SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Unit Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Unit Exercise

Question 1.
Answer the following.
(i) The exponential form 149 should read as
Answer:
14 is raised to the power of 9.

(ii) When base is 12 and exponent is 17, it’s exponential form is _________
Answer:
1217.

(iii) The value of (14 × 21)0 is
Answer:
We know a0 = 1
So, (14 × 21)0 = 1

Question 2.
Express the following numbers as a product of powers of prime factors :
(i) 648
Answer:

Given 648 = 2 × 324
= 2 × 2 × 162
= 2 × 2 × 2 × 81
= 2 × 2 × 2 × 3 × 27
= 2 × 2 × 2 × 3 × 3 × 9
= 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ 648 = 23 × 34

(ii) 1600
Answer:
Given 1600 = 2 × 800
= 2 × 2 × 400
= 2 × 2 × 2 × 200
= 2 × 2 × 2 × 2 × 100
= 2 × 2 × 2 × 2 × 2 × 2 × 25
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

∴ 1600 = 26 × 52

(iii) 3600
Answer:

Given 3600 = 2 × 1800
= 2 × 2 × 900
= 2 × 2 × 2 × 450
= 2 × 2 × 2 × 2 × 225
= 2 × 2 × 2 × 2 × 3 × 75
= 2 × 2 × 2 × 2 × 3 × 3 × 25
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
∴ 3600 = 24 × 32 × 52

Question 3.
Simplify the following using laws of exponents.
(i) a4 × a10
Answer:
a4 × a10
We know am × an = am+n
= a4+10
∴ a4 × a10 = a14

(ii) 1818 ÷ 1814
Answer:
1818 ÷ 1814
We Know am ÷ an = am-n
= 1818-14
∴ 1818 ÷ 1814 = 184

(iii) (xm)0
Answer:
(xm)0
We Know (am)n = am.n
= xm×n = x0(∵ a0 = 1)
∴ (xm)0 = 1

(iv) (62 × 64) ÷ 63
Answer:
(62 X 64) ÷ 63
We Know am × an = am+n
= (62+4) ÷ 63
= 66 ÷ 63
We Know am ÷ an = am-n
= 66-3
∴ (62 × 64) ÷ 63 = 63

(v) $$\left(\frac{2}{3}\right)^{p}$$
Answer:
$$\left(\frac{2}{3}\right)^{p}$$
We Know $$\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}=\frac{2^{p}}{3^{p}}$$
∴ $$\left(\frac{2}{3}\right)^{\mathrm{p}}=\frac{2^{\mathrm{p}}}{3^{\mathrm{p}}}$$

Question 4.
Identify the greater number in each of the following andjustify your answer.
(i) 210 or 102
Answer:
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
210 = 1024
102 = 10 × 10
102 = 100
1024 > 100
So, 210 > 102
∴ 210 is greater.

(ii) 54 or 45
Answer:
54 = 5 × 5 × 5 × 5 = 625
45 =4 × 4 × 4 × 4 × 4 = 1024
1024 > 625
So, 45 > 54
∴ 45 is greater number.

Question 5.
If $$\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}$$, then find the value of ‘k’
Answer:
Given $$\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}$$
We Know am × an = am+n
⇒ $$\left(\frac{4}{5}\right)^{2+5}=\left(\frac{4}{5}\right)^{\mathrm{k}}$$
⇒ $$\left(\frac{4}{5}\right)^{7}=\left(\frac{4}{5}\right)^{\mathrm{k}}$$
If the bases are equal powers should be equal
⇒ 7 = k
∴ k = 7

Question 6.
If 52p+1 ÷ 52 = 125, then find the value of ‘p’.
Answer:
Given 52p+1 ÷ 52 = 125
We Know am ÷ an = am-n
⇒ 52p+1-2 = 5 × 5 × 5
⇒ 52p-1 = 53
If the bases are equal, powers should be equal.
⇒ 2p – 1 =3
⇒ 2p = 3 + 1
⇒ 2p = 4
⇒ $$\frac{2 \mathrm{p}}{2}=\frac{4}{2}$$
∴ p = 2

Question 7.
Prove that $$\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}$$ = 1
Answer:
Given $$\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}$$ = 1

