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CCCCCHHHHHAAAAAPPPPPTTTTTEEEEERRRRR ––––– 11111
RATIO AND
PROPORTION,
INDICES,
LOGARITHMS
Copyright -The Institute of Chartered Accountants of India
RATIO AND PROPORTION, INDICES, LOGARITHMS
LEARNING OBJECTIVES
After reading this unit a student will learn –
(cid:139) How to compute and compare two ratios;
(cid:139) Effect of increase or decrease of a quantity on the ratio;
(cid:139) The concept and application of inverse ratio.
We use ratio in many ways in practical fields. For example, it is given that a certain sum of money
is divided into three parts in the given ratio. If first part is given then we can find out total
amount and the other two parts.
In the case when ratio of boys and girls in a school is given and the total no. of student is also
given, then if we know the no. of boys in the school, we can find out the no. of girls of that school
by using ratios.
1.1 RATIO
A ratio is a comparison of the sizes of two or more quantities of the same kind by division.
If a and b are two quantities of the same kind (in same units), then the fraction a/b is called the
ratio of a to b. It is written as a : b. Thus, the ratio of a to b = a/b or a : b. The quantities a and b are
called the terms of the ratio, a is called the first term or antecedent and b is called the second
term or consequent.
For example, in the ratio 5 : 6, 5 & 6 are called terms of the ratio. 5 is called first term and 6 is called
second term.
1.1.2 REMARKS
(cid:122) Both terms of a ratio can be multiplied or divided by the same (non–zero) number. Usually
a ratio is expressed in lowest terms (or simplest form).
Illustration I:
12 : 16 = 12/16 = (3 × 4)/(4 × 4) = 3/4 = 3 : 4
(cid:122) The order of the terms in a ratio is important.
Illustration II:
3 : 4 is not same as 4 : 3.
(cid:122) Ratio exists only between quantities of the same kind.
Illustration III:
(i) There is no ratio between no. of students in a class and the salary of a teacher.
(ii) There is no ratio between the weight of one child and the age of another child.
(cid:122) Quantities to be compared (by division) must be in the same units.
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Illustration IV:
(i) Ratio between 150 gm and 2 kg = Ratio between 150 gm and 2000 gm
= 150/2000 = 3/40 = 3 : 40
(ii) Ratio between 25 minutes and 45 seconds. = Ratio between (25 × 60) sec and 45 sec.
= 1500/45 = 100/3 = 100 : 3
Illustration V:
(i) Ratio between 3 kg & 5 kg. = 3/5
(cid:122) To compare two ratios, convert them into equivalent like fractions.
Illustration VI: To find which ratio is greater ——
1 1
2 : 3 ; 3.6 : 4.8
3 3
1 1
Solution: 2 : 3 = 7/3 : 10/3 = 7 : 10 = 7/10
3 3
3.6 : 4.8 = 3.6/4.8 = 36/48 = 3/4
L.C.M of 10 and 4 is 20.
So, 7/10 = (7 × 2)/(10 × 2) = 14/20
And 3/4 = (3 × 5)/(4 × 5) = 15/20
As 15 > 14 so, 15/20 > 14/20 i. e. 3/4 > 7/10
Hence, 3.6 : 4.8 is greater ratio.
(cid:122) If a quantity increases or decreases in the ratio a : b then new quantity = b of the original
quantity/a
The fraction by which the original quantity is multiplied to get a new quantity is called the
factor multiplying ratio.
Illustration VII: Rounaq weighs 56.7 kg. If he reduces his weight in the ratio 7 : 6, find his new
weight.
Solution: Original weight of Rounaq = 56.7 kg.
He reduces his weight in the ratio 7 : 6
His new weight = (6 × 56.7)/7 = 6 × 8.1 = 48.6 kg.
Example 1:Simplify the ratio 1/3 : 1/8 : 1/6
Solution: L.C.M. of 3, 8 and 6 is 24.
1/3 : 1/8 : 1/6= 1 × 24/3 : 1 × 24/8 : 1 × 24/6
= 8 : 3 : 4
Example 2: The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18
new girls are admitted in the school, find how many new boys may be admitted so that the ratio
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RATIO AND PROPORTION, INDICES, LOGARITHMS
of the no. of boys to the no. of girls may change to 2 : 3.
Solution: The ratio of the no. of boys to the no. of girls = 3 : 5
Sum of the ratios = 3 + 5 = 8
So, the no. of boys in the school = (3 × 720)/8 = 270
And the no. of girls in the school = (5 × 720)/8 = 450
Let the no. of new boys admitted be x, then the no. of boys become (270 + x).
After admitting 18 new girls, the no. of girls become 450 + 18 = 468
According to given description of the problem, (270 + x)/468 = 2/3
or, 3 (270 + x) = 2 x 468
or, 810 + 3x = 936 or, 3x = 126 or, x = 42.
Hence the no. of new boys admitted = 42.
1.1.3 INVERSE RATIO
One ratio is the inverse of another if their product is 1. Thus a : b is the inverse of b : a and vice–
versa.
1. A ratio a : b is said to be of greater inequality if a>b and of less inequality if a<b.
2. The ratio compound of the two ratios a : b and c : d is ac : bd.
For example compound ratio of 3 : 4 and 5 : 7 is 15 : 28.
Compound ratio of 2 : 3, 5 : 7 and 4 : 9 is 40 : 189.
3. A ratio compounded of itself is called its duplicate ratio.
Thus a2 : b2 is the duplicate ratio of a : b. Similarly, the triplicate ratio of a : b is a3 : b3.
For example, duplicate ratio of 2 : 3 is 4 : 9. Triplicate ratio of 2 : 3 is 8 : 27.
4. The sub–duplicate ratio of a : b is √a : √b and the sub triplicate ratio of a : b is 3 a : 3 b .
For example sub duplicate ratio of 4 : 9 is √4 : √9 = 2 : 3
And sub triplicate ratio of 8 : 27 is 3 8 : 3 27 = 2 : 3.
5. If the ratio of two similar quantities can be expressed as a ratio of two integers, the quantities
are said to be commensurable; otherwise, they are said to be incommensurable. √3 : √2 cannot
be expressed as the ratio of two integers and therefore, √3 and √2 are incommensurable
quantities.
6. Continued Ratio is the relation (or compassion) between the magnitudes of three or more
quantities of the same kind. The continued ratio of three similar quantities a, b, c is written
as a: b: c.
Illustration I: The continued ratio of Rs. 200, Rs. 400 and Rs. 600 is Rs. 200 : Rs. 400 :
Rs. 600 = 1 : 2 : 3.
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Example 1: The monthly incomes of two persons are in the ratio 4 : 5 and their monthly
expenditures are in the ratio 7 : 9. If each saves Rs. 50 per month, find their monthly incomes.
Solution: Let the monthly incomes of two persons be Rs. 4x and Rs. 5x so that the ratio is
Rs. 4x : Rs. 5x = 4 : 5. If each saves Rs. 50 per month, then the expenditures of two persons are Rs.
(4x – 50) and Rs. (5x – 50).
4x−50 7
= , or, 36x−450=35x−350
5x−50 9
or, 36x – 35x = 450 – 350, or, x = 100
Hence, the monthly incomes of the two persons are Rs. 4 × 100 and Rs. 5 × 100
i.e. Rs. 400 and Rs. 500.
Example 2 : The ratio of the prices of two houses was 16 : 23. Two years later when the price
of the first has increased by 10% and that of the second by Rs. 477, the ratio of the prices
becomes 11 : 20. Find the original prices of the two houses.
Solution: Let the original prices of two houses be Rs. 16x and Rs. 23x respectively. Then
by the given conditions,
16x+10%of16x 11
=
23x+477 20
16x+1.6x 11
=
or, , or, 320x + 32x = 253x + 5247
23x+477 20
or, 352x – 253x = 5247, or, 99x = 5247; ∴ x = 53
Hence, the original prices of two houses are Rs. 16 × 53 and Rs. 23 × 53 i.e. Rs. 848 and
Rs. 1,219.
Example 3 : Find in what ratio will the total wages of the workers of a factory be increased or
decreased if there be a reduction in the number of workers in the ratio 15 : 11 and an increment in
their wages in the ratio 22 : 25.
Solution: Let x be the original number of workers and Rs. y the (average) wages per workers.
Then the total wages before changes = Rs. xy.
After reduction, the number of workers = (11 x)/15
After increment, the (average) wages per workers = Rs. (25 y)/22
11 25 5xy
∴ The total wages after changes = ( x) × (Rs. y) = Rs.
15 22 6
Thus, the total wages of workers get decreased from Rs. xy to Rs. 5xy/6
5xy
Hence, the required ratio in which the total wages decrease is xy: =6:5.
