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

## RD Sharma Solutions Class 9 Chapter 2 Exercise 2.1

### RD Sharma Class 9 Solutions Chapter 2 Ex 2.1 Free Download

Level 1

1. Simplify the following:

(i) 3(a4 b3)10 x 5 (a2 b2)3

Solution:

= 3(a40 b30) x 5 (a6 b6)

= 15 (a46 b36)

(ii) (2x -2 y3)3

Solution:

= (23 x -2×3 y3×3)

= 8x -6 y9

(iii) $\frac{(4\times 10^{7})(6\times 10^{-5})}{8 \times 10^{4}}$

Solution:

$\frac{(4\times 10^{7})(6\times 10^{-5})}{8 \times 10^{4}}$

= $\frac{(24\times 10^{7}\times 10^{-5})}{8 \times 10^{4}}$

= $\frac{(24\times 10^{7-5})}{8 \times 10^{4}}$

= $\frac{(24\times 10^{2})}{8 \times 10^{4}}$

= $\frac{(3\times 10^{2})}{10^{4}}$

= $\frac{3}{100}$

(iv) $\frac{4ab^{2}(-5ab^{3})}{10a^{2}b^{2}}$

Solution:

= $\frac{-20a^{2}b^{5}}{10a^{2}b^{2}}$

= $-2b^{3}$

(v) $(\frac{x^{2}y^{2}}{a^{2}b^{3}})^{n}$

Solution:

= $\frac{x^{2n}y^{2n}}{a^{2n}b^{3n}}$

(vi) $\frac{(a^{3n-9})^{6}}{a^{2n-4}}$

Solution:

= $\frac{a^{18n-54}}{a^{2n-4}}$

= $a^{18n-2n-54+4}$

= $a^{16n-50}$

2. If a = 3 and b = -2, find the values of:

(i) $a^{a} + b^{b}$

(ii) $a^{b} + b^{a}$

(iii) $a^{b} + b^{a}$

Solution:

(i) We have,

$a^{a} + b^{b}$

= $3^{3} + (-2)^{-2}$

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

= $27 + \frac{1}{4}$

= $\frac{109}{4}$

(ii) $a^{b} + b^{a}$

= $3^{-2} + (-2)^{3}$

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

= $\frac{1}{9} – 8$

= $-\frac{71}{9}$

(iii) We have,

$a^{b} + b^{a}$

= $(3 + (-2))^{3(-2)}$

= $(3 – 2))^{-6}$

= $1^{-6}$

= 1

3.Prove that:

(i) $(\frac{x^{a}}{x^{b}})^{a^{2}+ab+b^{2}}\times (\frac{x^{b}}{x^{c}})^{b^{2}+bc+c^{2}}\times (\frac{x^{c}}{x^{a}})^{c^{2}+ca+a^{2}}$ = 1

(ii) $(\frac{x^{a}}{x^{-b}})^{a^{2}-ab+b^{2}}\times (\frac{x^{b}}{x^{-c}})^{b^{2}-bc+c^{2}}\times (\frac{x^{c}}{x^{-a}})^{c^{2}-ca+a^{2}}$ = $x^{2(a^{3}+b^{3}+c^{3})}$

(iii) $(\frac{x^{a}}{x^{b}})^{c}\times (\frac{x^{b}}{x^{c}})^{a}\times (\frac{x^{c}}{x^{a}})^{b}$ = 1

Solution:

(i) To prove

$(\frac{x^{a}}{x^{b}})^{a^{2}+ab+b^{2}}\times (\frac{x^{b}}{x^{c}})^{b^{2}+bc+c^{2}}\times (\frac{x^{c}}{x^{a}})^{c^{2}+ca+a^{2}}$ = 1

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

• $\frac{x^{a^{3}+a^{2}b+ab^{2}}}{x^{a^{2}b+ab^{2}+b^{3}}}\times\frac{x^{b^{3}+b^{2}c+bc^{2}}}{x^{b^{2}c+bc^{2}+c^{3}}}\times\frac{x^{c^{3}+c^{2}a+ca^{2}}}{x^{c^{2}a+ca^{2}+a^{3}}}$
• $x^{a^{3}+a^{2}b+ab^{2}-(b^{3}+a^{2}b+ab^{2})}\times x^{b^{3}+b^{2}c+bc^{2}-(c^{3}+b^{2}c+bc^{2})}\times x^{c^{3}+c^{2}a+ca^{2}-(a^{3}+c^{2}a+ca^{2})}$
• $x^{a^{3}-b^{3}}\times x^{b^{3}-c^{3}}\times x^{c^{3}-a^{3}}$
• $x^{a^{3}-b^{3} + b^{3}-c^{3}+c^{3}-a^{3}}$
• $x^{0}$
• 1

Or,

Therefore, LHS = RHS

Hence proved

(ii) To prove,

$(\frac{x^{a}}{x^{-b}})^{a^{2}-ab+b^{2}}\times (\frac{x^{b}}{x^{-c}})^{b^{2}-bc+c^{2}}\times (\frac{x^{c}}{x^{-a}})^{c^{2}-ca+a^{2}}$ = $x^{2(a^{3}+b^{3}+c^{3})}$

