NCERT Solutions for Class 9 Maths Chapter 1 Number System Ex 1.5

Exercise 1.5

1.  Find:

(i) $ {6}{{ {4}}^{\dfrac{ {1}}{ {2}}}}$

Ans: The given number is \[{{64}^{\dfrac{1}{2}}}\].

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where$a>0$.

Therefore,

$64^{\frac{1}{2}}=\sqrt[2]{64}$

$=\sqrt[2]{8\times 8}$

$=8$

Hence, the value of ${{64}^{\dfrac{1}{2}}}$ is $8$.

(ii) $ {3}{{ {2}}^{\dfrac{ {1}}{ {5}}}}$

Ans: The given number is ${{32}^{\dfrac{1}{5}}}$.

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[m]{{{a}^{m}}}$, where $a>0$

$32^{\frac{1}{5}}=\sqrt[5]{32}$

$=\sqrt[5]{2\times 2\times 2\times 2\times 2}$

$=\sqrt[5]{2^{5}}$

$=2$

Alternative Method:

By the law of indices ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

$=32^{\frac{1}{5}}=(2\times 2\times 2\times 2\times 2)^{\frac{1}{5}}$

$=(2^{5})^{\frac{1}{5}}$

$=2^{\frac{5}{5}}$

$=2$

Hence, the value of the expression ${{32}^{\dfrac{1}{5}}}$ is $2$.

(iii) $ {12}{{ {5}}^{\dfrac{ {1}}{ {5}}}}$

Ans: The given number is ${{125}^{\dfrac{1}{3}}}$.

By the laws of indices

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where$a>0$.

Therefore,

$125^{\frac{1}{3}}=\sqrt[3]{125}$

$\sqrt[3]{5\times 5\times 5}$

$=5$

Hence, the value of the expression ${{125}^{\dfrac{1}{3}}}$ is $5$.

2.  Find:

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

Ans: The given number is ${{9}^{\dfrac{3}{2}}}$.

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where$a>0$.

Therefore,

$9^{\frac{3}{2}}=\sqrt[2]{(9)^{3}}$

$=\sqrt[2]{9\times 9\times 9}$

$=\sqrt[2]{3\times 3\times 3\times 3\times 3\times 3}$

$=3\times 3\times 3$

$=27$

Alternative Method:

By the laws of indices, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

$9^{\frac{3}{2}}=(3\times 3)^{\frac{3}{2}}$

$=(3^{2})^{\frac{3}{2}}$

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

$=3^{3}$

That is,

${{9}^{\dfrac{3}{2}}}=27$.

Hence, the value of the expression ${{9}^{\dfrac{3}{2}}}$ is $27$.

(ii) $ {3}{{ {2}}^{\dfrac{ {2}}{ {5}}}}$

Ans: We know that ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where $a>0$.

We conclude that ${{32}^{\dfrac{2}{5}}}$ can also be written as

$\sqrt[5]{(32)^{2}}=\sqrt[5]{(2\times 2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2\times 2)}$

$=2\times 2$

$=4$

Therefore, the value of ${{32}^{\dfrac{2}{5}}}$ is $4$.

(iii) $ {1}{{ {6}}^{\dfrac{ {3}}{ {4}}}}$

Ans: The given number is ${{16}^{\dfrac{3}{4}}}$.

By the laws of indices, 

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where $a>0$.

Therefore,

$16^{\frac{3}{4}}=\sqrt[4]{(16)^{3}}$

$=\sqrt[4]{(2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2)}$

$=2\times 2\times 2$

$=8$

Hence, the value of the expression ${{16}^{\dfrac{3}{4}}}$ is $8$.

Alternative Method:

By the laws of indices,

${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

Therefore,

$16^{\frac{3}{4}}=(4\times 4)^{\frac{3}{4}}$

$=(4^{2})^{\frac{3}{4}}$

$=(4)^{2\times \frac{3}{4}}$

$=(2^{2})^{2\times \frac{3}{4}}$

$=2^{2\times 2\times \frac{3}{4}}$

$=2^{3}$

$=8$

Hence, the value of the expression is ${{16}^{\dfrac{3}{4}}}=8$.

(iv) $ {12}{{ {5}}^{ {-}\dfrac{ {1}}{ {3}}}}$

Ans: The given number is ${{125}^{-\dfrac{1}{3}}}$.

By the laws of indices, it is known that 

${{a}^{-n}}=\dfrac{1}{{{a}^{^{n}}}}$, where $a>0$.

Therefore, 

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

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

$\sqrt[3]{(\frac{1}{125})}$

$\sqrt[3]{(\frac{1}{5}\times \frac{1}{5}\times \frac{1}{5})}$

$=\frac{1}{5}$

Hence, the value of the expression ${{125}^{-\dfrac{1}{3}}}$ is  $\dfrac{1}{5}$.

3. Simplify:

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

Ans: The given expression is ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$.

By the laws of indices, it is known that

${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$, where $a>0$.

Therefore,

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

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

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

Hence, the value of the expression ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$ is ${{2}^{\dfrac{13}{15}}}$.

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

Ans: The given expression is  ${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}}$.

It is known by the laws of indices that,

 ${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

Therefore,

${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}} =\left ( \dfrac{1}{3^{21}} \right )$

Hence, the value of the expression  ${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}}$ is  $\left ( \dfrac{1}{3^{21}} \right )$

(iii) $\dfrac{ {1}{{ {1}}^{\dfrac{ {1}}{ {2}}}}}{ {1}{{ {1}}^{\dfrac{ {1}}{ {4}}}}}$

Ans: The given number is $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$.

It is known by the Laws of Indices that

$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, where $a>0$.

Therefore,

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

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

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

Hence, the value of the expression $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$ is  ${{11}^{\dfrac{1}{4}}}$.

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

Ans: The given expression is ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$.

It is known by the Laws of Indices that

${{a}^{m}}\cdot {{b}^{m}}={{(a\cdot b)}^{m}}$, where $a>0$.

Therefore,

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

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

Hence, the value of the expression ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$ is ${{(56)}^{\dfrac{1}{2}}}$.

Conclusion

NCERT Solutions for Maths Exercise 1.5 Class 9 Chapter 1 - Number System helps you understand real numbers and their properties. Focus on rational and irrational numbers and their decimal expansions. These solutions show clear, step-by-step methods to solve each problem. Practicing these exercises will strengthen your understanding of number systems and help you do better in exams. Vedantu’s solutions are here to support your learning and build confidence in maths.

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