RD Sharma Solutions for Class 7 Maths Exercise 6.2 Chapter 6 Exponents, in PDF, are available here. At BYJU’S, we have a set of expert faculty trying their best to provide exercise-wise solutions to students according to their level of understanding. To learn exercise-wise solutions, students can download the PDF available here. Solving RD Sharma Solutions for Class 7 help in boosting students’ confidence, which plays a crucial role in the final examination. This exercise explains the laws of exponents. It includes:
- Multiplying powers with the same base (First law)
- Dividing powers with the same base (Second law)
- Power with exponent zero and power of a power (Third law)
- Multiplying powers with the same exponents (Fourth law)
- Dividing the powers with the same exponents (Fifth law)
RD Sharma Solutions for Class 7 Maths Chapter 6 – Exponents Exercise 6.2
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 6 – Exponents Exercise 6.2
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 23 × 24 × 25
(ii) 512 ÷ 53
(iii) (72)3
(iv) (32)5 ÷ 34
(v) 37 × 27
(vi) (521 ÷ 513) × 57
Solution:
(i) Given 23 × 24 × 25
We know that first law of exponents states that am × an × ap = a(m+n+p)
Therefore above equation can be written as 23 x 24 x 25 = 2(3 + 4 + 5)
= 212
(ii) Given 512 ÷ 53
According to the law of exponents we have am ÷ an = am-n
Therefore given question can be written as 512 ÷ 53 = 512 – 3 = 59
(iii) Given (72)3
According to the law of exponents we have (am)n = amn
Therefore given question can be written as (72)3 = 76
(iv) Given (32)5 ÷ 34
According to the law of exponents we have (am)n = amn
Therefore (32)5 ÷ 34 = 3 10 ÷ 34
According to the law of exponents we have am ÷ an = am-n
3 10 ÷ 34 = 3(10 – 4) = 36
(v) Given 37 × 27
We know that law of exponents states that am x bm = (a x b)m
37 × 27 = (3 x 2)7 = 67
(vi) Given (521 ÷ 513) × 57
According to the law of exponents we have am ÷ an = am-n
= 5(21 -13) x 57
= 58 x 57
According to the law of exponents we have am x an = a(m +n)
= 5(8+7) = 515
2. Simplify and express each of the following in exponential form:
(i) {(23)4 × 28} ÷ 212
(ii) (82 × 84) ÷ 83
(iii) (57/52) × 53
(iv) (54× x10y5)/ (54 × x7y4)
Solution:
(i) Given {(23)4 × 28} ÷ 212
{(23)4 x 28} ÷ 212= {212 x 28} ÷ 212 [According to the law of exponents we have (am)n = amn]
= 2(12 + 8) ÷ 212[According to the law of exponents we have am x an = a(m +n)]
= 220 ÷ 212 [According to the law of exponents we have am ÷ an = am-n]
= 2 (20 – 12)
= 28
(ii) Given (82 × 84) ÷ 83
(82 × 84) ÷ 83 [According to the law of exponents we have am x an = a(m +n)]
= 8(2 + 4) ÷ 83
= 86 ÷ 83[According to the law of exponents we have am ÷ an = am-n]
= 8(6-3) = 83 = (23)3 = 29
(iii) Given (57/52) × 53
= 5(7-2) x 53[According to the law of exponents we have am ÷ an = am-n]
= 55 x 53[According to the law of exponents we have am x an = a(m +n)]
= 5(5 + 3) = 58
(iv) Given (54× x10y5)/ (54 × x7y4)
= (54-4× x10-7y5-4) [According to the law of exponents we have am ÷ an = am-n]
= 50x3y1 [since 50 = 1]
= 1x3y
3. Simplify and express each of the following in exponential form:
(i) {(32)3 × 26} × 56
(ii) (x/y)12 × y24 × (23)4
(iii)(5/2)6 × (5/2)2
(iv) (2/3)5× (3/5)5
Solution:
(i) Given {(32)3 × 26} × 56
= {36 × 26} × 56[According to the law of exponents we have (am)n = amn]
= 66 × 56 [since law of exponents states that am x bm = (a x b)m]
= 306
(ii) Given (x/y)12 × y24 × (23)4
= (x12/y12) × y24 × 212
= x12 × y24-12 × 212[According to the law of exponents we have am ÷ an = am-n]
= x12 × y12 × 212
= (2xy)12
(iii) Given (5/2)6 × (5/2)2
= (5/2)6+2[According to the law of exponents we have am x an = a(m +n)]
= (5/2)8
(iv) Given (2/3)5× (3/5)5
= (2/5)5[since law of exponents states that am x bm = (a x b)m]
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution:
Given 9 × 9 × 9 × 9 × 9 = (9)5 = (32)5
= 310
5. Simplify and write each of the following in exponential form:
(i) (25)3 ÷ 53
(ii) (81)5 ÷ (32)5
(iii) 98 × (x2)5/ (27)4 × (x3)2
(iv) 32 × 78 × 136/ 212 × 913
Solution:
(i) Given (25)3 ÷ 53
= (52)3 ÷ 53[According to the law of exponents we have (am)n = amn]
= 56 ÷ 53 [According to the law of exponents we have am ÷ an = am-n]
= 56 – 3
= 53
(ii) Given (81)5 ÷ (32)5[According to the law of exponents we have (am)n = amn]
= (81)5 ÷ 310[81 = 34]
= (34)5 ÷ 310 [According to the law of exponents we have (am)n = amn]
= 320 ÷ 310
= 320-10 [According to the law of exponents we have am ÷ an = am-n]
= 310
(iii) Given 98 × (x2)5/ (27)4 × (x3)2
= (32)8 × (x2)5/ (33)4× (x3)2[According to the law of exponents we have (am)n = amn]
= 316 × x10/312 × x6
= 316-12 × x10-6[According to the law of exponents we have am ÷ an = am-n]
= 34 × x4
= (3x)4
(iv) Given (32 × 78 × 136)/ (212 × 913)
= (32 × 7276 × 136)/(212× 133 × 73)[According to the law of exponents we have (am)n = amn]
= (212 × 76 × 136)/(212× 133 × 73)
= (76 × 136)/(133 × 73)
= 916/913[According to the law of exponents we have am ÷ an = am-n]
= 916-3
= 913
6. Simplify:
(i) (35)11 × (315)4 – (35)18 × (35)5
(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
(iv) (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
Solution:
(i) Given (35)11 × (315)4 – (35)18 × (35)5
= (3)55 × (3)60 – (3)90 × (3)25[According to the law of exponents we have (am)n = amn]
= 3 55+60 – 390+25
= 3115 – 3115
= 0
(ii) Given (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
= (24 × 2(n+1) -22 × 2n)/(24 × 2(n+2) -2n+1 × 22) [According to the law of exponents we have (am)n = amn]
= 22 × 2(n+3-2n)/)22× 2(n+4-2n+1)
= 2n × 23 – 2n/ 2n × 24 – 2n × 2
= 2n(23 – 1)/ 2n(24 – 1) [According to the law of exponents we have am ÷ an = am-n]
= 8 -1 /16 -2
= 7/14
= (1/2)
(iii) Given (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
= (10 × 5n+1 + 52 × 5n)/(3 × 5n+2 + (2 × 5) × 5n+1)
= (10 × 5n+1 + 5 × 5n+1)/(3 × 5n+2 + (2 × 5) × 5n+1) [According to the law of exponents we have (am)n = amn]
= 5n+1 (10+5)/ 5n+1 (10+15)[According to the law of exponents we have am ÷ an = am-n]
= 15/25
= (3/5)
(iv) Given (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
= (16)7 ×(52)5× (34)3/(3 × 5 )7 ×(3 × 8)5× (16 × 5)3
= (16)7 ×(52)5× (34)3/37 × 57 × 35 × 85× 163× 53
= (16)7/ 85 × 16 3
= (16)4/85
= (2 × 8)4/85
= 24/8
= (16/8)
= 2
7. Find the values of n in each of the following:
(i) 52n × 53 = 511
(ii) 9 x 3n = 37
(iii) 8 x 2n+2 = 32
(iv) 72n+1 ÷ 49 = 73
(v) (3/2)4 × (3/2)5 = (3/2)2n+1
(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2
Solution:
(i) Given 52n x 53 = 511
= 52n+3 = 511
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n = (8/2)
⇒ n = 4
(ii) Given 9 x 3n = 37
= (3)2 x 3n = 37
= (3)2+n = 37
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 – 2 = 5
(iii) Given 8 x 2n+2 = 32
= (2)3 x 2n+2 = (2)5 [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0
(iv) Given 72n+1 ÷ 49 = 73
= 72n+1 ÷ 72 = 73 [since 49 = 72]
= 72n+1-2 = 73
= 72n-1=73
On equating the coefficients, we get
2n – 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n =4/2 =2
(v) Given (3/2)4 × (3/2)5 = (3/2)2n+1
= (3/2)4+5 = (3/2)2n+1
= (3/2)9 = (3/2)2n+1
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 – 1
⇒ 2n = 8
⇒ n =8/2 =4
(vi) Given (2/3)10× {(3/2)2}5 = (2/3)2n – 2
= (2/3)10 × (3/2)10 = (2/3)2n – 2
= 2 10 × 310/310 × 210 = (2/3)2n – 2
= 1 = (2/3)2n – 2
= (2/3)0 = (2/3)2n – 2
On equating the coefficients, we get
0 =2n -2
2n -2 =0
2n =2
n = 1
8. If (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27), find the value of n.
Solution:
Given (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27)
= (32)n × 33 × 3n – (33)n/ (315 × 23) = (1/27)
= 3(2n+2+n) – (33)n/ (315 × 23) = (1/27)
= 3(3n+2)– (33)n/ (315 × 23) = (1/27)
= 33n × 32 – 33n/ (315 × 23) = (1/27)
= 33n × (32 – 1)/ (315 × 23) = (1/27)
= 33n × (9 – 1)/ (315 × 23) = (1/27)
= 33n × (8)/ (315 × 23) = (1/27)
= 33n × 23/ (315 × 23) = (1/27)
= 33n/315 = (1/27)
= 33n-15 = (1/27)
= 33n-15 = (1/33)
= 33n-15 = 3-3
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n = 12/3 = 4
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