SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.2 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.2

Question 1.

Simplify the following by using Laws of Exponents.

(i) 3^{7} × 3^{8}

Answer:

3^{7} × 3^{8}

We know a^{m} × a^{n} = a^{m+n}

3^{7} × 3^{8} = 3^{7} + 8 = 3^{15}

∴ 3^{7} × 3^{8} =3^{15}

(ii) 9^{2} × 9^{0} × 9^{3}

Answer:

9^{2} × 9^{0} × 9^{3}

We know a^{p} × a^{q} × a^{r} = p + q + r

9^{2} × 9^{0} × 9^{3} = 9^{2 + 0 + 3} = 9^{5}

∴ 9^{2} × 9^{0} × 9^{3} = 9^{5}

(iii) (2^{8})^{3}

Answer:

(2^{8})^{3}

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

(2^{8})^{3} = 2^{8×3} =2^{24}

∴ (2^{8})^{3} = 2^{24}

(iv) (a^{5})^{4}

Answer:

(a^{5})^{4}

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

(a^{5})^{4} = a^{5×4} = a^{20}

(a^{5})^{4} = a^{20}

(v) \(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{3} \times\left(\frac{2}{5}\right)^{8}\)

Answer:

(vi) 7^{5} ÷ 7^{8}

Answer:

7^{5} ÷ 7^{8}

We know a^{m} = \(\frac{1}{a^{n-m}}\) (n>m)

7^{5} ÷ 7^{8} = \(\frac{1}{7^{8-5}}=\frac{1}{7^{3}}\)

∴ 7^{5} ÷ 7^{8} = \(\frac{1}{7^{3}}\)

(vii) \(\frac{(-6)^{9}}{(-6)^{5}}\)

Answer:

\(\frac{(-6)^{9}}{(-6)^{5}}\)

We Know a^{m} ÷ a^{n} = a^{m-n}(m > n)

\(\frac{(-6)^{9}}{(-6)^{5}}\) = (-6)^{9-5} = (-6)^{4}

∴ \(\frac{(-6)^{9}}{(-6)^{5}}\) = (-6)^{4}

(viii) (6^{4} × 6^{2}) ÷ 6^{5}

Answer:

(6^{4} × 6^{2}) ÷ 6^{5}

We know atm X a = a”

= (6^{4+2}) ÷ 6^{5}

= 6^{6} ÷ 6^{5}

We know a^{m} ÷ a^{n} = a^{m-n} (m> n)

= 6^{6-5} = 6^{1} = 6

∴ (6^{4} × 6^{2}) ÷ 6^{5} = 6

(ix) \(\frac{5^{3}}{2^{3}}\)

Answer:

\(\frac{5^{3}}{2^{3}}\)

We know \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)

∴ \(\frac{5^{3}}{2^{3}}=\left(\frac{5}{2}\right)^{3}\)

(x) (-3)^{3} × (-3)^{10} × (-3)^{7}

Answer:

(-3)^{3} × (-3)^{10} × (-3)^{7}

we know a^{p} . a^{q}. a^{r} = a^{p+q+r}

(-3)^{3} × (-3)^{0} × (-3)^{7} = (-3)^{3+10+7}

∴ (- 3)^{3} × (- 3)^{10} × (- 3)^{7} = (- 3)^{20}

Question 2.

Simplify and express each of the following in Exponential form.

(i) \(\left(\frac{a^{5}}{a^{3}}\right)\) × a^{8}

Answer:

\(\left(\frac{a^{5}}{a^{3}}\right)\) × a^{8}

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

= (a^{5-3}) × a^{8}

= a^{2} × a^{8}

We know a^{m} × a^{n} = a^{m+n}

= a^{2+8} = a^{10}

∴ \(\left(\frac{a^{5}}{a^{3}}\right)\) × a^{8} = a^{10}

(ii) 2^{0} + 3^{0} – 4^{0}

Answer:

2^{0} + 3^{0} – 4^{0}

We know a^{0} = 1

∴ 2^{0} + 3^{0} – 4^{0} = 1

(iii) (2^{3} × 2)^{2}

Answer:

(2^{3} × 2)^{2}

We know a^{m}.a^{n} = a^{m+n}

(2^{3} × 2)^{2} = (2^{3+1})^{2} = (2^{4})^{2}

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

= 2^{4×2} = 2^{8}

∴(2^{3} × 2)^{2} = 2^{8}

(iv) [(5^{2})^{3} × 5^{4}] ÷ 5^{7}

Answer:

[(5^{2})^{3} × 5^{4}] ÷ 5^{7}

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

= [5^{2×3} × 5^{4}] ÷ 5^{7}

= [5^{6} < 5^{4}] ÷ 5^{7}

We know a^{m} .a^{n} = a^{m+n}

= [5^{6+4}] ÷ 5^{7}

= 5^{10} ÷ 5^{7}

We know a^{m} ÷ a^{n} = a^{m-n} (m > n)

= 5^{10-7} = 5^{3}

∴ [(5^{2})^{3} × 5^{4}] ÷ 5^{7} = 5^{3}

Question 3.