6
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RATIO AND PROPORTION, INDICES, LOGARITHMS
Exercise 1(A)
Choose the most appropriate option (a) (b) (c) or (d)
1. The inverse ratio of 11 : 15 is
(a) 15 : 11 (b) √11 : √15 (c) 121 : 225 (d) none of these
2. The ratio of two quantities is 3 : 4. If the antecedent is 15, the consequent is
(a) 16 (b) 60 (c) 22 (d) 20
3. The ratio of the quantities is 5 : 7. If the consequent of its inverse ratio is 5, the antecedent is
(a) 5 (b) √5 (c) 7 (d) none of these
4. The ratio compounded of 2 : 3, 9 : 4, 5 : 6 and 8 : 10 is
(a) 1 : 1 (b) 1 : 5 (c) 3 : 8 (d) none of these
5. The duplicate ratio of 3 : 4 is
(a) √3 : 2 (b) 4 : 3 (c) 9 : 16 (d) none of these
6. The sub duplicate ratio of 25 : 36 is
(a) 6 : 5 (b) 36 : 25 (c) 50 : 72 (d) 5 : 6
7. The triplicate ratio of 2 : 3 is
(a) 8 : 27 (b) 6 : 9 (c) 3 : 2 (d) none of these
8. The sub triplicate ratio of 8 : 27 is
(a) 27 : 8 (b) 24 : 81 (c) 2 : 3 (d) none of these
9. The ratio compounded of 4 : 9 and the duplicate ratio of 3 : 4 is
(a) 1 : 4 (b) 1 : 3 (c) 3 : 1 (d) none of these
10. The ratio compounded of 4 : 9, the duplicate ratio of 3 : 4, the triplicate ratio of 2 : 3 and 9 : 7 is
(a) 2 : 7 (b) 7 : 2 (c) 2 : 27 (d) none of these
11. The ratio compounded of duplicate ratio of 4 : 5, triplicate ratio of 1 : 3, sub duplicate ratio
of 81 : 256 and sub triplicate ratio of 125 : 512 is
(a) 4 : 512 (b) 3 : 32 (c) 1 : 12 (d) none of these
12. If a : b = 3 : 4, the value of (2a+3b) : (3a+4b) is
(a) 54 : 25 (b) 8 : 25 (c) 17 : 24 (d) 18 : 25
13. Two numbers are in the ratio 2 : 3. If 4 be subtracted from each, they are in the ratio 3 : 5.
The numbers are
(a) (16,24) (b) (4,6) (c) (2,3) (d) none of these
14. The angles of a triangle are in ratio 2 : 7 : 11. The angles are
(a) (20°, 70°, 90°) (b) (30°, 70°, 80°) (c) (18°, 63°, 99°) (d) none of these
15. Division of Rs. 324 between X and Y is in the ratio 11 : 7. X & Y would get Rupees
(a) (204, 120) (b) (200, 124) (c) (180, 144) (d) none of these
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16. Anand earns Rs. 80 in 7 hours and Promode Rs. 90 in 12 hours. The ratio of their earnings is
(a) 32 : 21 (b) 23 : 12 (c) 8 : 9 (d) none of these
17. The ratio of two numbers is 7 : 10 and their difference is 105. The numbers are
(a) (200, 305) (b) (185, 290) (c) (245, 350) (d) none of these
18. P, Q and R are three cities. The ratio of average temperature between P and Q is 11 : 12 and that
between P and R is 9 : 8. The ratio between the average temperature of Q and R is
(a) 22 : 27 (b) 27 : 22 (c) 32 : 33 (d) none of these
19. If x : y = 3 : 4, the value of x2y + xy2 : x3 + y3 is
(a) 13 : 12 (b) 12 : 13 (c) 21 : 31 (d) none of these
20. If p : q is the sub duplicate ratio of p–x2 : q–x2 then x2 is
p q pq
(a) (b) (c) (d) noneofthese
p+q p+q p-q
21. If 2s : 3t is the duplicate ratio of 2s – p : 3t – p then
(a) p2 = 6st (b) p = 6st (c) 2p = 3st (d) none of these
22. If p : q = 2 : 3 and x : y = 4 : 5, then the value of 5px + 3qy : 10px + 4qy is
(a) 71 : 82 (b) 27 : 28 (c) 17 : 28 (d) none of these
23. The number which when subtracted from each of the terms of the ratio 19 : 31 reducing it to
1 : 4 is
(a) 15 (b) 5 (c) 1 (d) none of these
24. Daily earnings of two persons are in the ratio 4:5 and their daily expenses are in the ratio
7 : 9. If each saves Rs. 50 per day, their daily incomes in Rs. are
(a) (40, 50) (b) (50, 40) (c) (400, 500) (d) none of these
25. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 Kms. in 5
hours, the speed of the first train is
(a) 10 Km/hr (b) 50 Km/hr (c) 71 Km/hr (d) none of these
1.2 PROPORTION
LEARNING OBJECTIVES
After reading this unit, a student will learn –
(cid:139) What is proportion?
(cid:139) Properties of proportion and how to use them.
If the income of a man is increased in the given ratio and if the increase in his income is given
then to find out his new income, Proportion problem is used.
Again if the ages of two men are in the given ratio and if the age of one man is given, we can find
out the age of another man by Proportion.
An equality of two ratios is called a proportion. Four quantities a, b, c, d are said to be in proportion
if a : b = c : d (also written as a : b :: c : d) i.e. if a/b = c/d i.e. if ad = bc.
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The quantities a, b, c, d are called terms of the proportion; a, b, c and d are called its first,
second, third and fourth terms respectively. First and fourth terms are called extremes (or
extreme terms). Second and third terms are called means (or middle terms).
If a : b = c : d then d is called fourth proportional.
If a : b = c : d are in proportion then a/b = c/d i.e. ad = bc
i.e. product of extremes = product of means.
This is called cross product rule.
Three quantities a, b, c of the same kind (in same units) are said to be in continuous proportion
if a : b = b : c i.e. a/b = b/c i.e. b2 = ac
If a, b, c are in continuous proportion, then the middle term b is called the mean proportional
between a and c, a is the first proportional and c is the third proportional.
Thus, if b is mean proportional between a and c, then b2 = ac i.e. b = ac .
When three or more numbers are so related that the ratio of the first to the second, the ratio of the
second to the third, third to the fourth etc. are all equal, the numbers are said to be in continued
proportion. We write it as
x/y = y/z = z/w = w/p = p/q = ................................................. when
x, y, z, w, p and q are in continued proportion. If a ratio is equal to the reciprocal of the other, then
either of them is in inverse (or reciprocal) proportion of the other. For example 5/4 is in inverse
proportion of 4/5 and vice–versa.
Note: In a ratio a : b, both quantities must be of the same kind while in a proportion
a : b = c : d, all the four quantities need not be of the same type. The first two quantities should be
of the same kind and last two quantities should be of the same kind.
Illustration I:
Rs. 6 : Rs. 8 = 12 toffees : 16 toffees are in a proportion.
Here 1st two quantities are of same kind and last two are of same kind.
Example 1: The nos. 2.4, 3.2, 1.5, 2 are in proportion because these nos. satisfy the property
the product of extremes = product of means.
Here 2.4 × 2 = 4.8 and 3.2 × 1.5 = 4.8
Example 2: Find the value of x if 10/3 : x : : 5/2 : 5/4
Solution: 10/3 : x = 5/2 : 5/4
Using cross product rule, x × 5/2 = (10/3) × 5/4
Or, x = (10/3) × (5/4) × (2/5) = 5/3
Example 3:Find the fourth proportional to 2/3, 3/7, 4
Solution: Let the fourth proportional be x then 2/3, 3/7, 4, x are in proportion.
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Using cross product rule, (2/3) × x = (3 × 4)/7
or, x = (3 × 4 × 3)/(7 × 2) = 18/7.
Example 4: Find the third proportion to 2.4 kg, 9.6 kg
Solution: Let the third proportion to 2.4 kg, 9.6 kg be x kg.
Then 2.4 kg, 9.6 kg and x kg are in continued proportion since b2 = ac
So, 2.4/9.6 = 9.6/x or, x = (9.6 × 9.6)/2.4 = 38.4
Hence the third proportional is 38.4 kg.
Example 5: Find the mean proportion between 1.25 and 1.8
Solution: Mean proportion between 1.25 and 1.8 is (1.25×1.8) = 2.25 = 1.5.
1.2.1 PROPERTIES OF PROPORTION
1. If a : b = c : d, then ad = bc
a c
Proof. = ; ∴ad=bc(Bycross-multiplication)
b d
2. If a : b = c : d, then b : a = d : c (Invertendo)
a c a c b d
Proof. = or 1 =1 , or, =
b d b d a c
Hence, b : a = d : c.
3. If a : b = c : d, then a : c = b : d (Alternendo)
a c
Proof. = or, ad=bc
b d
Dividing both sides by cd, we get
ad bc a b
= , or = , i.e. a:c = b:d.
cd cd c d
4. If a : b = c : d, then a + b : b = c + d : d (Componendo)
a c a c
Proof. = , or, +1= +1
b d b d
a+b c+d
or, = , i.e. a+b:b=c+d:d.
b d
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RATIO AND PROPORTION, INDICES, LOGARITHMS
5. If a : b = c : d, then a – b : b = c – d : d (Dividendo)
a c a c
Proof. = , ∴ −1= −1
b d b d
a−b c−d
= , i.e. a−b:b=c−d:d.
b d
6. If a : b = c : d, then a + b : a – b = c + d : c – d (Componendo and Dividendo)
a c a c a+b c+d
Proof. = , or +1= +1, or = ......................................1
b d b d b d
a c a−b c−d
Again −1,= −1, or = ...................................................2
b d b d
Dividing (1) by (2) we get
a+b c+d
= , i.e. a+b:a−b=c+d:c−d
a−b c−d
7. If a : b = c : d = e : f = ………………..….., then each of these ratios (Addendo) is equal
(a + c + e + ……..) : (b + d + f + …….)
a c e
Proof. = ,= = ............(say)k,
b d f
∴a = bk, c = dk, e = fk, .............
a+c+e.....
