# RD Sharma Class 9 Exponents of Real Numbers Exercise 2.1 Solutions

On this page you will find Maths RD Sharma Class 9 Exponents of Real Numbers Exercise 2.1 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 2 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Exponents of Real Numbers Exercise 2.1 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 2 Exponents of Real Numbers Exercise 2.1 below and it will be beneficial for them.

## RD Sharma Class 9 Exponents of Real Numbers Exercise 2.1 Solutions

1. Simplify the following:
(i) $3\left(a^{4} b^{3}\right)^{10} \times 5\left(a^{2} b^{2}\right)^{3}$
Solution:
$=3\left(a^{40} b^{30}\right) \times 5\left(a^{6} b^{6}\right)$
$=15\left(a^{46} b^{36}\right)$
(ii) $\left(2 x^{-2} y^{3}\right)^{3}$
Solution:
$=\left(2^{3} x^{-2 \times 3} y^{3 \times 3}\right)$
$=8 x^{-6} y^{9}$
(iii) $\frac{\left(4 \times 10^{7}\right)\left(6 \times 10^{-5}\right)}{8 \times 10^{4}}$
Solution:
$\frac{\left(4 \times 10^{7}\right)\left(6 \times 10^{-5}\right)}{8 \times 10^{4}}$
$=\frac{\left(24 \times 10^{7} \times 10^{-5}\right)}{8 \times 10^{4}}$
$=\frac{\left(24 \times 10^{7-5}\right)}{8 \times 10^{4}}$
$=\frac{\left(24 \times 10^{2}\right)}{8 \times 10^{4}}$
$=\frac{\left(3 \times 10^{2}\right)}{10^{4}}$
$=\frac{3}{100}$
(iv) $\frac{4 a b^{2}\left(-5 a b^{3}\right)}{10 a^{2} b^{2}}$
Solution:
$=\frac{-20 a^{2} b^{5}}{10 a^{2} b^{2}}$
$=-2 b^{3}$
(v) $\left(\frac{x^{2} y^{2}}{a^{2} b^{3}}\right)^{n}$
Solution:
$=\frac{x^{2 n} y^{2 n}}{a^{2 n} b^{3 n}}$
(vi) $\frac{\left(a^{3 n-9}\right)^{6}}{a^{2 n-4}}$
Solution:
$=\frac{\mathrm{a}^{18 \mathrm{n}-54}}{\mathrm{a}^{2 \mathrm{n}-4}}$
$=\mathrm{a}^{18 \mathrm{n}-2 \mathrm{n}-54+4}$
$=\mathrm{a}^{16 \mathrm{n}-50}$

2. If $\mathrm{a}=3$ and $\mathrm{b}=-2$, find the values of:
(i) $\mathrm{a}^{\mathrm{a}}+\mathrm{b}^{\mathrm{b}}$
(ii) $a^{b}+b^{a}$
(iii) $a^{b}+b^{a}$
Solution:
(i) We have,
$a^{a}+b^{b}$
$=3^{3}+(-2)^{-2}$
$=3^{3}+\left(-\frac{1}{2}\right)^{2}$
$=27+\frac{1}{4}$
$=\frac{109}{4}$
(ii) $\mathrm{a}^{\mathrm{b}}+\mathrm{b}^{\mathrm{a}}$
$=3^{-2}+(-2)^{3}$
$=\left(\frac{1}{3}\right)^{2}+(-2)^{3}$
$=\frac{1}{9}-8$
$=-\frac{71}{9}$
(iii) We have,
$a^{b}+b^{a}$
$=(3+(-2))^{3(-2)}$
$=(3-2))^{-6}$
$=1^{-6}$
$=1$

3. Prove that:
(i) $\left(\frac{x^{a}}{x^{b}}\right)^{a^{2}+a b+b^{2}} \times\left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+b c+c^{2}} \times\left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+c a+a^{2}}=1$
(ii) $\left(\frac{x^{a}}{x^{-b}}\right)^{a^{2}-a b+b^{2}} \times\left(\frac{x^{b}}{x^{-c}}\right)^{b^{2}-b c+c^{2}} \times\left(\frac{x^{c}}{x^{-a}}\right)^{c^{2}-c a+a^{2}}=x^{2\left(a^{3}+b^{3}+c^{3}\right)}$
(iii) $\left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1$
Solution:
(i) To prove
$\left(\frac{x^{a}}{x^{b}}\right)^{a^{2}+a b+b^{2}} \times\left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+b c+c^{2}} \times\left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+c a+a^{2}}=1$
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
$\frac{x^{a^{3}+a^{2} b+a b^{2}}}{x^{a^{2} b+a b^{2}+b^{3}}}$ $\times \frac{x^{b^{3}+b^{2} c+b c^{2}}}{x^{b^{2} c+b c^{2}+c^{3}}}$ $\times \frac{x^{c^{3}+c^{2} a+c a^{2}}}{x^{c^{2} a+c a^{2}+a^{3}}}$
$x^{a^{3}+a^{2} b+a b^{2}-\left(b^{3}+a^{2} b+a b^{2}\right)}$ $\times x^{\hat{b}^{3}+b^{2} c+b c^{2}-\left(c^{3}+b^{2} c+b c^{2}\right)}$ $\times x^{c^{3}+c^{2} a+c a^{2}-\left(a^{3}+c^{2} a+c a^{2}\right)}$
$x^{a^{3}-b^{3}} \times x^{b^{3}-c^{3}} \times x^{c^{3}-a^{3}}$
$x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}}$
$\mathrm{X}^{0}$
1
Or,
Therefore, $L H S=R H S$
Hence proved
(ii) To prove,
$\left(\frac{x^{a}}{x^{-b}}\right)^{a^{2}-a b+b^{2}} \times\left(\frac{x^{b}}{x^{-c}}\right)^{b^{2}-b c+c^{2}} \times\left(\frac{x^{c}}{x^{-a}}\right)^{c^{2}-c a+a^{2}}=x^{2\left(a^{3}+b^{3}+c^{3}\right)}$
Left hand side ( $(\mathrm{LHS})=$ Right hand side (RHS)
Considering $\mathrm{LHS}$,
$x^{(a+b)\left(a^{2}-a b+b^{2}\right)} \times x^{(b+c)\left(b^{2}-b c+c^{2}\right)} \times x^{(c+a)\left(c^{2}-c a+a^{2}\right)}$
$x^{a^{3}+b^{3}} \times x^{b^{3}+c^{3}} \times x^{c^{3}+a^{3}}$
$x^{a^{3}+b^{3}+b^{3}+c^{3}+c^{3}+a^{3}}$
$x^{2\left(a^{3}+b^{3}+c^{3}\right)}$
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved
(iii) To prove,
$\left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1$
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
$\left(\frac{x^{a c}}{x^{b c}}\right) \times\left(\frac{x^{b a}}{x^{c a}}\right) \times\left(\frac{x^{b c}}{x^{a b}}\right)$
$x^{a c-b c} \times x^{b a-c a} \times x^{b c-a b}$
$x^{a c-b c+b a-c a+b c-a b}$
$x^{0}$
1
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved

4. Prove that:
(i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$
(ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}$
Solution:
(i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$
Left hand side $(\mathrm{LHS})=$ Right hand side (RHS)
Considering LHS,
$\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$
$\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$
$\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$
1
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved
(ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}$
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
$\frac{1}{1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}}+\frac{1}{1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}}$
$\frac{x^{a}}{x^{a}+x^{b}+x^{c}}+\frac{x^{b}}{x^{b}+x^{a}+x^{c}}+\frac{x^{c}}{x^{c}+x^{b}+x^{a}}$
$\frac{x^{a}+x^{b}+x^{c}}{x^{a}+x^{b}+x^{c}}$
1
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved

5. Prove that:
(i) $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=a b c$
(ii) $\left(a^{-1}+b^{-1}\right)^{-1}$
Solution:
(i) To prove,
$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=a b c$
Left hand side $(\mathrm{LHS})=$ Right hand side (RHS)
Considering LHS,
$\frac{a+b+c}{\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}}$
$\frac{a+b+c}{\frac{a+b+c}{a b c}}$
Abc
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved
(ii) To prove,
$\left(a^{-1}+b^{-1}\right)^{-1}=\frac{a b}{a+b}$
Left hand side $(\mathrm{LHS})=$ Right hand side (RHS)
Considering LHS,
$\frac{1}{\left(a^{-1}+b^{-1}\right)}$
$\frac{1}{\left(\frac{1}{a}+\frac{1}{b}\right)}$
$\frac{1}{\left(\frac{a+b}{a b}\right)}$
$\frac{a b}{a+b}$
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved

6. If $a b c=1$, show that $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$
Solution:
To prove,
$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$
Left hand side $(\mathrm{LHS})=$ Right hand side (RHS)
Considering LHS,
$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$
$\frac{b}{b+a b+1}+\frac{c}{c+b c+1}+\frac{a}{a+a c+1}$     ….(1)
We know $a b c=1$
$c=\frac{1}{a b}$
By substituting the value $c$ in equation $(1)$, we get
$\frac{\mathrm{b}}{\mathrm{b}+\mathrm{ab}+1}+\frac{\frac{1}{\mathrm{ab}}}{\frac{1}{\mathrm{ab}}+\mathrm{b}\left(\frac{1}{\mathrm{ab}}\right)+1}+\frac{\mathrm{a}}{\mathrm{a}+\mathrm{a}\left(\frac{1}{\mathrm{ab}}\right)+1}$
$\frac{b}{b+a b+1}+\frac{\frac{1}{a b} \times a b}{1+b+a b}+\frac{a b}{1+a b+b}$
$\frac{b}{b+a b+1}+\frac{1}{1+b+a b}+\frac{a b}{1+a b+b}$
$\frac{1+a b+b}{b+a b+1}$
1
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved

7. Simplify:
(i) $\frac{3^{\mathrm{n}} \times 9^{\mathrm{n}+1}}{3^{\mathrm{n}-1} \times 9^{\mathrm{n}-1}}$
Solution:
$=\frac{3^{\mathrm{n}} \times 9^{\mathrm{n}} \times 9}{\frac{3^{n}}{3} \times \frac{9^{n}}{9}}$
$=9 \times 3 \times 9$
$=243$
(ii) $\frac{\left(5 \times 25^{\mathrm{n}+1}\right)\left(25 \times 5^{2 \mathrm{n}}\right)}{\left(5 \times 5^{2 \mathrm{n}+3}\right)-(25)^{\mathrm{n}+1}}$
Solution:
$=\frac{\left(5 \times 25^{n} \times 25\right)-\left(25 \times 25^{n}\right)}{\left(5 \times 25^{n} \times 125\right)\left(25^{n} \times 25\right)}$
$=\frac{25^{\mathrm{n}} \times 25(5-1)}{25^{\mathrm{n}} \times 25(25-1)}$
$=\frac{4}{24}$
$=\frac{1}{6}$
(iii) $\frac{\left(5^{n+3}\right)-\left(6 \times 5^{n+1}\right)}{\left(9 \times 5^{n}\right)-\left(2^{2} \times 5^{n}\right)}$
Solution:
$=\frac{\left(5^{n+3}\right)-\left(6 \times 5^{n+1}\right)}{\left(9 \times 5^{n}\right)-\left(2^{2} \times 5^{n}\right)}$
$=\frac{\left(5^{n} \times 5^{3}\right)-\left(6 \times 5^{n} \times 5\right)}{\left(9 \times 5^{n}\right)-\left(2^{2} \times 5^{n}\right)}$
$=\frac{5^{n}(125-30)}{5^{n}(9-4)}$
$=\frac{95}{5}$
$=19$
(iv) $\frac{\left(6 \times 8^{n+1}\right)+\left(16 \times 2^{3 n-2}\right)}{\left(10 \times 2^{3 n+1}\right)-7 \times(8)^{n}}$
Solution:
$=\frac{\left(6 \times 8^{\mathrm{n}} \times 8\right)+\left(16 \times 8^{\mathrm{n}} \times \frac{1}{4}\right)}{\left(10 \times 8^{\mathrm{n}} \times 2\right)-\left(7 \times(8)^{\mathrm{n}}\right)}$
$=\frac{8^{n}(48+4)}{8^{n}(20-7)}$
$=\frac{52}{13}$
$=4$

Level 2
8. Solve the following equations for $\mathrm{x}$ :
(i) $7^{2 x+3}=1$
(ii) $2^{x+1}=4^{x-3}$
(iii) $2^{5 x+3}=8^{x+3}$
(iv) $4^{2 \mathrm{x}}=\frac{1}{32}$
(v) $4^{\mathrm{x}-1} \times(0.5)^{3-2 \mathrm{x}}=\left(\frac{1}{8}\right)^{\mathrm{x}}$
(vi) $2^{3 x-7}=256$
Solution:
(i) We have,
$\Rightarrow 7^{2 x+3}=1$
$\Rightarrow 7^{2 x+3}=7^{0}$
$\Rightarrow 2 x+3=0$
$\Rightarrow 2 x=-3$
$\Rightarrow x=-\frac{3}{2}$
(ii) We have,
$2^{x+1}=4^{x-3}$
$2^{x+1}=2^{2 x-6}$
$x+1=2 x-6$
$x=7$
(iii) We have,
$2^{5 x+3}=8^{x+3}$
$2^{5 x+3}=2^{3 x+9}$
$5 x+3=3 x+9$
$2 x=6$
$x=3$
(iv) We have,
$4^{2 x}=\frac{1}{32}$
$2^{4 x}=\frac{1}{2^{5}}$
$2^{4 x}=2^{-5}$
$4 x=-5$
$x=\frac{-5}{4}$
(v) We have,
$4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x}$
$2^{2 x-2} \times\left(\frac{1}{2}\right)^{3-2 x}=\left(\frac{1}{2}\right)^{3 x}$
$2^{2 x-2} \times 2^{2 x-3}=\left(\frac{1}{2}\right)^{3 x}$
$2^{2 x-2+2 x-3}=\left(\frac{1}{2}\right)^{3 x}$
$2^{4 x-5}=2^{-3 x}$
$4 x-5=-3 x$
$7 x=5$
$x=\frac{5}{7}$
(vi) $2^{3 x-7}=256$
$2^{3 x-7}=2^{8}$
$3 x-7=8$
$3 x=15$
$x=5$

