# RS Aggarwal Solutions for Class 7 Maths Exercise 5A Chapter 5 Exponents

RS Aggarwal Solutions for Class 7 Maths Exercise 5A Chapter 5 Exponents, Students can download the PDF which is given here. Students are going to learn about Exponents, the product of a rational number multiplied several times by itself can be expressed in the same notation. It contains other topics like Laws of Exponents. Expert tutors at BYJUâ€™S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 5 Exponents. Students gain more knowledge by referring to RS Aggarwal Class 7 Solutions.

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1. Write each of the following in power notation:

(i) (5/7) Ã—(5/7) Ã— (5/7) Ã— (5/7)

Solution:-

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(5/7)4

(ii) (-4/3) Ã—(-4/3) Ã— (-4/3) Ã— (-4/3) Ã— (-4/3)

Solution:-

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-4/3)5

(iii) (-1/6) Ã—(-1/6) Ã— (-1/6)

Solution:-

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-1/6)3

(iv) (-8) Ã—(-8) Ã— (-8) Ã— (-8) Ã— (-8)

Solution:-

The product of rational number multiplied several times by itself can be expressed in the power notations as,

(-8)5

2. Express each of the following in power notation:

(i) (25/36)

Solution:-

We have,

25 = 5 Ã— 5 = (5)2

36 = 6 Ã— 6 = (6)2

Then,

= (52/62)

âˆ´ (5/6)2

(ii)(-27/64)

Solution:-

We have,

-27 = -3 Ã— -3 Ã— -3 = (-3)3

64 = 4 Ã— 4 Ã— 4 = (4)3

Then,

= (-33/43)

âˆ´ (-3/4)3

(iii) (-32/243)

Solution:-

We have,

-32 = -2 Ã— -2 Ã— -2 Ã— -2 Ã— -2 = (-2)5

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = (3)5

Then,

= (-25/35)

âˆ´ (-2/3)5

(iv) (-1/128)

Solution:-

We have,

-1 = -1 Ã— -1 Ã— -1 Ã— -1 Ã— -1 Ã— -1 Ã— -1 = (-1)7

128 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = (2)7

Then,

= (-17/27)

âˆ´ (-1/2)7

3. Express each of the following a rational number:

(i) (2/3)5

Solution:-

We have,

(2/3)5 = (25/35)

= (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2) / (3 Ã— 3 Ã— 3 Ã— 3 Ã— 3)

= (32/243)

(i) (-8/5)3

Solution:-

We have,

(-8/5)3 = (-83/53)

= (-8 Ã— -8 Ã— -8) / (5 Ã— 5 Ã— 5)

= (-512/125)

(ii) (-13/11)2

Solution:-

We have,

(-13/11)2 = (-132/112)

= (-13 Ã— -13) / (11 Ã— 11)

= (169/121)

(iii) (1/6)3

Solution:-

We have,

(1/6)3 = (13/63)

= (1 Ã— 1 Ã— 1) / (6 Ã— 6 Ã— 6)

= (1/216)

(v) (-1/2)5

Solution:-

We have,

(-1/2)5 = (-15/25)

= (-1 Ã— -1 Ã— -1Ã— -1 Ã— -1) / (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2)

= (-1/32)

(vi) (-4/7)3

Solution:-

We have,

(-4/7)3 = (-43/73)

= (-4 Ã— -4 Ã— -4) / (7 Ã— 7 Ã— 7)

= (-64/343)

(vii) (-1)9

Solution:-

We have,

(-1)9 = (-19)

= (-1 Ã— -1 Ã— -1Ã— -1 Ã— -1 Ã— -1 Ã— -1Ã— -1 Ã— -1)

= (-1)

4. Express each of the following as a rational number:

(i) (4)-1

Solution:-

We have:

(4)-1 = (4/1)-1

= (1/4)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (1/4)

(ii) (-6)-1

Solution:-

We have:

(-6)-1 = (-6/1)-1

= (1/-6)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-1/6)