Now a+c+e.......=k(b+d+f)............ or =k
b+d+f.....
Hence, (a + c + e + ……..) : (b + d + f + …….)
Example 1: If a : b = c : d = 2.5 : 1.5, what are the values of ad : bc and a+c : b+d?
a c 2.5
Solution: we have = ,= ...........................(1)
b d 1.5
ad
From (1) ad = bc, or, =1, i.e. ad : bc = 1:1
bc
a c a+c
Again from (1) = =
b d b+d
a+c 2.5 25 5
∴ = = = , i.e. a + c : b + d = 5 : 3
b+d 1.5 15 3
Hence, the values of ad : bc and a + c : b + d are 1 : 1 and 5 : 3 respectively.
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a b c a +b+c
Example 2: If = = , then prove that = 2
3 4 7 c
a b c a+b+c a+b+c
Solution: We have = = = =
3 4 7 3+4+7 14
a+b+c c a+b+c 14
∴ = or = = 2
14 7 c 7
Example 3: A dealer mixes tea costing Rs. 6.92 per kg. with tea costing Rs. 7.77 per kg. and sells
the mixture at Rs. 8.80 per kg. and earns a profit of 171 % on his sale price. In what proportion
2
does he mix them?
Solution: Let us first find the cost price (C.P.) of the mixture. If S.P. is Rs. 100, profit is
171 ∴ C.P. = Rs. (100 - 171 ) = Rs. 821 = Rs. 165/2
2 2 2
If S.P. is Rs. 8.80, C.P. is (165 × 8.80)/(2 × 100) = Rs. 7.26
∴ C.P. of the mixture per kg = Rs. 7.26
2nd difference = Profit by selling 1 kg. of 2nd kind @ Rs. 7.26
= Rs. 7.77 – Rs. 7.26 = 51 paise
1st difference = Rs. 7.26 – Rs. 6.92 = 34 paise
We have to mix the two kinds in such a ratio that the amount of profit in the first case must
balance the amount of loss in the second case.
Hence, the required ratio = (2nd diff) : (1st diff.) = 51 : 34 = 3 : 2.
1.2.2 LAWS ON PROPORTION AS DERIVED EARLIER
(i) p : q = r : s => q : p = s : r (Invertendo)
(p/q = r/s) => (q/p = s/r)
(ii) a : b = c : d => a : c = b : d (Alternendo)
(a/b = c/d) => (a/c = b/d)
(iii) a : b = c : d => a+b : b = c+d : d (Componendo)
(a/b = c/d) => (a+b)/b = (c+d)/d
(iv) a : b = c : d => a–b : b = c–d : d (Dividendo)
(a/b = c/d) => (a–b)/b = (c–d)/d
(v) a : b = c : d => a+b : a–b = c+d : c–d (Componendo & Dividendo)
(a+b)/(a–b) = (c+d)/(c–d)
(vi) a : b = c : d = a+c : b+d (Addendo)
(a/b = c/d = a+c/b+d)
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RATIO AND PROPORTION, INDICES, LOGARITHMS
(vii) a : b = c : d = a–c : b–d (Subtrahendo)
(a/b = c/d = a–c/b–d)
(viii) If a : b = c : d = e : f = ............ then each of these ratios = (a – c – e – .......) : (b – d – f – .....)
Proof: The reader may try it as an exercise (Subtrahendo) as the proof is similar to that
derival in 7 above
Exercise 1(B)
Choose the most appropriate option (a) (b) (c) or (d)
1. The fourth proportional to 4, 6, 8 is
(a) 12 (b) 32 (c) 48 (d) none of these
2. The third proportional to 12, 18 is
(a) 24 (b) 27 (c) 36 (d) none of these
3. The mean proportional between 25, 81 is
(a) 40 (b) 50 (c) 45 (d) none of these
4. The number which has the same ratio to 26 that 6 has to 13 is
(a) 11 (b) 10 (c) 21 (d) none of these
5. The fourth proportional to 2a, a3, c is
(a) ac/2 (b) ac (c) 2/ac (d) none of these
6. If four numbers 1/2, 1/3, 1/5, 1/x are proportional then x is
(a) 6/5 (b) 5/6 (c) 15/2 (d) none of these
7. The mean proportional between 12x2 and 27y2 is
(a) 18xy (b) 81xy (c) 8xy (d) none of these
(Hint: Let z be the mean proportional and z = (12x2 x 27y2)
8. If A = B/2 = C/5, then A : B : C is
(a) 3 : 5 : 2 (b) 2 : 5 : 3 (c) 1 : 2 : 5 (d) none of these
9. If a/3 = b/4 = c/7, then a+b+c/c is
(a) 1 (b) 3 (c) 2 (d) none of these
10. If p/q = r/s = 2.5/1.5, the value of ps:qr is
(a) 3/5 (b) 1 (c) 5/3 (d) none of these
11. If x : y = z : w = 2.5 : 1.5, the value of (x+z)/(y+w) is
(a) 1 (b) 3/5 (c) 5/3 (d) none of these
12. If (5x–3y)/(5y–3x) = 3/4, the value of x : y is
(a) 2 : 9 (b) 7 : 2 (c) 7 : 9 (d) none of these
13. If A : B = 3 : 2 and B : C = 3 : 5, then A:B:C is
(a) 9 : 6 : 10 (b) 6 : 9 : 10 (c) 10 : 9 : 6 (d) none of these
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14. If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y is
(a) 6/23 (b) 23/6 (c) 3/2 (d) none of these
15. If x : y = 2 : 3, y : z = 4 : 3 then x : y : z is
(a) 2 : 3 : 4 (b) 4 : 3 : 2 (c) 3 : 2 : 4 (d) none of these
16. Division of Rs. 750 into 3 parts in the ratio 4 : 5 : 6 is
(a) (200, 250, 300) (b) (250, 250, 250) (c) (350, 250, 150) (d) none of these
17. The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio
7 : 8 : 9. Their present ages are
(a) (45, 50, 55) (b) (40, 60, 50) (c) (35, 45, 70) (d) none of these
18. The numbers 14, 16, 35, 42 are not in proportion. The fourth term for which they will be in
proportion is
(a) 45 (b) 40 (c) 32 (d) none of these
19. If x/y = z/w, implies y/x = w/z, then the process is called
(a) Dividendo (b) Componendo (c) Alternendo (d) none of these
20. If p/q = r/s = p–r/q–s, the process is called
(a) Subtrahendo (b) Addendo (c) Invertendo (d) none of these
21. If a/b = c/d, implies (a+b)/(a–b) = (c+d)/(c–d), the process is called
(a) Componendo (b) Dividendo (c) Componendo (d) none of these
and Dividendo
22. If u/v = w/p, then (u–v)/(u+v) = (w–p)/(w+p). The process is called
(a) Invertendo (b) Alternendo (c) Addendo (d) none of these
23. 12, 16, *, 20 are in proportion. Then * is
(a) 25 (b) 14 (c) 15 (d) none of these
24. 4, *, 9, 13½ are in proportion. Then * is
(a) 6 (b) 8 (c) 9 (d) none of these
25. The mean proportional between 1.4 gms and 5.6 gms is
(a) 28 gms (b) 2.8 gms (c) 3.2 gms (d) none of these
a b c a+b+c
26. If = = then is
4 5 9 c
(a) 4 (b) 2 (c) 7 (d) none of these.
27. Two numbers are in the ratio 3 : 4; if 6 be added to each terms of the ratio, then the new ratio
will be 4 : 5, then the numbers are
(a) 14, 20 (b) 17, 19 (c) 18 and 24 (d) none of these
a b
28. If = then
4 5
a + 4 b - 5 a + 4 b + 5 a - 4 b + 5
(a) = (b) = (c) = (d) none of these
a - 4 b + 5 a - 4 b - 5 a + 4 b - 5
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a b
29. If a : b = 4 :1then + is
b a
(a) 5/2 (b) 4 (c) 5 (d) none of these
x y z
30. If = = then
b+c−a c+a−b a+b−c
(b – c)x + (c – a)y + (a – b)z is
(a) 1 (b) 0 (c) 5 (d) none of these
1.3 INDICES
LEARNING OBJECTIVES
After reading this unit, a student will learn –
(cid:139) A meaning of indices and their application;
(cid:139) Laws of indices which facilitates their easy applications.
We are aware of certain operations of addition and multiplication and now we take up certain
higher order operations with powers and roots under the respective heads of indices.
We know that the result of a repeated addition can be held by multiplication e.g.
4 + 4 + 4 + 4 + 4 = 5(4) = 20
a + a + a + a + a = 5(a) = 5a
Now, 4 × 4 × 4 × 4 × 4 = 45;
a × a × a × a × a = a5.