9. Solve the following equations for $\mathrm{X}$ :
(i) $2^{2 x}-2^{x+3}+2^{4}=0$
(ii) $3^{2 x+4}+1=2 \times 3^{x+2}$
Solution:
(i) We have,
$\Rightarrow 2^{2 x}-2^{x+3}+2^{4}=0$
$\Rightarrow 2^{2 x}+2^{4}=2^{x} \cdot 2^{3}$
$\Rightarrow$ Let $2^{\mathrm{x}}=\mathrm{y}$
$\Rightarrow \mathrm{y}^{2}+2^{4}=\mathrm{y} \times 2^{3}$
$\Rightarrow y^{2}-8 y+16=0$
$\Rightarrow y^{2}-4 y-4 y+16=0$
$\Rightarrow y(y-4)-4(y-4)=0$
$\Rightarrow y=4$
$\Rightarrow x^{2}=2^{2}$
$\Rightarrow x=2$
(ii) We have,
$3^{2 x+4}+1=2 \times 3^{x+2}$
$\left(3^{x+2}\right)^{2}+1=2 \times 3^{x+2}$
Let $3^{\mathrm{x}+2}=\mathrm{y}$
$\mathrm{y}^{2}+1=2 \mathrm{y}$
$y^{2}-2 y+1=0$
$y^{2}-y-y+1=0$
$y(y-1)-1(y-1)=0$
$(y-1)(y-1)=0$
$y=1$

10. If $49392=\mathrm{a}^{4} \mathrm{~b}^{2} \mathrm{c}^{3}$, find the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$, where $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$, where $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ are different positive primes.
Solution:
Taking out the LCM, the factors are $2^{4}, 3^{2}$ and $7^{3}$
$a^{4} b^{2} c^{3}=2^{4}, 3^{2}$ and $7^{3}$
$a=2, b=3$ and $c=7$ [Since, $a, b$ and $c$ are primes]

11. If $1176=2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 7^{\mathrm{c}}$, Find $\mathrm{a}, \mathrm{b}$, and $\mathbf{c}$.
Solution:
Given that 2,3 and 7 are factors of 1176 .
Taking out the LCM of 1176 , we get
$2^{3} \times 3^{1} \times 7^{2}=2^{a} \times 3^{b} \times 7^{c}$
By comparing, we get
$a=3, b=1$ and $c=2$

12. Given $4725=3^{\mathrm{a}} \times 5^{\mathrm{b}} \times 7^{\mathrm{c}}$, find
(i) The integral values of $\mathrm{a}, \mathbf{b}$ and $\mathbf{c}$
Solution:
Taking out the LCM of 4725 , we get
$3^{3} \times 5^{2} \times 7^{1}=3^{\mathrm{a}} \times 5^{\mathrm{b}} \times 7^{\mathrm{c}}$
By comparing, we get
$a=3, b=2$ and $c=1$
(ii) The value of $2^{-a} \times 3^{b} \times 7^{c}$
Solution:

13. If $\mathrm{a}=\mathrm{xy}^{\mathrm{p}-1}, \mathrm{~b}=\mathrm{xy}^{\mathrm{q}-1}$ and $\mathrm{c}=\mathrm{xy}^{\mathrm{r}-1}$, prove that $\mathrm{a}^{\mathrm{q}-\mathrm{r}} \mathrm{b}^{\mathrm{r}-\mathrm{p}} \mathrm{c}^{\mathrm{p}-\mathrm{q}}=1$
Solution:
Given,
$a=x y^{p-1}, b=x y^{q-1}$ and $c=x y^{r-1}$
To prove, $a^{q-r} b^{r-p} c^{p-q}=1$
Left hand side $(\mathrm{LHS})=$ Right hand side $(\mathrm{RHS})$
Considering LHS,
$=a^{q-r} b^{r-p} c^{p-q}$    ….(i)
By substituting the value of $a, b$ and $c$ in equation (i), we get
$=\left(\mathrm{xy}^{\mathrm{p}-1}\right)^{\mathrm{q} \mathrm{r}}\left(\mathrm{xy}^{\mathrm{q}-1}\right)^{\mathrm{r} \mathrm{p}}\left(\mathrm{xy}^{\mathrm{r}-1}\right)^{\mathrm{p}-\mathrm{q}}$
$=x y^{p q-p r-q+r} x y^{q r-p q-r+p} x y^{r p-r q-p+q}$
$=x y^{p q-p r-q+r+q r-p q-r+p+r p-r q-p+q}$
$=\mathrm{xy}^{0}$
$=1$

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