(iii) (1/3)-1

Solution:-

We have:

(1/3)-1 = (-6/1)-1

= (3/1)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= 3

(iv) (–2/3)-1

Solution:-

We have:

(-2/3)-1 = (-2/3)-1

= (3/-2)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-3/2)

5. Find the reciprocal of each of the following:

(i) (3/8)4

Solution:-

We know that the reciprocal of (a/b) m is (b/a) m

Then,

Reciprocal of (3/8)4 is (8/3)4

(ii) (-5/6)11

Solution:-

We know that the reciprocal of (a/b) m is (b/a) m

Then,

Reciprocal of (-5/6)11 is (-6/5)11

(iii) (6)7

Solution:-

We know that the reciprocal of (a/b) m is (b/a) m

Then,

Reciprocal of (6)7 is (1/6)7

(iv) (-4)3

Solution:-

We know that the reciprocal of (a/b) m is (b/a) m

Then,

Reciprocal of (-4)3 is (-1/4)3

6. Find the value of each of the following:

(i) 80

Solution:-

By definition, we have a0= 1 for every integer.

âˆ´ 80 = 1

(ii) (-3)0

Solution:-

By definition, we have a0= 1 for every integer.

âˆ´ (-3)0 = 1

(iii) 40 + 50

Solution:-

By definition, we have a0= 1 for every integer.

âˆ´ 40 + 50

= 1 + 1

= 2

(iv) 60 Ã— 70

Solution:-

By definition, we have a0= 1 for every integer.

âˆ´ 60 Ã— 70

= 1 Ã— 1

= 1

7. Simplify each of the following and express each as a rational number:

(i) (3/2)4 Ã— (1/5)2

Solution:-

We have,

(34/24) = (3Ã—3Ã—3Ã—3)/ (2Ã—2Ã—2Ã—2) = (81/16)

(12/52) = (1Ã—1)/ (5Ã—5) = (1/25)

Then,

= (81/16) Ã— (1/25)

= (81Ã—1) / (16/25)

= (81/400)

(ii) (-2/3)5 Ã— (-3/7)3

Solution:-

We have,

(-25/35) = (-2Ã—-2Ã—-2Ã—-2Ã—-2)/ (3Ã—3Ã—3Ã—3Ã—3) = (-32/243)

(-33/73) = (-3Ã—-3Ã—-3)/ (7Ã—7Ã—7) = (-27/343)

Then,

= (-32/243) Ã— (-27/343)

= (-32Ã—-27) / (243Ã—343)

By simplifying,

= (-32Ã—-1) / (9/343)

= (32/3087)

(iii) (-1/2)5 Ã— 23 Ã— (3/4)2

Solution:-

We have,

(-15/25) = (-1Ã—-1Ã—-1Ã—-1Ã—-1)/ (2Ã—2Ã—2Ã—2Ã—2) = (-1/32)

(2)3= (2Ã—2Ã—2) = 8

(32/42) = (3Ã—3)/ (4Ã—4) = (9/16)

Then,

= (-1/32) Ã— 8 Ã— (9/16)

= (-1Ã—8Ã—9) / (32Ã— 1Ã—16)

By simplifying,

= (-1Ã—1Ã—9) / (32Ã—1Ã—2)

= (-9/64)

(iv) (2/3)2 Ã— (-3/5)3 Ã— (7/2)2

Solution:-

We have,

(22/32) = (2Ã—2)/ (3Ã—3) = (4/9)

(-3/5)3= (-3Ã—-3Ã—-3) / (5Ã—5Ã—5) = (-27/125)

(72/22) = (7Ã—7)/ (2Ã—2) = (49/4)

Then,

= (4/9) Ã— (-27/125) Ã— (49/4)

= (4Ã—-27Ã—49) / (9Ã— 125Ã—4)

On simplifying,

= (1Ã—-3Ã—49) / (1Ã—125Ã—1)

= (-147/125)