It may be noticed that in the first case 4 is multiplied 5 times and in the second case ‘a’ is multiplied
5 times. In all such cases a factor which multiplies is called the “base” and the number of times it is
multiplied is called the “power” or the “index”. Therefore, “4” and “a” are the bases and “5” is the
index for both. Any base raised to the power zero is defined to be 1; i.e. ao = 1. We also define
r a=a 1 r .
If n is a positive integer, and ‘a’ is a real number, i.e. n ∈ N and a ∈ R (where N is the set of
positive integers and R is the set of real numbers), ‘a’ is used to denote the continued product of
n factors each equal to ‘a’ as shown below:
an = a × a × a ………….. to n factors.
Here an is a power of “a“ whose base is “a“ and the index or power is “n“.
For example, in 3 × 3 × 3 × 3 = 34 , 3 is base and 4 is index or power.
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Law 1
am × an = am+n , when m and n are positive integers; by the above definition, am = a × a
………….. to m factors and an = a × a ………….. to n factors.
∴am × an = (a × a ………….. to m factors) (a × a ……….. to n factors)
= a × a ………….. to (m + n) factors
= am+n
Now, we extend this logic to negative integers and fractions. First let us consider this for negative
integer, that is m will be replaced by –n. By the definition of am × an = am+n ,
We get a–n×an = a–n+n = a0 = 1
For example 34 × 3 5 = (3 × 3 × 3 × 3) × (3 × 3 × 3 × 3 × 3) = 3 4 + 5 = 39
Again, 3–5 = 1/35 = 1/(3 × 3 × 3 × 3 × 3) = 1/243
Example 1: Simplify 2x1/23x-1 if x =4
Solution: We have 2x1/23x-1
= 6x1/2x-1=6x1/2-1
= 6x−1/2
6 6 6 6
= = = =3
=
x1/2 41/2 (22)1/2 2
Example 2: Simplify 6ab2c3 × 4b–2c–3d
Solution: 6ab2c3 × 4b–2c–3d
= 24 × a × b2 × b–2 × c3 × c–3 d
= 24 × a × b2+(–2) × c3+(–3) × d
= 24 × a × b2–2 × c3–3 × d
= 24 a b0 × c0 × d
= 24ad
Law 2
am/an = am–n, when m and n are positive integers and m > n.
By definition, am = a × a ………….. to m factors
am a×a.................tomfactors
Therefore, am÷an= =
an a×a...................tonfactors
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= a × a ……… to m–n factors
= am–n
Now we take a numerical and check the validity of this Law
27 2×2.............to7factors
27÷24= =
24 2×2................to4factors
= 2 × 2 × 2 ……….. to (7–4) factors.
= 2 × 2 × 2 ……….. to 3 factors
= 23 = 8
27 2×2×2×2×2×2×2
or 27÷24= =
24 2×2×2×2
= 2 × 2 × 2 = 21+1+1 = 23
= 8
4 x -1
Example 3: Find the value of
x-1/3
4 −1
x
Solution:
x −1/3
= 4x-1 - (-1/3)
= 4x-1 + 1/3
4
= 4x-2/3 or
x2/3
1 2 7
−
2a2×a3×6a 3
ifa=4
Example 4: Simplify - 5 3
9a 3 ×a2
1 2 7
−
2a2×a3×6a 3
ifa=4
Solution: - 5 3
9a 3 ×a2
1 2 7
2.2.3.a2 + 3 − 3 4 a (3 + 4 - 14) / 6
= 5 3 = 3 (-10 + 9) / 6
- + a
3.3a 3 2
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−7 1
4 a- 7/6 4 +
= . = a 6 6
3 a-1/6 3
4 4 1 4 1 1
= a-1 = . = . =
3 3 a 3 4 3
Law 3
(am)n = amn. where m and n are positive integers
By definition (am)n = am × am × am ……….. to n factors
= (a × a ……….. to m factors)………. to n times
= a × a …………. to mn factors
= amn
Following above, (am)n = (am)p/q
(We will keep m as it is and replace n by p/q, where p and q are positive integers)
Now the qth power of (am)p/q is
{ (am)p/q}q
= (am)(p/q)x q
= amp
If we take the qth root of the above we obtain
( amp)1/q = q amp
Now with the help of a numerical let us verify this law.
(24)3 = 24 × 24 × 24
= 24+4+4
= 212 = 4096
Law 4
(ab)n = an.bn when n can take all of the values.
For example 63 = (2 × 3) 3 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33
First, we look at n when it is a positive integer. Then by the definition, we have
(ab)n = ab × ab ……………. to n factors
= (a × a ……..……. to n factors) (b × b …………. n factors)
= an × bn
When n is a positive fraction, we will replace n by p/q.
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Then we will have (ab)n = (ab)p/q
The qth power of (ab)p/q = {(ab)(p/q)}q = (ab)p
Example 5: Simplify (xa.y–b)3 . (x3 y2)–a
Solution: (xa.y–b)3 . (x3 y2)–a
= (xa)3 . (y–b)3 . (x3)–a . (y2)–a
= x3a–3a . y–3b–2a.
= x0. y–3b–2a.
1
=
y3b+2a
Example 6: 6 a4b x6 .(a2/3x-1)-b
Solution: 6 a4bx6 .(a2/3x-1)-b
1 2
= (a4bx6)6.(a3)-b.(x-1)-b
1 1 2
= (a4b)6.(x6)6.a −
3
b.x-1×-b
2 2b
= a3 b .x.a − 3 .xb
2 2
= a3 b− 3 b .x1+b
= a0 . x1+b = x1+b
Example 7: Find x, if x x= (x x)x
Solution: x (x) 1/2 = xx.xx/2
or, x1+ 1/2 = xx + x/2
or, x3/2 = x3x/2
[If base is equal, then power is also equal]
3 3x 3 2
i.e. = or, x = × = 1
2 2 2 3
∴ X = 1
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Example 8: Find the value of k from (√9)–7 × (√3)–5 = 3k
Solution: (√9)–7 × (√3)–5 = 3k
or, (32 × 1/2) –7 × (3½) –5 = 3k
or, 3−7−5/2 = 3k
or, 3 –19/2 = 3k or, k = –19/2
1.3.1 LAWS OF INDICES
(i) am × an = am+n (base must be same)
Ex. 23 × 22 = 23+2 = 25
(ii) am ÷ an = am–n
Ex. 25 ÷ 23 = 25–3 = 22
(iii) (am)n = amn
Ex. (25)2 = 25×2 = 210
(iv) ao = 1
Example : 20 = 1, 30 = 1
(v) a–m = 1/am and 1/a–m = am
Example: 2–3 = 1/23 and 1/2–5 = 25
(vi) If ax = ay, then x=y
(vii)If xa = ya, then x=y
(viii) ma = a1/m , √x = x½ , √4 = (22)1/2 = 21/2 × 2= 2
Example: 3 8 = 81/3 = (23)1/3 = 23×1/3= 2
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Exercise 1(C)
Choose the most appropriate option (a) (b) (c) or (d)
