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.