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

• $x^{(a+b)(a^{2}-ab+b^{2})}\times x^{(b+c)(b^{2}-bc+c^{2})}\times x^{(c+a)(c^{2}-ca+a^{2})}$
• $x^{a^{3}+b^{3}} \times x^{b^{3}+c^{3}}\times x^{c^{3}+a^{3}}$
• $x^{a^{3}+b^{3}+b^{3}+c^{3}+c^{3}+a^{3}}$
• $x^{2(a^{3}+b^{3}+c^{3})}$

Therefore, LHS = RHS

Hence proved

(iii) To prove,

$(\frac{x^{a}}{x^{b}})^{c}\times (\frac{x^{b}}{x^{c}})^{a}\times (\frac{x^{c}}{x^{a}})^{b}$ =1

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

• $(\frac{x^{ac}}{x^{bc}})\times (\frac{x^{ba}}{x^{ca}})\times (\frac{x^{bc}}{x^{ab}})$
• $x^{ac – bc}\times x^{ba-ca}\times x^{bc-ab}$
• $x^{ac – bc + ba – ca + bc-ab}$
• $x^{0}$
• 1

Therefore, LHS = 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 (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, LHS = 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, LHS = RHS

Hence proved

5.Prove that:

(i) $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$ = abc

(ii) $(a^{-1}+b^{-1})^{-1}$

Solution:

(i) To prove,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$ = abc

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

• $\frac{a+b+c}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}$
• $\frac{a+b+c}{\frac{a+b+c}{abc}}$
• abc

Therefore, LHS = RHS

Hence proved

(ii) To prove,

$(a^{-1}+b^{-1})^{-1}$ = $\frac{ab}{a+b}$

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

• $\frac{1}{(a^{-1}+b^{-1})}$
• $\frac{1}{(\frac{1}{a}+\frac{1}{b})}$
• $\frac{1}{(\frac{a+b}{ab})}$
• $\frac{ab}{a+b}$

Therefore, LHS = RHS

Hence proved

6. If abc = 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 (LHS) = Right hand side (RHS)

Considering LHS,

• $\frac{1}{1+a+\frac{1}{b}} + \frac{1}{1+b+\frac{1}{c}} + \frac{1}{1+c+\frac{1}{a}}$
• $\frac{b}{b+ab+1}+\frac{c}{c+bc+1}+\frac{a}{a+ac+1}$ ….(1)

We know abc = 1

• c = $\frac{1}{ab}$

By substituting the value c in equation (1), we get

• $\frac{b}{b+ab+1}+\frac{\frac{1}{ab}}{\frac{1}{ab}+b(\frac{1}{ab})+1}+\frac{a}{a+a(\frac{1}{ab})+1}$
• $\frac{b}{b+ab+1}+\frac{\frac{1}{ab}\times ab}{1+b+ab}+\frac{ab}{1+ab+b}$
• $\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{1+ab+b}$
• $\frac{1+ab+b}{b+ab+1}$
• 1

Therefore, LHS = RHS

Hence proved

7. Simplify:

(i) $\frac{3^{n}\times 9^{n+1}}{3^{n-1}\times 9^{n-1}}$

Solution:

= $\frac{3^{n}\times 9^{n}\times 9}{\frac{3^{n}}{3}\times \frac{9^{n}}{9}}$

= 9 x 3 x 9

= 243

(ii) $\frac{(5\times 25^{n+1})(25\times 5^{2n})}{(5\times 5^{2n+3})-(25)^{n+1}}$

Solution:

= $\frac{(5\times 25^{n}\times 25)-(25\times 25^{n})}{(5\times 25^{n}\times 125)(25^{n}\times 25)}$

= $\frac{25^{n}\times 25(5-1)}{25^{n}\times 25(25-1)}$

= $\frac{4}{24}$

= $\frac{1}{6}$

(iii) $\frac{(5^{n+3})-(6\times 5^{n+1})}{(9\times 5^{n})-(2^{2}\times 5^{n})}$

Solution:

= $\frac{(5^{n+3})-(6\times 5^{n+1})}{(9\times 5^{n})-(2^{2}\times 5^{n})}$

= $\frac{(5^{n}\times 5^{3})-(6\times 5^{n}\times 5)}{(9\times 5^{n})-(2^{2}\times 5^{n})}$

= $\frac{5^{n}(125-30)}{5^{n}(9-4)}$

= $\frac{95}{5}$

= 19

(iv) $\frac{(6\times 8^{n+1})+(16\times 2^{3n-2})}{(10\times 2^{3n+1})-7\times (8)^{n}}$

Solution:

= $\frac{(6\times 8^{n}\times 8)+(16\times 8^{n}\times \frac{1}{4})}{(10\times 8^{n}\times 2)-(7\times (8)^{n})}$