1. 4x–1/4 is expressed as
(a) –4x1/4 (b) x–1 (c) 4/x1/4 (d) none of these
2. The value of 81/3 is
(a) 3√2 (b) 4 (c) 2 (d) none of these
3. The value of 2 × (32) 1/5 is
(a) 2 (b) 10 (c) 4 (d) none of these
4. The value of 4/(32)1/5 is
(a) 8 (b) 2 (c) 4 (d) none of these
5. The value of (8/27)1/3 is
(a) 2/3 (b) 3/2 (c) 2/9 (d) none of these
6. The value of 2(256)–1/8 is
(a) 1 (b) 2 (c) 1/2 (d) none of these
7. 2½ .4¾ is equal to
(a) a fraction (b) a positive integer (c) a negative integer (d) none of these
1
81x44
8. has simplified value equal to
y-8
(a) xy2 (b) x2y (c) 9xy2 (d) none of these
9. xa–b × xb–c × xc–a is equal to
(a) x (b) 1 (c) 0 (d) none of these
0
2p2q3
10. The value of is equal to
3xy
(a) 0 (b) 2/3 (c) 1 (d) none of these
11. {(33)2 × (42)3 × (53)2} / {(32)3 × (43)2 × (52)3} is
(a) 3/4 (b) 4/5 (c) 4/7 (d) 1
12. Which is True ?
(a) 20 > (1/2)0 (b) 20 < (1/2)0 (c) 20 = (1/2)0 (d) none of these
13. If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r is
(a) 1 (b) 0 (c) 1/2 (d) none of these
14. The value of ya–b × yb–c × yc–a × y–a–b is
(a) ya+b (b) y (c) 1 (d) 1/ya+b
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15. The True option is
(a) x2/3 = 3√x2 (b) x2/3 = √x3 (c) x2/3 > 3√x2 (d) x2/3 < 3√x2
16. The simplified value of 16x–3y2 × 8–1x3y–2 is
(a) 2xy (b) xy/2 (c) 2 (d) none of these
17. The value of (8/27)–1/3 × (32/243)–1/5 is
(a) 9/4 (b) 4/9 (c) 2/3 (d) none of these
18. The value of {(x+y)2/3 (x–y)3/2/√x+y × √ (x–y)3}6 is
(a) (x+y)2 (b) (x–y) (c) x+y (d) none of these
19. Simplified value of (125)2/3 × √25 × 3√53 × 51/2 is
(a) 5 (b) 1/5 (c) 1 (d) none of these
20. [{(2)1/2 . (4)3/4 . (8)5/6 . (16)7/8 . (32)9/10}4]3/25 is
(a) A fraction (b) an integer (c) 1 (d) none of these
21. [1–{1–(1–x2)–1}–1]–1/2 is equal to
(a) x (b) 1/x (c) 1 (d) none of these
22. {(xn)n–1/n}1/n+1 is equal to
(a) xn (b) xn+1 (c) xn–1 (d) none of these
23. If a3–b3 = (a–b) (a2 + ab + b2), then the simplified form of
x l l2+lm+m2 xm m2+mn+n2 xn l2+ln+n2
× ×
xm
xn
xl
(a) 0 (b) 1 (c) x (d) none of these
24. Using (a–b)3 = a3–b3–3ab(a–b) tick the correct of these when x = p1/3 – p–1/3
(a) x3+3x = p + 1/p (b) x3 + 3x = p – 1/p (c) x3 + 3x = p + 1 (d) none of these
25. On simplification, 1/(1+am–n+am–p) + 1/(1+an–m+an–p) + 1/(1+ap–m+ap–n) is equal to
(a) 0 (b) a (c) 1 (d) 1/a
a+b b+c c+a
xa xb xc
26. The value of × ×
xb
xc
xa
(a) 1 (b) 0 (c) 2 (d) none of these
1 1
-
27. If x=33+3 3 ,then3x3-9xis
(a) 15 (b) 10 (c) 12 (d) none of these
28. If ax = b, by = c, cz = a, then xyz is
(a) 1 (b) 2 (c) 3 (d) none of these
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(a2+ab+b2) (b2+bc+c2) (c2+ca+a2)
xa xb xc
29. The value of × ×
xb
xc
xa
(a) 1 (b) 0 (c) –1 (d) none of these
1 1 1
+ + is
30. If 2x = 3y = 6z,
x y z
(a) 1 (b) 0 (c) 2 (d) none of these
1.4 LOGARITHM
LEARNING OBJECTIVE
(cid:139) After reading this unit, a student will get fundamental knowledge of logarithm and its
application for solving business problems.
The logarithm of a number to a given base is the index or the power to which the base must be
raised to produce the number, i.e. to make it equal to the given number. If there are three quantities
indicated by say a, x and n, they are related as follows:
If ax = n
Then x is said to be the logarithm of the number n to the base ‘a’ symbolically it can be expressed
as follows:
log n = x
a
i.e. the logarithm of n to the base ‘a’ is x, we give some illustrations below:
(i) 24 = 16 ⇒ log 16 = 4
2
i.e. the logarithm of 16 to the base 2 is equal to 4
(ii) 103 = 1000 ⇒ log 1000 = 3
10
i.e. the logarithm of 1000 to the base 10 is 3
1 1
(iii) 5-3 = ⇒ log = - 3
5
125 125
1
i.e. the logarithm of to the base 5 is –3
125
(iv) 23 = 8 ⇒ log 8 = 3
2
i.e. the logarithm of 8 to the base 2 is 3
1. Two equations ax = n and x = logn are only transformations of each other and should be
a
remembered to change one form of the relation into the other.
2. The logarithm of 1 to any base is zero. This is because any number raised to the power zero
is one.
Since a0 = 1 , log 1 = 0
a
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3. The logarithm of any quantity to the same base is unity. This is because any quantity
raised to the power 1 is that quantity only.
Since a1 = a , log a = 1
a
Illustrations:
1
1. If log 2 = find the value of a.
a ,
6
We have a1/6 = 2 ⇒ a = ( 2 )6 = 23 = 8
2. Find the logarithm of 5832 to the base 3√ 2.
Let us take log 5832 = x
3 2
We may write, (3 2 )x=5832=8X729=23X36=( 2 )6(3)6=(3 2 )6
Hence, x = 6
Logarithms of numbers to the base 10 are known as common logarithm.
1.4.1 FUNDAMENTAL LAWS OF LOGARITHM
1. Logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers
to the same base, i.e.
Log mn = log m + log n
a a a
Proof:
Let log m = x so that ax = m – (I)
a
Log n = y so that ay = n – (II)
a
Multiplying (I) and (II), we get
m × n = ax × ay = ax+y
log mn = x + y (by definition)
a
∴ log mn = log m + log n
a a a
2. The logarithm of the quotient of two numbers is equal to the difference of their logarithms to
the same base, i.e.
m
log = log m – log n
a n a a
Proof:
Let log m = x so that ax = m ————(I)
a
log n = y so that ay = n ———————(II)
a
Dividing (I) by (II) we get
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m ax
= =ax-y
n ay
Then by the definition of logarithm, we get
m
log = x – y = log m – log n
a n a a
1
Similarly, log = log 1-log n = 0-log n= – log n [∵ log 1 = 0]
a n a a a a a
Illustration I: log ½ = log 1 – log 2 = –log 2
3. Logarithm of the number raised to the power is equal to the index of the power multiplied
by the logarithm of the number to the same base i.e.
log mn = n log m
a a
Proof:
Let log m = x so that ax = m
a
Raising the power n on both sides we get
(ax)n = (m)n
axn = mn (by definition)
log mn = nx
a
i.e. log mn = n log m
a a
Illustrations II: 1(a) Find the logarithm of 1728 to the base 2√3
Solution: We have 1728 = 26 × 33 = 26 × (√3)6 = (2√3)6; and so, we may write
log 1728 = 6
2√3
1
1(b) Solve log 25-2log 3+log 18
10 10 10
2
Solution: The given expression
1
2
=log 252 -log 3 +log 18
10 10
10
=log 5-log 9+log 18
10 10 10
5x18
=log =log 10=1
10 10
9
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1.4.2 CHANGE OF BASE
If the logarithm of a number to any base is given, then the logarithm of the same number to
any other base can be determined from the following relation
log m
log m= a
log a m=log b m×log a b ⇒ b log b
a
Proof:
Let log m = x, log m = y and log b = z
a b a
Then by definition,
ax = m, by = m and az = b
Also ax = by = (az)y = ayz
Therefore, x = yz
⇒ log m = log m × log b
a b a
log m
log m= a
b
log b
a
Putting m = a, we have
log a = log a × log b
a b a
⇒log a × log b = 1, since log a = 1.
b a a
Example 1: Change the base of log 31 into the common logarithmic base.
5
log x
Since log x= b
Solution: a
log a
b
log 31
∴log 31= 10
5 log 5
10
log 8
Provethat 3 =3log 2
Example 2:
log 16 log 10
10
9 4
Solution: Change all the logarithms on L.H.S. to the base 10 by using the formula.
log x
log x= a
b
log b
, We may write
a
log 8 log 23 3log 2
log 8 = 10 = 10 = 10
3 log 3 log 3 log 3
10 10 10
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log 16 log 24 4log 2
log 16 = 10 = 10 = 10
9 log 9 log 32 2log 3
10 10 10
log 10 1 1
log 10 = 10 = = [log 10=1]
4 log 4 log 22 2log 2 10
10 10 10
3 log 2 2 log 3 2 log 2
∴L.H.S.= 10 × 10 × 10 ∴[log 10=1]
log 3 4log 2 1
10
10 10
= 3 log 2 = R.H.S.
10
Logarithm Tables:
The logarithm of a number consists of two parts, the whole part or the integral part is called
the characteristic and the decimal part is called the mantissa where the former can be known
by mere inspection, the latter has to be obtained from the logarithm tables.
Characteristic:
The characteristic of the logarithm of any number greater than 1 is positive and is one less than
the number of digits to the left of the decimal point in the given number. The characteristic of the
logarithm of any number less than one (1) is negative and numerically one more than the number
of zeros to the right of the decimal point. If there is no zero then obviously it will
be –1. The following table will illustrate it.
Number Characteristic
3 7 1 One less than the
4 6 2 3 3 number of digits to
6.21 0 the left of the decimal point
Number Characteristic
.8 –1 One more than the
.07 –2 number of zeros on
.00507 –3 the right immediately
.000670 –4 after the decimal point.
Zero on positive characteristic when the number under consideration is greater than unity:
Since 100 = 1 , log 1 = 0
101 = 10 , log 10 = 1
102 = 100 , log 100 = 2
103 = 1000 , log 1000 = 3
All numbers lying between 1 and 10 i.e. numbers with 1 digit in the integral part have their
logarithms lying between 0 and 1. Therefore, their integral parts are zero only.
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All numbers lying between 10 and 100 have two digits in their integral parts. Their logarithms lie
between 1 and 2. Therefore, numbers with two digits have integral parts with 1 as characteristic.
In general, the logarithm of a number containing n digits only in its integral parts is (n – 1) + a
fraction. For example, the characteristics of log 75, log 79326, log 1.76 are 1, 4 and 0 respectively.
Negative characteristics
1
Since10−1= =0.1→log0.1=−1
10
1
10−2= =0.01→log0.01=−2
100
All numbers lying between 1 and 0.1 have logarithms lying between 0 and –1, i.e. greater than –
1 and less than 0. Since the decimal part is always written positive, the characteristic is –1.
All numbers lying between 0.1 and 0.01 have their logarithms lying between –1 and –2 as
characteristic of their logarithms.
