Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

Question 1.

Evaluate:

(i) 3^{-2}

(ii) (-4)^{-2}

(iii) (\(\frac{1}{2}\))^{-5}

Solution:

Question 2.

Simplify and express the result in power notation with positive exponent?

(i) (-4)^{5} + (-4)^{8}

(ii) \(\left(\frac{1}{2^{3}}\right)^{2}\)

(iii) (-3)^{4} × (\(\frac{5}{3}\))^{4}

(iv) (3^{-7} + 3^{-10}) × 3^{-5}

(v) 2^{-3} × (-7)^{-3}

Solution:

(i) (-4)^{5} + (-4)^{8}

∵ a^{m} + a^{n} = a^{m-n}

∴ (-4)^{5} + (-4)^{8} = (-4)^{5-8}

= (-4)^{-3} = \(\frac{1}{(-4)^{3}}\)

(ii)

(iii)

= [(-1) × 5]^{4} = [(-1)^{4} × (+5)^{4}]

= 1 × (5)^{4} = (5)^{4}

(iv) (3^{-7} ÷ 3^{-10}) × 3^{-5}

∵ a^{m} + a^{n} = a^{m-n} and a^{m} × a^{n} = a^{m+n}

∴ (3^{-7} ÷ 3^{-10} × 3^{-5} = [3^{-7-(-10)}] × 3^{-5}

= 3^{3+(-5)} = 3^{-2} = \(\frac{1}{(3)^{2}}\)

(v) 2^{-3} × (-7)^{-3}

∵ a^{m} × b^{m} = (ab)^{m}

∴ 2^{-3} × (-7)^{-3} = [2 × (-7)]^{-3}

= [-14]^{-3} = \(\frac{1}{(-14)^{3}}\)

Question 3.

Find the value of:

(i) (3^{0} + 4^{-1}) × 2^{2}

(ii) (2^{-1} × 4^{-1}) + 2^{-2}

(iii) (\(\frac{1}{2}\))^{-2} + (\(\frac{1}{3}\))^{-2} + (\(\frac{1}{4}\))^{-2}

(iv) (3^{-1} + 4^{-1} + 5^{-1})^{0}

(v) {(\(\frac{-2}{3}\))^{-2}}^{2}

Solution:

(i) (3^{0} + 4^{-1}) × 2^{2}

∵ a^{0} = 1 and a^{-1} = \(\frac{1}{a}\)

∴ (3^{0} + 4^{-1}) × 2^{2}

= (1 + \(\frac{1}{2}\)) 2^{2} = \(\frac{5}{4}\) × 4 = 5

(ii) (2^{-1} × 4^{-1}) ÷ 2^{-2}

∵ a^{m} × b^{m} = (ab)^{m}

∴ (2^{-1} × 4^{-1}) + 2^{-2} = (2 × 4^{-1}) ÷ 2^{-2}

= (2^{1} × 2^{2})^{-1} ÷ 2^{-2}

= (2^{1+2})^{-1}) ÷ 2^{-2} = (2^{3})^{-1} ÷ 2^{-2}

= 2^{-3} ÷ 2^{-2} = (2)^{(-3)-(-2)}

= (-2)^{-3+2} = 2^{-1} = \(\frac{1}{2}\)

(iii)

(iv)

(v)

Question 4.

Evaluate

- \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\) and
- (5
^{-1}× 2^{-1}) × 6^{-1}

Solution:

1. \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\) = \(\frac{8^{-1} \times(5 \times 5 \times 5)}{2^{-4}}\)

= \(\frac{1}{8}\) × 2^{4} × 125 = \(\frac{1}{8}\) × 2 × 2 × 2 × 2 × 125

= 2 × 125 = 250

2. (5^{-1} × 2^{-1}) × 6^{-1}) = (5 × 2^{-1}) × 6^{-1})

= 10^{-1} × 6^{-1}

= (10 × 6^{-1}) = 60^{-1} = \(\frac{1}{60}\)

Question 5.

Find the value of m for which 5^{m} + 5^{-3} = 5^{5}.

Solution:

∵ a^{m} × b^{m} = (ab)^{m}

∴ L.H.S = 5^{m} + 5^{-3}

= 5^{m-(-3)}

= 5^{m+3}

Now, 5^{m+3} = 5^{5}

Since, bases 5 are equal, therefore exponents are also equal.

i.e; m + 3 = 5

or m = 5 – 3 = 2

Thus, the required value of m is 2.

Question 6.

Evaluate:

(i) \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)

(ii) \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)

Solution:

Question 7.

Simplify:

(i) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\) (t ≠ 0)

(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)

Solution: