Class 7 Maths

Exponents and Powers NCERT Exercise 13.2

Question 1: Using laws of exponents simplify and write the answer in exponential form:

(a) 32 × 34 × 38

Answer: 32 × 34 × 38

= 3(2 + 4 + 8) = 314

(b) 615 ÷ 610

Answer: 615 ÷ 610 = 6(15 – 10) = 65

(c) `a^3 xx a^2`

Answer: `a^3 xx a^2 = a^(3 + 2) = a^5`

(d) `7^x \xx 7^2`

Answer: `7^x \xx 7^2 = 7^(x + 2)`

(e) `(5^2)^3 xx 5^3`

Answer: `(5^2)^3 xx 5^3 = 5^6 xx 5^3`

`= 5^(6 + 3)= 5^9`

(f) `2^5 xx 5^5`

Answer: `2^5 xx 5^5x = (2 xx 5)^5 = 10^5`

(g) `a^4 xx b^4`

Answer: `a^4 xx b^4 = (ab)^4`

(h) `(3^4)^4`

Answer: `(3^4)^4 = 3^16`

(i) `(2^20 ÷ 2^15) xx 2^3`

Answer: `(2^20 ÷ 2^15) xx 2^3 = [2^(20 – 15)] xx 2^3`

`= 2^5 xx 2^3 = 2^(5 + 3) = 2^8`

(j) `8^t ÷8^2`

Answer: `8^t ÷ 8^2 = 8^(t - 2)`

Question 2: Simplify and express each of the following in exponential form:

(a) `(2^3xx3^4xx4)/(3xx32)`

Answer: `=(2^3xx3^4xx2^2)/(3xx2^5)`

`=(2^(3+2)xx3^4)/(3xx2^5)`

`=(2^5xx3^(4-1))/(2^5)`

`=2^(5-5)xx3^3=2^0xx3^3=3^3`

(b) `((5^2)^3xx5^4)÷5^7`

Answer: `=(5^6xx5^4)÷5^7`

`=(5^(6+4))÷5^7=5^(10)÷5^7`

`=5^(10-7)=5^3`

(c) `(3xx7^2xx11^8)/(21xx11^3)`

Answer: `=(3xx7^2xx11^8)/(3xx7xx11^3)`

`=3^(1-1)xx7^(2-1)xx11^(8-3)`

`=3^0xx7^1xx11^5=7xx11^5`

(d) `(3^7)/(3^4xx3^3)`

Answer: `=(3^7)/(3^(4+3))=(3^7)/(3^7)`

`=3^(7-7)=3^0=1`

(e) `2^0+3^0+4^0`

Answer: `=1+1+1=3`

(f) `2^0xx3^0xx4^0`

Answer: `=1xx1xx1=1`

(g) `(3^0+2^0)xx5^0`

Answer: `=(1+1)xx1=2xx1=2`

(h) `(2^8xxa^5)/(4^3xxa^3)`

Answer: `=(2^8xxa^(5-3))/((2^2)^3)`

`=(2^8xxa^2)/(2^6)=2^(8-6)xxa^2`

`=2^2xxa^2=(2a)^2`

(i) `(4^5xxa^8b^3)/(4^5xxa^5b^2)`

Answer: `=4^(5-5)xxa^(8-5)xxb^(3-2)`

`=4^0xxa^3xxb^1=a^3b`

(j) `(2^3xx2)^2`

Answer: `=2^(3xx2)xx2^2=2^6xx2^2`

`=2^(6+2)=2^8`

Question 3: Say true or false and justify your answer:

  1. `10 xx 10^11 = 100^11`

    Answer: False, because first number will have 12 zeroes while the second number will have 22 zeroes at the end.
  2. `2^3 < 5^2`

    Answer: False, because first number is 8 while second number is 25.
  3. `2^3 xx 3^2 = 6^5`

    Answer: False, because `6^5 = (2 xx 3)^5 = 2^5 xx 3^5`
  4. `3^0 = 1000^0`

    Answer: True, because both are equal to 1.

Question 4: Express each of the following as a product of prime factors only in exponential form:

(a) `108 xx 192`

Answer: `108 xx 192 ``= (2 xx 2 xx 3 xx 3 xx 3) xx (2 xx 2 xx 2 xx 2 xx 2 xx 2 xx 3)`

`= 2^2xx 3^3 xx 2^6 xx 3 ``= 2^(2 + 6) xx 3^(3 + 1) = 2^8 xx 3^4`

(b) 270

Answer: `270 = 2 xx 3^3 xx 5`

(c) `729 xx 64`

Answer: `729 xx 64 = 3^6 xx 2^8`

(d) 768

Answer: `768 = 2^8 xx 3`

Question 5: Simplify

(a) `((2^5)^2xx7^3)/(8^3xx7)`

Answer: `=(2^(5xx2)xx7^(3-1))/((2^3)^3)`

`=(2^(10)xx7^2)/(2^9)=2^(10-9)xx7^2=2xx7^2`

(b) `(25xx5^2xxt^8)/(8^3xxt^4)`

Answer: `=(5^2xx5^2xxt^(8-4))/((2^3)^3)`

`=(5^(2+2)xxt^4)/(2^9)=(5^4xxt^4)/(2^9)`

(c) `(3^5xx10^5xx25)/(5^7xx6^5)`

Answer: `=(3^5xx(2xx5)^5xx5^2)/(5^7xx(2xx3)^5)`

`=(3^5xx2^5xx5^5xx5^2)/(5^7xx2^5xx3^5)`

`=3^(5-5)xx2^(5-5)xx5^(5+2-7)`

`=3^0xx2^0xx5^0=1`