Question 8.
Express the following numbers in the expanded form.
(i) 20068
Answer:
20068 = (2 × 10,000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)
∴ 20068 = (2 × 104) + (6 × 101) + (8 × 1)

(ii) 120718
Answer:
120718 = (1 × 1,00,000) + (2 × 10,000) + (0 × 1000) + (7 × 100) + (1 × 10) + (8 × 1)
∴ 120718 = (1 × 105) + (2 × 104) + (7 × 102) + (1 × 101) + (8 × 1)

Question 9.
Express the number appearing in the following statements in standard form :
(i) The Moon is 384467000 meters away from the Earth approximately.
Answer:
Distance of Moon from the Earth = 384467000 metres
= 3.84467000 × 100000000

Decimal is shifted eight places to the left.
= 3.84 4 67 × 108 m

Distance of Moon from the earth = 3.84 4 67 × 108 m

(ii) Mass of the Sun is 1 ,989,000,000,000,000,000,00000,000000 kg.
Answer:
Mass of Sun = 1 ,989,000,000,000,000,000,00000,000000 kg
= 1.989 × 1 ,000,000,000,000,000,00000,000000
= 1.989 × 1030 kg
∴ Mass of Sun = 1.989 × 1030 kg.

Question 10.
Lasya solved some problems of exponents and powers in the following way. Do you agree with the solution ? If not why? Justify your answer.
Answer:
(i) x3 × x2 = x6
Answer:
No. I won’t agree with this solution.
Given, x3 × x2
We know am × an = am+n
= a3+2
= x5 which is ≠ x6
so, x3 × x2 ≠ x6
We have to add to powers. But, Lasya multiplied the powers. That’s why Lasya’s solution is wrong.

(ii) (63)10 = 613
Answer:
No, (63)10 is not equal to 6.
We know (am)n = amn
= 63×10
= 630 which is ≠ 613
so, (63)10 ≠ 613
We have to multiply. the powers. But, Lasya added the powers. That’s why Lasya’s solution is wrong.

(iii) $$\frac{4 x^{6}}{2 x^{2}}$$ = 2x3
Answer:
No, I won’t agree with this solution.
$$\frac{4 x^{6}}{2 x^{2}}=\frac{2^{2}}{2^{1}} \times \frac{x^{6}}{x^{2}}$$
We know am ÷ an = am-n
= 22-1 × x6-2
= 21.x4
= 2x4 which is ≠ 2x3
so, $$\frac{4 x^{6}}{2 x^{2}}$$ ≠ 2x3
We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

(iv) $$\frac{3^{5}}{9^{5}}=\frac{1}{3}$$
Answer:
No. I won’t agree with this solution.
$$\frac{3^{5}}{9^{5}}=\frac{3^{5}}{\left(3^{2}\right)^{5}}$$

We Know (am)n = amn

We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

Question 11.
Is – 22 is equal to 4? Justify your answer.
Answer:
– 22 = – (2 × 2) = – 4 ≠ 4
∴ – 22 = – 4

Question 12.
Beulah computed 25 × 210 = 250. Has she done it correctly? Give the reason.
Answer:
Given 25 × 210
We know am × an = am+n
= 25+10 = 215 ≠ 250
∴ 25 × 210 ≠ 250
Beulah did wrong.
Here we have to add the powers.
But, he multiplied the powers.

Question 13.
Rafi computed $$\frac{3^{9}}{3^{3}}$$ as 33. Has he done
Answer:
Given $$\frac{3^{9}}{3^{3}}$$
We know $$\frac{a^{m}}{a^{n}}$$ = am-n
= 39-3 = 36 ≠ 33
So, $$\frac{3^{9}}{3^{3}}$$ ≠ 33
Rafi computed wrong.
Here we have to subtract the powers. But, he divided the powers.

Question 14.
Is (a2)3 equal to a8? Give the reason.
Answer:
Given (a2)3 equal to a8
We Know (am)n = amn
= (a2)3 = a2×3 = a6 ≠ a8
∴ (a2)3 = a6

We have to do 2 × 3 = 6, But not
23 = 2 × 2 × 2 = 8