Exercise 1.5 NCERT Class 9 Number Systems Best Mathematics Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4

Exercise 1.5 NCERT Class 9 contains a total of three questions. The questions are based on the following topic.

Laws of Exponents:

$a^{m} \times a^{n} = a^{m + n}$
$\frac{a^{m}}{a^{n}} = a^{m - n}$
${(a^{m})}^{n} = a^{m n}$
$a^{m} \times b^{m} = {(a b)}^{m}$
${(a b)}^{m} = a^{m} \times b^{m}$
${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$
$a^{- n} = \frac{1}{a^{n}}$
$a^{0} = 1$

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Exercise 1.5 NCERT Class 9 Solutions

Q.1 Q.2 Q.3

The first question in Exercise 1.5 NCERT Class 9 asks us to find the value of expressions given in exponential forms.

1. Find.

(i) $64^{\frac{1}{2}}$

Answer

We have

$64^{\frac{1}{2}}$

= ${(8^{2})}^{\frac{1}{2}}$ ... $[∵ 64 = 8 \times 8]$

= $8^{2 \times \frac{1}{2}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= 8.

(ii) $32^{\frac{1}{5}}$

Answer

We have

$32^{\frac{1}{5}}$

= ${(2^{5})}^{\frac{1}{5}}$ ... $[∵ 32 = 2 \times 2 \times 2 \times 2 \times 2]$

= $2^{5 \times \frac{1}{5}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= 2.

(iii) $125^{\frac{1}{3}}$

Answer

We have

$125^{\frac{1}{3}}$

= ${(5^{3})}^{\frac{1}{3}}$ ... $[∵ 125 = 5 \times 5 \times 5]$

= $5^{3 \times \frac{1}{3}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $5$ .

The second question in Exercise 1.5 NCERT Class 9 asks us to find the value of expressions given in exponential forms.

2. Find.

(i) $9^{\frac{3}{2}}$

Answer

We have
$9^{\frac{3}{2}}$

= ${(3^{2})}^{\frac{3}{2}}$ ... $[∵ 9 = 3 \times 3]$

3^{2 \times \frac{3}{2}}

...

[∵ {(a^{m})}^{n} = a^{m n}]

= $3^{3}$

= 27.

(ii) $32^{\frac{2}{5}}$

Answer

We have
$32^{\frac{2}{5}}$

= ${(2^{5})}^{\frac{2}{5}}$ ... $[∵ 32 = 2 \times 2 \times 2 \times 2 \times 2]$

= $2^{5 \times \frac{2}{5}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $2^{2}$

= 4.

(iii) $16^{\frac{3}{4}}$

Answer

We have
$16^{\frac{3}{4}}$

= ${(2^{4})}^{\frac{3}{4}}$ ... $[∵ 16 = 2 \times 2 \times 2 \times 2]$

= $2^{4 \times \frac{3}{4}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $2^{3}$

= $8$

(iv) $125^{\frac{- 1}{3}}$

Answer

We have
$125^{\frac{- 1}{3}}$

= ${(5^{3})}^{\frac{- 1}{3}}$ ... $[∵ 125 = 5 \times 5 \times 5]$

= $5^{3 \times \frac{- 1}{3}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $5^{- 1}$

= $\frac{1}{5}$ ... $[∵ a^{- 1} = \frac{1}{a}]$ .

In Exercise 1.5 NCERT Class 9, the third question asks us to simplify a given expression.

3. Simplify.

(i) $2^{\frac{2}{3}} . 2^{\frac{1}{5}}$

Answer

We have
$2^{\frac{2}{3}} . 2^{\frac{1}{5}}$

= $2^{\frac{2}{3} + \frac{1}{5}}$ ... [ $∵ a^{m} \times a^{n} = a^{m + n}$ ]

= $2^{\frac{10 + 3}{15}}$

= $2^{\frac{13}{15}}$ .

(ii) ${(\frac{1}{3^{3}})}^{7}$

Answer

We have
${(\frac{1}{3^{3}})}^{7}$

= ${(3^{- 3})}^{7}$ ... $[∵ a^{- n} = \frac{1}{a^{n}}]$

= $3^{- 3 \times 7}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $3^{- 21}$ .

(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Answer

We have
$\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

= $11^{\frac{1}{2} - \frac{1}{4}}$ ... $[∵ \frac{a^{m}}{a^{n}} = a^{m - n}]$

= $11^{\frac{2 - 1}{4}}$

= $11^{\frac{1}{4}}$ .

(iv) $7^{\frac{1}{2}} . 8^{\frac{1}{2}}$

Answer

We have
$7^{\frac{1}{2}} . 8^{\frac{1}{2}}$

= ${(7 \times 8)}^{\frac{1}{2}}$ ... $[∵ a^{m} \times b^{m} = {(a b)}^{m}]$

= $56^{\frac{1}{2}}$ .

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Laws of Exponents:

Exercise 1.5 NCERT Class 9 Solutions

1. Find.

(i) 6412

Answer

(ii) 3215

Answer

(iii) 12513

Answer

2. Find.

(i) 932

Answer

(ii) 3225

Answer

(iii) 1634

Answer

(iv) 125−13

Answer

3. Simplify.

(i) 223.215

Answer

(ii) (133)7

Answer

(iii) 11121114

Answer

(iv) 712.812

Answer

Important Links

(i) $64^{\frac{1}{2}}$

(ii) $32^{\frac{1}{5}}$

(iii) $125^{\frac{1}{3}}$

(i) $9^{\frac{3}{2}}$

(ii) $32^{\frac{2}{5}}$

(iii) $16^{\frac{3}{4}}$

(iv) $125^{\frac{- 1}{3}}$

(i) $2^{\frac{2}{3}} . 2^{\frac{1}{5}}$

(ii) ${(\frac{1}{3^{3}})}^{7}$

(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

(iv) $7^{\frac{1}{2}} . 8^{\frac{1}{2}}$