Simplify \(\left(\frac{x^{a}}{x^{b}}\right) \times\left(\frac{x^{b}}{x^{c}}\right) \times\left(\frac{x^{c}}{x^{a}}\right)\)

Answer:

\(\left(\frac{x^{a}}{x^{b}}\right) \times\left(\frac{x^{b}}{x^{c}}\right) \times\left(\frac{x^{c}}{x^{a}}\right)\)

Question 4.

Find the value of the following.

(i) (-1)^{1000}

Answer:

(-1)^{1000}

(-1)^{1} = – 1

(-1)^{2} = – 1 × – 1 = + 1

(-1)^{3} = – 1 × -1 × -1 = – 1

(-1)^{4} = -1 × -1 × – 1 × – 1 = + 1

(-1)^{5} = -1 × -1 × -1 × -1 × -1 = -1

(-1)^{100} = -1 × -1 × -1 × ………… 100 times = + 1

So, (-1)^{odd number} = -1 and

(-1)^{even number} = + 1

Here 1000 is even number

So, (-1)^{1000} = + 1

(ii) (1)^{250}

Answer:

1^{250} = 1 × 1 × 1 ……… 250 times = 1

∴ 1^{250} = 1

(iii) (1)^{121}

Answer:

1^{121} = -1 × -1 × -1 ……..(121 times)

Here 121 is an odd number

So, (1)^{121} = – 1

(iv) (10000)^{0}

Answer:

(10000)^{0}

we know a^{0} = 1

So, (10000)^{0} = 1

Question 5.

If 7^{5} × 7^{3x} = 7^{20} then find the value of ’x’.

Answer:

Given, 7^{5} × 7^{3x} = 7^{20}

W know a^{m} X a^{n} = a^{m+n}

7^{5 + 3x} = 7^{20}

If the bases are equal, then the powers should also be equal.

⇒ 5 + 3x = 20

⇒ 5 + 3x – 5 = 20 – 5

⇒ 3x = 15

⇒ \(\frac{3 x}{3}=\frac{15}{3}\) = 5

∴ x = 5

Question 6.

If 10^{y} = 10000 then 5y =?

Answer:

Given, 10^{y} = 10000

10^{y} = 10 × 10 × 10 × 10

10^{y} = 10^{4}

If the bases are equal, then the powers should also be equal.

⇒ y = 4

Multiply by 5 on bothsides,

5 × y = 5 × 4

∴ 5y = 20

Question 7.

If 5^{x} = 100 then find the following values.

(i) 5^{x+y}

Answer:

Given 5^{x+y} = 100

5^{x} = 10^{2}

Multiply by 5^{2}

5^{x} × 5^{2} = 100 × 5^{2}

5^{x+2} = 100 × 25 (∵ a^{m} × a^{n} = a^{m+n})

∴ 5^{x+2} = 2500

(ii) 5^{x-2}

Answer:

Given 5^{x} = 100

5^{x} = 10^{2}

Divide by 5^{2} on both sides

5^{x} ÷ 5^{2} = 100 ÷ 5^{2}

∴ 5^{x-2} = 4

Question 8.

By what number should 3^{4} be multiplied so that the product is 243?

Answer:

Given number is 3^{4}

If we multiply 3^{4} by ‘a’ the product is 243.

i.e. 3^{4} × a = 243

= 3 × 3 × 3 × 3 × 3

3^{4} × a = 35

Divide by 3^{4} on both sides

If we multiply 3^{4} by 3 we get 243.

Question 9.

Arushi computed (5^{2})^{4} as 5^{16} Has she done it correctly or not’? Justify your answer.

Answer:

Given, (5^{2})^{4}

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

= 5^{2×4} = 5^{8}

∴ (5^{2})^{4} = 5^{8}

But Arushi got 5^{16} So, (5^{2})^{4} ≠ 5^{16}

Therefore Arushi did wrong.

Question 10.

Is 3^{5} × 4^{5} equal to 12^{25}? If not why? Justify your answer.

Answer:

Given, 3^{5} × 4^{5}

We know a^{m} × b^{m} = (a × b)^{m}

So, 3^{5} × 4^{5} = 12 but not 12^{5×5}

= (3 × 4)^{5}

= (12)^{5}

3^{5} × 4^{5}= 12^{5}

But given 3^{5} × 4^{5} = 12^{25}.

So, this is wrong.

So, 3^{5} × 4^{5} ≠ 12^{25}