Chapter 8: Exponents and Powers

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$ and ${a^m} \times {a^n} = {a^{m + n}}$

$x = {\left( {100} \right)^{ - 3}} \div {\left( {100} \right)^0}$

$\Rightarrow x = {\left( {100} \right)^{ - 3 - 0}}$

$\Rightarrow x = {\left( {100} \right)^{ - 3}}$

Thus,

$\Rightarrow {x^{ - 3}} = {\left( {{{\left( {100} \right)}^{ - 3}}} \right)^{ - 3}} = {\left( {100} \right)^{ 9}}$

Ans: Let x be multiplied with (–29)0 to get (+29)0

So, 

$x \times {\left( { - 29} \right)^0} = {\left( {29} \right)^0}$

$\Rightarrow x \times 1 = 1$

$\Rightarrow x = 1$

Ans: Let (–15)–1 be divided by x to get quotient (–15)–1.

So,

$\dfrac{{{{\left( { - 15} \right)}^{ - 1}}}}{x} = {\left( { - 15} \right)^{ - 1}}$

$\Rightarrow x = \dfrac{{{{\left( { - 15} \right)}^{ - 1}}}}{{{{\left( { - 15} \right)}^{ - 1}}}}$

$\Rightarrow x = 1$ 

Ans: a is called multiplicative inverse of b, if a x b= 1.

${\left( { - 7} \right)^{ - 2}} \div {\left( {90} \right)^{ - 1}} = \dfrac{1}{{{{\left( 7 \right)}^2}}} \div \dfrac{1}{{{{\left( {90} \right)}^1}}} = \dfrac{1}{{49}} \times 90 = \dfrac{{90}}{{49}}$

Put $b = \dfrac{{90}}{{49}}$

$\Rightarrow a \times \dfrac{{90}}{{49}} = 1$

$\Rightarrow a = \dfrac{{49}}{{90}}$

Ans: Given, 53x–1 ÷ 25 = 125

Since, 25 = 5 x 5 = 52

And 125 = 5 x 5 x 5 = 53

Therefore, 53x–1 ÷ (5)2 = (5)3

So, (5)3x–1-2 = (5)3 [As am ÷ an = (a)m-n ] 

(5)3x–3 = (5)3

On comparing both sides, we get

⇒3x – 3 = 3

⇒3x = 6

⇒x = 2

Ans: For standard form,

39,00,00,000 = 39 x 10 x 10 x 10 x 10 x 10 x 10 x 10

= 39 x 107 = 3.9 x 101 x 107

Use the law of exponent am × an = am+n

Thus, 39,00,00,000= 3.9 x 108

Therefore, the standard form of 39,00,00,000 = 3.9 x 108

Ans: For standard form, 

0.000005678 = 0.5678 x 10-5 = 5.678 x 10-5 x 10-1

= 5.678 x 10-6 

Hence, 5.678 x 10-6 is the standard form of 0.000005678.

Ans: Product of 3.2 × 106 and 4.1 × 10–1 = (3.2 x 106) (4.1x 10-1)

= (3.2 x 4.1) x 106 x 10-1

= 13.12 x 105 = 1.312 x 105 x 101 [As am × an = am+n]

= 1.312 x 106

Ans: Use am ÷ an = (a)m-n,

$\dfrac{{1.5 \times {{10}^6}}}{{2.5 \times {{10}^{ - 4}}}} = \dfrac{{15}}{{25}} \times {10^{6 + 4}} = \dfrac{3}{5} \times {10^{10}}$

= $0.6 \times {10^{10}}$

= $6 \times {10^{ - 1}} \times {10^{10}} = 6 \times {10^9}$

Ans: Total distance traveled by migratory bird = 15000 km.

= 15000 x 1000 m [As 1 km= 1000 m]

= 15000000 m

= 15 x 106 m

Scientific notation of 15 x 106 = 1.5 x 107 m.