In general, the logarithm of a number having n zeros just after the decimal point is –
(n + 1) + a fraction.
Hence, we deduce that the characteristic of the logarithm of a number less than unity is one more
than the number of zeros just after the decimal point and is negative.
Mantissa:
The mantissa is the fractional part of the logarithm of a given number
Number Mantissa Logarithm
Log 4597 = (……… 6625) = 3.6625
Log 459.7 = (……… 6625) = 2.6625
Log 45.94 = (……… 6625) = 1.6625
Log 4.594 = (……… 6625) = 0.6625
Log .4594 = (……… 6625) = 1.6625
Thus with the same figures there will be difference in the characteristic only. It should be
remembered, that the mantissa is always a positive quantity. The other way to indicate this is
Log .004594 = – 3 + .6625 = – 3.6625.
Negative mantissa must be converted into a positive mantissa before reference to a logarithm
table. For example
– 3.6872 = – 4 + (4–3.6872) = 4 + 0.3128 = 4.3128
It may be noted that 4.3128 is different from – 4.3128 as – 4.3128 is a negative number whereas,
in 4.3128, 4 is negative while .3128 is positive.
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RATIO AND PROPORTION, INDICES, LOGARITHMS
Illustration I: Add 4.74628 and 3.42367
– 4 + .74628
3 + .42367
–1 + 1.16995 – 0.16995
Antilogarithms:
If x is the logarithm of a given number n with a given base then n is called the antilogarithm
(antilog) of x to that base.
This can be expressed as follows:-
If log n = x then n = antilog x
a
For example, if log 61720 = 4.7904 then 61720 = antilog 4.7904
Number Mantissa Logarithm
206 2.3139 206.0
20.6 1.3139 20.60
2.06 0.3139 2.060
.206 –1.3139 .2060
.0206 –2.3139 .02060
Example 1: Find the value of log 5 if log 2 is equal to .3010
10
Solution : log5=log =log10-log2
2
= 1 – .3010
= .6990
Example 2: Find the number whose logarithm is 2.4678.
Solution: From the antilog table, for mantissa .467, the number = 2931
for mean difference 8, the number = 5
∴ for mantissa .4678, the number = 2936
The characteristic is 2, therefore, the number must have 3 digits in the integral part.
Hence, Antilog 2.4678 = 293.6
Example 3: Find the number whose logarithm is –2.4678.
Solution: -2.4678 = -3 + 3 − 2.4678 = −3 + .5322 = 3.5322
For mantissa .532, the number = 3404
For mean difference 2, the number = 2
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∴ for mantissa .5322, the number = 3406
The characteristic is –3, therefore, the number is less than one and there must be two zeros
just after the decimal point.
Thus, Antilog (–2.4678) = 0.003406
Properties of Logarithm
(I) logmn = logm + logn
a a a
Ex. log (2 × 3) = log 2 + log 3
(II) log(m/n) = logm – logn
a a a
Ex. log (3/2) = log3 – log2
(III) logmn = n logm
a a
Ex. log 23 = 3 log 2
(IV) loga = 1
a
Ex. log 10 = 1, log 2 = 1, log 3 = 1 etc.
10 2 3
(V) log1 = 0
a
Ex. log 1 = 0, log 1 = 0 etc.
2 10
(VI) loga × logb = 1
b a
Ex. log 2 × log 3 = 1
3 2
(VII)log a × logb = loga
b c c
Ex. log 2 × log 3 = log 2
3 5 5
(VIII)log a = log a/log b
b
Ex. log 2 = log2/log3
3
Note:
(A) If base is understood, base is taken as 10
(B) Thus log 10 = 1, log 1 = 0
(C) Logarithm using base 10 is called Common logarithm and logarithm using base e is called
Natural logarithm {e = 2.33 (approx.) called exponential number}.
Relation between Indices and Logarithm
Let x = log m and y = logn
a a
∴ ax = m and ay = n
so ax. ay = mn
or ax+y = mn
or x+y = logmn
a
or logm + logn= logmn [∵ log a = 1]
a a a a
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RATIO AND PROPORTION, INDICES, LOGARITHMS
or logmn = logm + logn
a a a
Also, (m/n) = ax/ay
or (m/n) = ax–y
or log (m/n) = (x–y)
a
or log (m/n) = logm – logn [Q log a = 1]
a a a a
Again mn = m.m.m. ———————— to n times
so logmn = log(m.m.m ——————— to n times)
a a
or logmn = logm + logm + logm + —————— + logm
a a a a a
or logmn = n log m
a a
Now a0 = 1 ⇒ 0 = log1
a
let log a = x and logb = y
b a
∴ a = bx and b=ay
∴so a = (ay)x
or axy = a
or xy = 1
or log a × logb = 1
b a
let log c = x & logb = y
b c
∴ c = bx & b = cy
soc = cxy or xy = l
log c × logb = l
b c
Example 1: Find the logarithm of 64 to the base 2 2
Solution: log 64=log 82 =2 log 8 =2log (2√2)2= 4 log 2√2 = 4x1= 4
2√2 2√2 2√2 2√2 2√2
Example 2: If logbc = x, log ca = y, logab = z, prove that
a b c
1 1 1
+ + =1
x+1 y+1 z+1
Solution: x+1 = log bc + log a = log abc
a a a
y+1 = log ca + log b = log abc
b b b
z+1 = logab + logc = logabc
c c c
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1 1 1 1 1 1
+ + = + +
Therefore
x+1 y+1 z+1 log abc log abc log abc
a b c
= log a +log b + log c
abc abc abc
= log abc = 1 (proved)
abc
Example 3: If a=log 12, b=log 24, and c=log 36 then prove that
24 36 48
1+abc = 2bc
Solution: 1+abc = 1+ log 12 × log 24 × log 36
24 36 48
= 1+ log 12 × log 36
36 48
= 1 + log 12
48
= log 48 + log 12
48 48
= log 48×12
48
= log (2×12)2
48
= 2 log 24
48
= 2 log 24 x log 36
36 48
= 2bc
Exercise 1(D)
Choose the most appropriate option. (a) (b) (c) and (d)
1. log 6 + log 5 is expressed as
(a) log 11 (b) log 30 (c) log 5/6 (d) none of these
2. log 8 is equal to
2
(a) 2 (b) 8 (c) 3 (d) none of these
3. log 32/4 is equal to
(a) log 32/log 4 (b) log 32 – log 4 (c) 23 (d) none of these
4. log (1 × 2 × 3) is equal to
(a) log 1 + log 2 + log 3 (b) log 3 (c) log 2 (d) none of these
5. The value of log 0.0001 to the base 0.1 is
(a) –4 (b) 4 (c) ¼ (d) none of these
6. If 2 log x = 4 log 3, the x is equal to
(a) 3 (b) 9 (c) 2 (d) none of these
7. log 64 is equal to
√2
(a) 12 (b) 6 (c) 1 (d) none of these
8. log 1728 is equal to
2√3
(a) 2√3 (b) 2 (c) 6 (d) none of these
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9. log (1/81) to the base 9 is equal to
(a) 2 (b) ½ (c) –2 (d) none of these
10. log 0.0625 to the base 2 is equal to
(a) 4 (b) 5 (c) 1 (d) none of these
11. Given log2 = 0.3010 and log3 = 0.4771 the value of log 6 is
(a) 0.9030 (b) 0.9542 (c) 0.7781 (d) none of these
12. The value of log 2 is
2
(a) 0 (b) 2 (c) 1 (d) none of these
13. The value of log 0.333 to the base 9 is
(a) – ½ (b) ½ (c) 1 (d) none of these
14. If log x + log y = log (x+y), y can be expressed as
(a) x–1 (b) x (c) x/x–1 (d) none of these
15. The value of log [log {log (log 273)}] is equal to
2 2 3 3
(a) 1 (b) 2 (c) 0 (d) none of these
16. If log x + log x + log x = 21/4, these x is equal to
2 4 16
(a) 8 (b) 4 (c) 16 (d) none of these
