Exercise 1.5 (Revised) - Chapter 1 - Number Systems - Ncert Solutions class 9 - Maths

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Chapter 1 - Number Systems | NCERT Solutions Class 9 Maths

Ex 1.5 Question 1.

Find:(i) $64^{\frac{1}{5}}$ (ii) $32^{\frac{1}{5}}$ (iii) $125^{\frac{1}{3}}$

Answer.

(i) $64^{\frac{1}{2}}$
We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We conclude that $64^{\frac{1}{2}}$ can also be written as $\sqrt[2]{64}=\sqrt[2]{8 \times 8}$
$
\sqrt[2]{64}=\sqrt[2]{8 \times 8}=8
$

Therefore, the value of $64^{\frac{1}{2}}$ will be 8 .
(ii) $32^{\frac{1}{5}}$

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We conclude that $32^{\frac{1}{5}}$ can also be written as $\sqrt[5]{32}=\sqrt[2]{2 \times 2 \times 2 \times 2 \times 2}$
$
\sqrt[5]{32}=\sqrt[2]{2 \times 2 \times 2 \times 2 \times 2}=2
$

Therefore, the value of $32^{\frac{1}{5}}$ will be 2 .
(iii) $125^{\frac{1}{3}}$

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We conclude that $125^{\frac{1}{3}}$ can also be written as $\sqrt[3]{125}=\sqrt[3]{5 \times 5 \times 5}$
$
\sqrt[3]{125}=\sqrt[3]{5 \times 5 \times 5}=5
$

Therefore, the value of $125^{\frac{1}{3}}$ will be 5 .

Ex 1.5 Question 2.

Find:(i) $9^{\frac{3}{2}}$
(ii) $32^{\frac{2}{5}}$
(iii) $16^{\frac{3}{4}}$
(iv) $125^{\frac{-1}{3}}$

Answer.

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

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.
$
\begin{aligned}
& \sqrt[2]{(9)^3}=\sqrt[2]{3 \times 3 \times 3 \times 3 \times 3 \times 3} \\
& =3 \times 3 \times 3 \\
& =27
\end{aligned}
$

Therefore, the value of $9^{\frac{3}{2}}$ will be 27 .
(ii) $32^{\frac{2}{5}}$

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We conclude that $32^{\frac{2}{5}}$ can also be written as $\sqrt[5]{(32)^2}=\sqrt[5]{(2 \times 2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2 \times 2)}=2 \times 2$ $=4$

Therefore, the value of $32^{\frac{2}{5}}$ will be 4 .
(iii) $16^{\frac{3}{4}}$

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We conclude that $16^{\frac{3}{4}}$ can also be written as $\sqrt[4]{(16)^3}=\sqrt[4]{(2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2)}$ $=2 \times 2 \times 2$
$
=8
$

Therefore, the value of $16^{\frac{3}{4}}$ will be 8 .
(iv) $125^{\frac{-1}{3}}$

We know that
$
a^{-n}=\frac{1}{a^n}
$

We conclude that $125^{\frac{-1}{3}}$ can also be written as $125^{\frac{1}{3}}$, or $\left(\frac{1}{125}\right)^{\frac{1}{3}}$.

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

We know that $\left(\frac{1}{125}\right)^{\frac{1}{3}}$ can also be written as $\sqrt[3]{\left(\frac{1}{125}\right)}=\sqrt[3]{\left(\frac{1}{5} \times \frac{1}{5} \times \frac{1}{5}\right)}$

$
=\frac{1}{5} \text {. }
$

Therefore, the value of $125^{\frac{-1}{3}}$ will be ${ }^{\frac{1}{5}}$.

Ex 1.5 Question 3.

Simplify:
(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
(ii) $\left(3^{\frac{1}{3}}\right)^7$
(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

Answer.

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

We know that $a^m \cdot a^n=a^{(m+n)}$.

We can conclude that $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{2}{3}+\frac{1}{5}}$.
$
2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{10+3}{15}}=(2)^{\frac{13}{15}}
$
(ii) $\left(3^{\frac{1}{3}}\right)^7$

We know that $a^m \times a^n=a^{m+n}$

We conclude that $\left(3^{\frac{1}{3}}\right)^7$ can also be written as $\left(3^{\frac{7}{3}}\right)$.
(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

We know that $\frac{a^m}{a^n}=a^{m-n}$

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

Therefore, the value of $11^{\frac{1}{4}}$ will be $11^{\frac{1}{2}}$.
(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

We know that $a^m \cdot b^m=(a \times b)^m$.

We can conclude that $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}$.
$
7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}
$

Therefore, the value of $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$ will be ${ }^{(56)^{\frac{1}{2}}}$