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 14^{9} 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:

12^{17}.

(iii) The value of (14 × 21)^{0} is

Answer:

We know a^{0} = 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 = 2^{3} × 3^{4}

(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 = 2^{6} × 5^{2}

(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 = 2^{4} × 3^{2} × 5^{2}

Question 3.

Simplify the following using laws of exponents.

(i) a^{4} × a^{10}

Answer:

a^{4} × a^{10}

We know a^{m} × a^{n} = a^{m+n}

= a^{4+10}

∴ a^{4} × a^{10} = a^{14}

(ii) 18^{18} ÷ 18^{14}

Answer:

18^{18} ÷ 18^{14}

We Know a^{m} ÷ a^{n} = a^{m-n}

= 18^{18-14}

∴ 18^{18} ÷ 18^{14} = 18^{4}

(iii) (x^{m})^{0}

Answer:

(x^{m})^{0}

We Know (a^{m})^{n} = a^{m.n}

= x^{m×n} = x^{0}(∵ a^{0} = 1)

∴ (x^{m})^{0} = 1

(iv) (6^{2} × 6^{4}) ÷ 6^{3}

Answer:

(6^{2} X 6^{4}) ÷ 6^{3}

We Know a^{m} × a^{n} = a^{m+n}

= (6^{2+4}) ÷ 6^{3}

= 6^{6} ÷ 6^{3}

We Know a^{m} ÷ a^{n} = a^{m-n}

= 6^{6-3}

∴ (6^{2} × 6^{4}) ÷ 6^{3} = 6^{3}

(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) 2^{10} or 10^{2}

Answer:

2^{10} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

2^{10} = 1024

10^{2} = 10 × 10

10^{2} = 100

1024 > 100

So, 2^{10} > 10^{2}

∴ 2^{10} is greater.

(ii) 5^{4} or 4^{5}

Answer:

5^{4} = 5 × 5 × 5 × 5 = 625

4^{5} =4 × 4 × 4 × 4 × 4 = 1024

1024 > 625

So, 4^{5} > 5^{4}

∴ 4^{5} 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 a^{m} × a^{n} = a^{m+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 5^{2p+1} ÷ 5^{2} = 125, then find the value of ‘p’.

Answer:

Given 5^{2p+1} ÷ 5^{2} = 125

We Know a^{m} ÷ a^{n} = a^{m-n}

⇒ 5^{2p+1-2} = 5 × 5 × 5

⇒ 5^{2p-1} = 5^{3}

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 × 10^{8} m

Distance of Moon from the earth = 3.84 4 67 × 10^{8} 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 × 10^{30} kg

∴ Mass of Sun = 1.989 × 10^{30} 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) x^{3} × x^{2} = x^{6}

Answer:

No. I won’t agree with this solution.

Given, x^{3} × x^{2}

We know a^{m} × a^{n} = a^{m+n}

= a^{3+2}

= x^{5} which is ≠ x^{6}

so, x^{3} × x^{2} ≠ x^{6}

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

(ii) (6^{3})^{10} = 6^{13}

Answer:

No, (6^{3})^{10} is not equal to 6.

We know (a^{m})^{n} = a^{mn}

= 6^{3×10}

= 6^{30} which is ≠ 6^{13}

so, (6^{3})^{10} ≠ 6^{13}

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

(iii) \(\frac{4 x^{6}}{2 x^{2}}\) = 2x^{3}

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 a^{m} ÷ a^{n} = a^{m-n}

= 2^{2-1} × x^{6-2}

= 2^{1}.x^{4}

= 2x^{4} which is ≠ 2x^{3}

so, \(\frac{4 x^{6}}{2 x^{2}}\) ≠ 2x^{3}

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 (a^{m})^{n} = a^{mn}

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

Question 11.

Is – 2^{2} is equal to 4? Justify your answer.

Answer:

– 2^{2} = – (2 × 2) = – 4 ≠ 4

∴ – 2^{2} = – 4

Question 12.

Beulah computed 2^{5} × 2^{10} = 2^{50}. Has she done it correctly? Give the reason.

Answer:

Given 2^{5} × 2^{10}

We know a^{m} × a^{n} = a^{m+n}

= 2^{5+10} = 2^{15} ≠ 2^{50}

∴ 2^{5} × 2^{10} ≠ 2^{50}

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 3^{3}. Has he done

Answer:

Given \(\frac{3^{9}}{3^{3}}\)

We know \(\frac{a^{m}}{a^{n}}\) = a^{m-n}

= 3^{9-3} = 3^{6} ≠ 3^{3}

So, \(\frac{3^{9}}{3^{3}}\) ≠ 3^{3}

Rafi computed wrong.

Here we have to subtract the powers. But, he divided the powers.

Question 14.

Is (a^{2})^{3} equal to a^{8}? Give the reason.

Answer:

Given (a^{2})^{3} equal to a^{8}

We Know (a^{m})^{n} = a^{mn}

= (a^{2})^{3} = a^{2×3} = a^{6} ≠ a^{8}

∴ (a^{2})^{3} = a^{6}

We have to do 2 × 3 = 6, But not

2^{3} = 2 × 2 × 2 = 8