17. Given that log 2 = x and log 3 = y, the value of log 60 is expressed as
10 10 10
(a) x – y + 1 (b) x + y + 1 (c) x – y – 1 (d) none of these
18. Given that log 2 = x, log 3 = y, then log 1.2 is expressed in terms of x and y as
10 10 10
(a) x + 2y – 1 (b) x + y – 1 (c) 2x + y – 1 (d) none of these
19. Given that log x = m + n and log y = m – n, the value of log 10x/y2 is expressed in terms of m
and n as
(a) 1 – m + 3n (b) m – 1 + 3n (c) m + 3n + 1 (d) none of these
20. The simplified value of 2 log 5 + log 8 – ½ log 4 is
10 10 10
(a) ½ (b) 4 (c) 2 (d) none of these
21. log [1 – {1 – (1 – x2)–1}–1]–1/2 can be written as
(a) log x2 (b) log x (c) log 1/x (d) none of these
22. The simplified value of log 6 729 3 9-1.27-4/3 is
(a) log 3 (b) log 2 (c) log ½ (d) none of these
23. The value of (log a × logb × log c)3 is equal to
b c a
(a) 3 (b) 0 (c) 1 (d) none of these
24. The logarithm of 64 to the base 2√2 is
(a) 2 (b) √2 (c) ½ (d) none of these
25. The value of log 25 given log 2 = 0.3010 is
8
(a) 1 (b) 2 (c) 1.5482 (d) none of these
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ANSWERS
Exercise 1(A)
1. a 2. d 3. c 4. a 5. c 6. d 7. a 8. c
9. a 10. c 11. d 12. d 13. a 14. c 15. d 16. a
17. c 18. b 19. b 20. d 21. a 22. c 23. a 24. c
25. d
Exercise 1(B)
1. a 2. b 3. c 4. d 5. d 6. c 7. a 8. c
9. c 10. b 11. c 12. d 13. a 14. d 15. d 16. a
17. a 18. b 19. d 20. a 21. c 22. d 23. c 24. a
25. b 26. b 27. c 28. b 29. a 30. b
Exercise 1(C)
1. c 2. c 3. c 4. b 5. a 6. a 7. b 8. d
9. b 10. c 11. d 12. c 13. b 14. d 15. a 16. c
17. a 18. c 19. d 20. b 21. a 22. d 23. b 24. b
25. c 26. a 27. b 28. a 29. a 30 d
Exercise 1(D)
1. b 2. c 3. b 4. a 5. b 6. b 7. a 8. c
9. c 10. d 11. c 12. c 13. a 14. c 15. c 16. a
17. b 18. c 19. a 20. c 21. b 22. d 23. c 24. d
25. c
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ADDITIONAL QUESTION BANK
7 -5
6 -1 7 2 2 6 -2 7 3 2
1. The value of × is
2 -4 3 -5
6 7 6 7
(A) 0 (B) 252 (C) 250 (D) 248
x2/7 x2/5 x-9/7 z5/6
2. The value of × × × is
z-1/2 z2/3 z2/3 x-3/5
(A) 1 (B) –1 (C) 0 (D) None
2
x+3×
3
2x−y×
5
x+y+3×
6
y+1
3. On simplification reduces to
6
x+1×
10
y+3×
15
x
(A) –1 (B) 0 (C) 1 (D) 10
9 y.3 2. ( 3 −y)−1 −27 y
1
4. If = then x–y is given by
3 3x.2 3 27
(A) –1 (B) 1 (C) 0 (D) None
1 1 1
1 a-c 1 b-a 1 c-b
5. Show that xa-b × xb-c × xc-a is given by
(A) 1 (B) –1 (C) 3 (D) 0
16(32)x −2 3x−2.4 x+1 5(5)x−1
6. Show that − is given by
15(2)x−1(16)x
5
2x
(A) 1 (B) –1 (C) 4 (D) 0
a+b b+c c+a
xa xb xc
× ×
7. Show that is given by
xb
xc
xa
(A) 0 (B) –1 (C) 3 (D) 1
xa2 xb2 xc2
8. Show that
(a+b) ×(b+c) ×(c+a)
reduces to
xb2 xc2 xa2
(A) 1 (B) 0 (C) –1 (D) None
1 1 1
b+c a-b c+a b-c a+bc-a
9. Show that xc-a × xa-b ×xb-c reduces to
(A) 1 (B) 3 (C) –1 (D) None
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a b c
xb xc xa
× ×
10. Show that reduces to
xc
xa
xb
(A) 1 (B) 3 (C) 0 (D) 2
1 1 1
xb bc xc ca xa ab
× ×
11. Show that reduces to
xc
xa
xb
(A) –1 (B) 0 (C) 1 (D) None
(a2+ab+b2) (b2+bc+c2) (c2+ca+a2)
xa xb xc
12. Show that × × is given by
xb
xc
xa
(A) 1 (B) –1 (C) 0 (D) 3
1 1 1
1 a-c 1 b-a 1 c-b
13. Show that xa-b × xb-c × xc-a is given by
(A) 0 (B) 1 (C) –1 (D) None
b+c-a c+a-b a+b-c
xb xc xa
× ×
14. Show that is given by
xc
xa
xb
(A) 1 (B) 0 (C) –1 (D) None
a2-ab+b2 b2-bc+c2 c2-ca+a2
xa xb xc
× ×
15. Show that is reduces to
x-b
x-c
x-a
(A) 1 (B) -2(a2+b2+c2) (C) 2(a3+b3+c3) (D) -2(a3+b3+c3)
x x x
16. xa2b-1c-1 .xb2c-1a-1 .xc2a-1b-1 would reduce to zero if a+b+c is given by
(A) 1 (B) –1 (C) 0 (D) None
( )z
17. The value of z is given by the following if zz z= z z
3 3 9
(A) 2 (B) (C) - (D)
2 2 4
1 1 1
18. + + would reduce to one if a+b+c is given by
xb+x-c+1 xc+x-a+1 xa+x-b+1
(A) 1 (B) 0 (C) –1 (D) None
1 1 1
19. On simplification would reduces to
1+za-b+za-c 1+zb-c+zb-a 1+zc-a+zc-b
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1 1
(A) (B) (C) 1 (D) 0
z2(a+b+c) z(a+b+c)
20. If(5.678)x =(0.5678)y =10z then
1 1 1 1 1 1 1 1 1
- + =1 - - =0 - + =-1
(A) (B) (C) (D) None
x y z x y z x y z
21. If x=4 1 3+4 -1 3 prove that 4x3-12x is given by
(A) 12 (B) 13 (C) 15 (D) 17
22. Ifx=5 1 3+5 -1 3 prove that 5x3-15x is given by
(A) 25 (B) 26 (C) 27 (D) 30
2 1
23. Ifax 3+bx 3+c=0 then the value of a3x2+b3x+c3 is given by
(A) 3abcx (B) –3abcx (C) 3abc (D) –3abc
24. Ifap=b bq=c cr=a the value of pqr is given by
(A) 0 (B) 1 (C) –1 (D) None
25. Ifap=bq=cr andb2=ac the value of q (p+r)/pris given by
(A) 1 (B) –1 (C) 2 (D) None
a+b
a b
xa-b xb-a
÷
26. On simplification a b reduces to
xa+b xb+a
(A) 1 (B) –1 (C) 0 (D) None
xab a+b xb2+c2 b+c xca c+a
27. On simplification xa2+b2
×
xbc
× xc2+a2
reduces to
(A) x-2a3 (B) x2a3 (C) x -2(a3+b3+c3) (D) x 2(a3+b3+c3)
a+b b+c c+a
xab xbc xca
× ×
28. On simplification reduces to
xa2+b2
xb2+c2
xc2+a2
(A) x-2a3 (B) x2a3 (C)
x
-2(a3+b3+c3) (D)
x
2(a3+b3+c3)
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29. On simplification
mx x+y
×
my y+z ÷3(mxmz)x-z
mreduces to
my
mz
1 1
(A) 3 (B) –3 (C) - (D)
3 3
1 1
30. The value of + is given by
1+ay-x 1+ax-y
(A) –1 (B) 0 (C) 1 (D) None
1 1 1
+ + is
31. Ifxyz =1 then the value of 1 + x + y-1 1 + x + z-1 1 + z + x-1
(A) 1 (B) 0 (C) 2 (D) None
1 1 2
32. If2a = 3b =(12)c then - - reduces to
c b a
(A) 1 (B) 0 (C) 2 (D) None
1 1 1
33. If2a=3b=6-c then the value of + + reduce to
a b c
(A) 0 (B) 2 (C) 3 (D) 1
34. If3a=5b=(75)c then the value of ab-c(2a+b) reduces to
(A) 1 (B) 0 (C) 3 (D) 5
35. If2a=3b=(12)cthen the value of ab-c(a+2b) reduces to
(A) 0 (B) 1 (C) 2 (D) 3
1 1 1
36. If2a=4b=8c andabc=288 then the value + + is given by
2a 4b 8c
1 1 11 11
(A) (B) - (C) (D) -
8 8 96 96 1
1 1 1 1
37. Ifap=bq=cr=ds and ab = cd then the value of
p
11 +
q
–
r
– –
s
reduces to
1 1
(A) (B) (C) 0 (D) 1
a b
a
ab a -1
38. Ifab=ba then the value of -ab reduces to
b
(A) a (B) b (C) 0 (D) None
39. If m=bx,n=by and ( mynx) =b2the value of xy is given by
(A) –1 (B) 0 (C) 1 (D) None
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40. Ifa=xym-1 b=xyn-1 c=xyp-1 then the value of an−p×bp−m×cm−nreduces to
(A) 1 (B) –1 (C) 0 (D) None
41. Ifa=xn+pym b=xp+myn c = xm+nyp then the value ofan-p×bp-m×cm-nreduces to
(A) 0 (B) 1 (C) –1 (D) None
42. If a=3 2+1-3 2-1 then the value ofa3+3a-2is
(A) 3 (B) 0 (C) 2 (D) 1
43. If a = x
1
3+x
-1
3 then a3-3ais
(A) x + x-1 (B) x - x-1 (C) 2x (D) 0
