ + ππ)(1) β΄ logππ ππ = 1
logππππππ = π₯π₯ + ππ
logππππππ = logππππ + logππ ππ
(ii) π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππβ π₯π₯π₯π₯π₯π₯ππππ (iv) π₯π₯π₯π₯π₯π₯πππππ₯π₯π₯π₯π₯π₯ππ ππ = π₯π₯π₯π₯π₯π₯ππ ππ
Proof:
Proof: πΏπΏπππ£π£ logππππ = 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
logππππ logππ ππ = logππ ππ
πΏπΏπππ£π£ logππππ = 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 = logππππβ 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₯ =3
2log10 10 π₯π₯ =
3
2(1) β΄ π₯π₯π₯π₯π₯π₯ππππ = ππ π₯π₯ =
3
2
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₯ =
1
2log7 7 π₯π₯ =
1
2(1) β΄ π₯π₯π₯π₯π₯π₯ππππ = ππ π₯π₯ =
1
2
(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₯ππππ = ππ π₯π₯ = 2
(ii) π₯π₯π₯π₯π₯π₯ππ ππππ
Solution:
log8 1
2
log8 1
2 πΏπΏπππ£π£ log8 1
2= π₯π₯
In exponential form:
8π₯π₯ =1
2
(23)π₯π₯ = 2β1
23π₯π₯ = 2β1
Now
3π₯π₯ = β1
Divide B.S by 3, we get π₯π₯ =β1
3
(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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ 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₯ππππβππππππππ
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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
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₯ππππ 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)
ππππ π₯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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ π¨π¨πππ¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππ+ π₯π₯π₯π₯π₯π₯ππππ
1
2log π₯π₯ β 2 log 3ππ + 3log π§π§ = log
βπ₯π₯π§π§39ππ2
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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
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₯ππππ
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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
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₯ππππ
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₯ππππ π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
logβ3 ππ = logβ3 5 Γ 8
2
logβ3 ππ = logβ3 40
2
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₯ππππ = 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 ππlog7 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₯ππππ
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₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
log6 25
5= 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.ππππππππππππ.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 653.7
0.07213 Γ 8456
πΏπΏπππ£π£ π₯π₯ =0.4932 Γ 653.7
0.07213 Γ 8456
Taking log on B.S
log π₯π₯ = log0.4932 Γ 653.7
0.07213 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.ππβππ.ππππππππππ
Solution:
(78.41)3β142.3β0.15624
πΏπΏπππ£π£ π₯π₯ =(78.41)3β142.3β0.1562
4
Taking log on B.S
log π₯π₯ = log(78.41)3β142.3β0.1562
4
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 +1
2log 142.3β 1
4log 0.1562
log π₯π₯ = 3 log(78.41) +1
2log(142.3) β 1
4log(0.1562)
log π₯π₯ = 3 (1 + .8944) +1
2( 2 + .1532) β 1
4(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) +1
2( 2.1532) β 1
4(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
=1
3log 72 ππππππ β72
3=
1
3(ππππππ2 Γ 2 Γ 2 Γ 3 Γ 3) ππππππ β72
3=
1
3(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
=1
3(ππππππ23 + ππππππ 32) ππππππ β72
3=
1
3(3 ππππππ2 + 2 ππππππ3) ππππππ β72
3=
1
3[3(0.3010) + 2(0.4771)] ππππππ β72
3=
1
3[0.9030 + 0.9542] ππππππ β72
3=
1
3[1.8572] ππππππ β72
3= 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 = log24
10 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
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 οΏ½ 3
10οΏ½4 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = ππ π₯π₯π₯π₯π₯π₯ππππ
log 0.0081 = 4 log3
10 π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
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₯ππππ π¨π¨π¨π¨ π₯π₯π₯π₯π₯π₯ππππππ = π₯π₯π₯π₯π₯π₯ππππβ π₯π₯π₯π₯π₯π₯ππππ
logβ5 3ππ = logβ5 9 Γ 2
3
logβ5 3ππ = logβ5 3 Γ 2
logβ5 3ππ = logβ5 6
Thus 3ππ = 6 ππ =6
2 ππ = 3
Q4: Write in logarithm form: ππβππ =ππππππ
3β3 =1
27
In logarithm form:
log3 1
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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