(v) {(-3/4)3-(-5/2)3} Ã— 42

Solution:-

We have,

= {(-33/43)-(53/23)} Ã— 16

= {(-27/64)-(-125/8)} Ã— 16

First we find the difference of {(-27/64)-(125/8)}

LCM of 64 and 8 is 64

= (-27Ã—1) / (64Ã—1) = (-27/64)

= (125Ã—8) / (8 Ã—8) = (1000/64)

= (1000- (-27))/64

= (1000+27)/64

= (973/64)

= (973/64) Ã— 16

= (973/4)

8. Simplify and express each as a rational number:

(i) (4/9)6Ã— (4/9)-4

Solution:-

We have,

= (4/9) (6+ (-4)) â€¦ [{(a/b) m Ã— (a/b) n} = (a/b) m-n

= (4/9) (6-4)

= (4/9)2

= (42/92)

= (16/81)

(ii) (-7/8)-3Ã— (-7/8)2

Solution:-

We have,

= (-7/8) (-3+ 2) â€¦ [{(a/b) m Ã— (a/b) n} = (a/b) m-n

= (-7/8) (-1)

= (-8/7)

(iii) (4/3)-3Ã— (4/3)-2

Solution:-

We have,

= (4/3) (-3+ (-2)) â€¦ [{(a/b) m Ã— (a/b) n} = (a/b) m-n

= (4/3) (-3-2)

= (4/3)-5

= (3/4)5

= (35/45)

=243/1024

9. Express each of the following as a rational number:

(i) 5-3

Solution:-

We know that,

= (5)-3 = (1/5)3 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (13/53)

= (1/125)

(ii) (-2)-5

Solution:-

We know that,

= (-2)-5 = (-1/2)3 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-13/23)

= (-1/8)

(iii) (1/4)-4

Solution:-

We know that,

= (1/4)-4 = (4/1)4 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (44/14)

= (256/1)

= 256

(iv) (-3/4)-3

Solution:-

We know that,

= (-3/4)-3 = (-4/3)3 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-43/33)

= (-64/27)

(v) (-3)-1Ã— (1/3)-1

Solution:-

We know that,

= (-3)-1= (-1/3)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (1/3)-1 = (3/1)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-1/3) Ã— (3/1)

= (-1Ã—3)/ (3Ã—1)

= (-3/3)

= -1

(vi) (5/7)-1Ã— (7/4)-1

Solution:-

We know that,

= (5/7)-1= (7/5)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (7/4)-1 = (4/7)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (7/5) Ã— (4/7)

= (7Ã—4)/ (5Ã—7)

On simplifying,

= (1Ã—4)/ (5Ã—1)

= 4/5

(vii) (5-1 7-1)-1

Solution:-

We know that,

= (5)-1= (1/5)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (7)-1 = (1/7)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

Now subtract,

= {(1/5) â€“ (1/7)}-1

= {(7-5)/35}-1 â€¦ [LCM of 5 and 7 is 35]

= {2/35}-1

= {35/2}

(viii) {(4/3)-1 â€“ (1/4)-1}-1

Solution:-

We know that,

= (4/3)-1= (3/4)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (1/4)-1 = (4/1)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

Now subtract,

= {(3/4) â€“ (4/1)}-1

= {(3-16)/4}-1 â€¦ [LCM of 4 and 1 is 4]

= {-13/4}-1

= {-4/13}

(ix) {(3/2)-1 Ã· (-2/5)-1}

Solution:-

We know that,

= (3/2)-1= (2/3)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

= (-2/5)-1 = (-5/2)1 â€¦ [âˆµ (a/b)-n = (b/a) n]

Now divide,

= {(2/3) Ã· (-5/2)}

= {(2/3) Ã— (-2/5}

= {(2Ã—-2) / (3Ã—5)

= {-4/15}

(x) (23/25)0

Solution:-

= (23/25)0 = 1

Because, by definition, we have a0= 1 for every integer.

### Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 5 – Exponents

Exercise 5C Solutions