44. Ifa = 3 1 4 + 3 -1 4 and b = 3 1 4 - 3-1 4 then the value of3 ( a2+b2)2 is
(A) 67 (B) 65 (C) 64 (D) 62
1 15 1
45. Ifx = 3+ is equal the value of x- × x-
3 5 2 3
x-
3
1 5
(A) 5 (B) 3 (C) (D)
3 6
4 6 a+2 2 a+2 3
46. If a = then the value of + is given by
2+ 3 a-2 2 a-2 3
(A) 1 (B) –1 (C) 2 (D) –2
1
47. If P + 3Q + 5R + 15S = then the value of P is
1+ 3+ 5
(A) 7/11 (B) 3/11 (C) -1/11 (D) -2/11
48. If a = 3 + 2 2 then the value of a 1 2 + a -1 2 is
(A) 2 (B) - 2 (C) 2 2 (D) -2 2
49. If a = 3+2 2 then the value of a 1 2 - a -1 2 is
(A) 2 2 (B) 2 (C) 2 2 (D) -2 2
1( )
50. If a = 5- 21 then the value ofa3 + a-3- 5a2- 5a-2+ a + a-1 is
2
(A) 0 (B) 1 (C) 5 (D) –1
51. If a =
7+4 3
then the value of
[a(a-14)]2
is
7-4 3
(A) 14 (B) 7 (C) 2 (D) 1
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52. If a = 3- 5 then the value of a4 – a3 – 20a2 – 16a + 24 is
(A) 10 (B) 14 (C) 0 (D) 15
3+ 2
53. Ifa = then the value of 2a4- 21a3+ 12a2-a + 1 is
3- 2
(A) 21 (B) 1 (C) 12 (D) None
54. The square root of 3+ 5 is
( )
(A) 5 + 1 (B) – 5 + 1 (C) Both the above (D) None
2 2 2 2
55. Ifx = 2- 2- 2 …∝ the value of X is given by
(A) –2 (B) 1 (C) 2 (D) 0
3+ 2 3- 2
56. If a = b = then the value of a + b is
3- 2 3+ 2
(A) 10 (B) 100 (C) 98 (D) 99
3+ 2 3- 2
57. If a = b = then the value of a2 + b2 is
3- 2 3+ 2
(A) 10 (B) 100 (C) 98 (D) 99
3+ 2 3- 2 1 1
58. If a = b = then the value of + is
3- 2 3+ 2 a2 b2
(A) 10 (B) 100 (C) 98 (D) 99
59. The square root of x + x2-y2 is given by
1 1
(A) 2 x+y + x-y (B) 2 x+y - x-y (C) x+y + x-y (D) x+y - x-y
9 3 + 11 2
60. The cube root of is given by
(A) 3 3 1+ 2 3 (B) 3 3 1- 2 3 (C) 3 1+ 2 3 (D) 3 1+ 2 3
61. log(1+2+3) is exactly equal to
(A) log 1 + log 2 + log 3 (B)log(1×2×3) (C) Both the above (D) None
62. The logarithm of 21952 to the base of 2 7 and 19683 to the base of 3 3 are
(A) Equal (B) Not equal (C) Have a difference of 2269 (D) None
64 50 81
63. The value of is 16 log + 12 log + 7 log + log 2
60 48 80
(A) 0 (B) 1 (C) 2 (D) –1
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RATIO AND PROPORTION, INDICES, LOGARITHMS
64. alogb-logc × blogc-loga × cloga-logb has a value of
(A) 1 (B) 0 (C) –1 (D) None
1 1 1
+ +
65. ( ) ( ) ( ) is equal to
log abc log abc log abc
ab bc ca
(A) 0 (B) 1 (C) 2 (D) –1
1 1 1
+ +
66. 1+log (bc) 1+log (ca) 1+log (ab) is equal to
a b c
(A) 0 (B) 1 (C) 3 (D) –1
1 1 1
+ +
67. log (x) log (x) log (x) is equal to
a b c
b c a
(A) 0 (B) 1 (C) 3 (D) –1
68. log (a).log (b).log (c) is equal to
b c a
(A) 0 (B) 1 (C) –1 (D) None
1 2
69. log a2 . log (b3) . log (c3) is equal to
b c a
(A) 0 (B) 1 (C) –1 (D) None
70. The value of is logb logc loga
a c.b a.c b
(A) 0 (B) 1 (C) –1 (D) None
71. The value of
(bc)logb
c
.(ca)logc
a
.(ab)loga
b is
(A) 0 (B) 1 (C) –1 (D) None
an bn cn
72. The value of log + log + log is
bn cn an
(A) 0 (B) 1 (C) –1 (D) None
a2 b2 c2
73. The value of log + log + log is
bc ca ab
(A) 0 (B) 1 (C) –1 (D) None
74. log (a9) + log a = 10if the value of a is given by
(A) 0 (B) 10 (C) –1 (D) None
loga logb logc
= =
75. If the value of abc is
y-z z-x x-y
(A) 0 (B) 1 (C) –1 (D) None
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loga logb logc
= =
76. If the value of ay+z.bz+x.cx+y is given by
y-z z-x x-y
(A) 0 (B) 1 (C) –1 (D) None
1 1
77. If loga = logb = logc the value of a4b3c-2 is
2 5
(A) 0 (B) 1 (C) –1 (D) None
1 1 1
78. If loga = logb = logcthe value of a4- bc is
2 3 5
(A) 0 (B) 1 (C) –1 (D) None
1 1 1
79. If
4
log
2
a =
6
log
2
b = -
24
log
2
c the value of a3b2c is
(A) 0 (B) 1 (C) –1 (D) None
1 1
+
80. The value of is
log (ab) log (ab)
a b
(A) 0 (B) 1 (C) –1 (D) None
1 1 1 1
+ + =
81. If then the value if z is given by
log t log t log t log t
a b c z
(A) abc (B) a + b + c (C) a(b + c) (D) (a + b)c
1 1 1
82. If l = 1+log bc, m = 1 + log ca, , n=1+log ab then the value of + + -1 is
a b c l m n
(A) 0 (B) 1 (C) –1 (D) 3
83. If a = b2 = c3 = d4 then the value of log (abcd) is
a
1 1 1 1 1 1
(A) 1 + + + (B) 1 + + + C) 1+2+3+4 (D) None
2 3 4 2! 3! 4!
84. The sum of the series log b + log b2 + log b3 +.......log bn is given by
a a2 a3 an
(A) log bn (B) log bn (C) nlog bn (D) None
a an an
1
85. has a value of
alog
b
a
(A) a (B) b (C) (a + b) (D) None
86. The value of the following expression alog a b.log b c.log c d.log d t is given by
(A) t (B) abcdt (C) (a + b +c + d + t) (D) None
87. For any three consecutive integers x y z the equation log(1+xz) - 2logy = 0 is
(A) True (B) False (C) Sometimes true
(D) cannot be determined in the cases of variables with cyclic order.
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RATIO AND PROPORTION, INDICES, LOGARITHMS
a+b 1 a b
88. If log = (loga+logb) then the value of + is
3 2 b a
(A) 2 (B) 5 (C) 7 (D) 3
a+b log a log b
89. If a2 + b2 = 7ab then the value of is log - -
3 2 2
(A) 0 (B) 1 (C) –1 (D) 7
1
90. If a3 + b3 = 0 then the value of log(a+b) - (loga + logb + log3) is equal to
2
(A) 0 (B) 1 (C) –1 (D) 3
91. If x = log bc y = log ca z = log ab then the value of xyz – x – y – z is
a b c
(A) 0 (B) 1 (C) –1 (D) 2
92. On solving the equation logt + log(t-3) = 1 we get the value of t as
(A) 5 (B) 2 (C) 3 (D) 0
93. On solving the equation log log (log t)=1 we get the value of t as
3 2 3
(A) 8 (B) 18 (C) 81 (D) 6561
94. On solving the equation log log (log 32)= 2 we get the value of t as
1 t 4
2
5 25 625
(A) (B) (C) (D) None
2 4 16
1 1
95. If
(4.8)x =(0.48)y
=1,000 then the value of
-
is
x y
1 1
(A) 3 (B) –3 (C) (D) -
3 3
(x )
96. If x2a-3y2a = x6-ay5a then the value of alog is
y
(A) 3logx (B) logx (C) 6logx (D) 5logx
en −e−n
97. If x= then the value of n is
en +e−n
1 1+x 1+x 1-x 1 1-x
(A) log (B) log (C) log (D) log
2 e 1-x e 1-x e 1+x 2 e 1+x
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ANSWERS
1) B 18) B 35) A 52) C 69) B 86) A
2) A 19) C 36) C 53) B 70) B 87) A
3) C 20) B 37) C 54) C 71) B 88) C
4) B 21) D 38) C 55) B 72) A 89) A
5) A 22) B 39) C 56) A 73) A 90) A
6) A 23) B 40) A 57) C 74) B 91) D
7) D 24) B 41) B 58) C 75) B 92) A
8) A 25) C 42) B 59) A 76) B 93) D
9) A 26) A 43) A 60) C 77) B 94) C
10) A 27) A 44) C 61) C 78) A 95) C
11) C 28) C 45) D 62) A 79) B 96) A
12) A 29) D 46) C 63) B 80) B 97) A
13) B 30) C 47) A 64) A 81) A
14) A 31) A 48) C 65) C 82) A
15) C 32) B 49) B 66) B 83) A
16) C 33) A 50) A 67) A 84) A
17) D 34) B 51) D 68) B 85) B
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