= $\frac{8^{n}(48+4)}{8^{n}(20-7)}$

= $\frac{52}{13}$

= 4

Level 2

8. Solve the following equations for x:

(i) $7^{2x+3}$ = 1

(ii) $2^{x+1}= 4^{x-3}$

(iii) $2^{5x+3}= 8^{x+3}$

(iv) $4^{2x}= \frac{1}{32}$

(v) $4^{x-1}\times (0.5)^{3-2x} = (\frac{1}{8})^{x}$

(vi) $2^{3x-7}$ = 256

Solution:

(i) We have,

=> $7^{2x+3}$ = 1

=> $7^{2x+3}$ = 70

=> 2x + 3 = 0

=> 2x = -3

=> x = $-\frac{3}{2}$

(ii) We have,

• $2^{x+1}= 4^{x-3}$
• $2^{x+1}= 2^{2x-6}$
• x + 1 = 2x – 6
• x = 7

(iii) We have,

• $2^{5x+3}= 8^{x+3}$
• $2^{5x+3}= 2^{3x+9}$
• 5x + 3 = 3x + 9
• 2x = 6
• x = 3

(iv) We have,

• $4^{2x}= \frac{1}{32}$
• $2^{4x}= \frac{1}{2^{5}}$
• $2^{4x}= 2^{-5}$
• 4x = -5
• x = $\frac{-5}{4}$

(v) We have,

• $4^{x-1}\times (0.5)^{3-2x} = (\frac{1}{8})^{x}$
• $2^{2x-2}\times (\frac{1}{2})^{3-2x} = (\frac{1}{2})^{3x}$
• $2^{2x-2}\times 2^{2x-3} = (\frac{1}{2})^{3x}$
• $2^{2x-2+2x-3} = (\frac{1}{2})^{3x}$
• $2^{4x-5} = 2^{-3x}$
• 4x-5 = -3x
• 7x = 5
• x = $\frac{5}{7}$

(vi) $2^{3x-7}$ = 256

• $2^{3x-7}$ = $2^{8}$
• 3x – 7 = 8
• 3x = 15
• x = 5

9. Solve the following equations for x:

(i) $2^{2x}-2^{x+3}+2^{4}=0$

(ii) $3^{2x+4}+1 = 2\times 3^{x+2}$

Solution:

(i) We have,

=> $2^{2x}-2^{x+3}+2^{4}=0$

=>  $2^{2x}+2^{4}=2^{x}.2^{3}$

=> Let $2^{x}= y$

=> $y^{2} + 2^{4} = y\times 2^{3}$

=>  $y^{2}-8y+16 = 0$

=> $y^{2}-4y-4y+16 = 0$

=> y(y-4) -4(y-4) = 0

=> y = 4

=> $x^{2} = 2^{2}$

=> x = 2

(ii) We have,

• $3^{2x+4}+1 = 2\times 3^{x+2}$
• $(3^{x+2})^{2}+1 = 2\times 3^{x+2}$
• Let $3^{x+2}$ = y
• $y^{2}+1=2y$
• $y^{2}-2y+1=0$
• $y^{2}-y-y+1=0$
• $y(y-1)-1(y-1)=0$
• $(y-1)(y-1)=0$
• y = 1

10. If 49392 = $a^{4}b^{2}c^{3}$, find the values of a, b and c, where a, b and c, where a, b, and 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 = $2^{a}\times 3^{b}\times 7^{c}$, Find a, b, and 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 = $3^{a}\times 5^{b}\times 7^{c}$, find

(i) The integral values of a, b and c

Solution:

Taking out the LCM of 4725, we get

• $3^{3}\times 5^{2}\times 7^{1}$ = $3^{a}\times 5^{b}\times 7^{c}$

By comparing, we get

• a = 3, b = 2 and c = 1.

(ii) The value of $2^{-a}\times 3^{b}\times 7^{c}$

Solution:

• $2^{-a} \times 3^{b}\times 7^{c} = \(2^{-3}\times 3^{2}\times 7^{1}$$2^{-3}\times 3^{2}\times 7^{1}\]”> • \(2^{-3}\times 3^{2}\times 7^{1}$ = $\frac{1}{8}\times 9\times 7$
• $\frac{63}{8}$

13. If a = $xy^{p-1}$, b = $xy^{q-1}$ and c = $xy^{r-1}$, prove that $a^{q-r}b^{r-p}c^{p-q}$ = 1

Solution:

Given,

a = $xy^{p-1}$, b = $xy^{q-1}$ and c = $xy^{r-1}$

To prove, $a^{q-r}b^{r-p}c^{p-q}$ = 1

Left hand side (LHS) = Right hand side (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

= $(xy^{p-1})^{q-r}(xy^{q-1})^{r-p}(xy^{r-1})^{p-q}$

= $xy^{pq-pr-q+r}xy^{qr-pq-r+p}xy^{rp-rq-p+q}$

= $xy^{pq-pr-q+r+qr-pq-r+p+rp-rq-p+q}$

= $xy^{0}$

= 1

#### Practise This Question

In the list of counting numbers, 1 is the smallest number. State true or false.