Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.1

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Question 1 (i). Simplify 3(a⁴b³)¹⁰× 5(a²b²)³

Solution:

Given 3(a⁴b³)¹⁰× 5(a²b²)³

= 3 × a⁴⁰× b³⁰× 5 × a⁶× b⁶

= 3 × a⁴⁶× b³⁶× 5 [a^m× aⁿ= a^m+n]

= 15 × a⁴⁶× b³⁶

= 15a⁴⁶b³⁶

Thus, 3(a⁴b³)¹⁰× 5(a²b²)³= 15a⁴⁶b³⁶

Question 1 (ii). Simplify (2x^-2y³)³

Solution:

Given (2x^-2y³)³

= 2³× x^-6× y⁹

= 8 × x^-6× y⁹ [a^m× aⁿ=a^m+n]

= 8x^-6y⁹

Thus, (2x^-2y³)³ = 8x^-6y⁹

Question 1 (iii). Simplify $\frac{(4×10^7)(6×10^{-5})}{8×10^4}$

Solution:

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

= $\frac{4×10^7×6×10^{-5}}{8×10^4}$

= $\frac{24×10^{7+(-5)}}{8×10^4}$ [a^m× aⁿ= a^m+n]

= $\frac{24×10^2}{8×10^4}$

= 3/10²

= 3/100

Thus, $\frac{(4×10^7)(6×10^{-5})}{8×10^4} =3/100$

Question 1 (iv). Simplify $\frac{4ab^2(-5ab^3)}{10a^2b^2}$

Solution:

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

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

= $\frac{-20×a×a×b^2×b^3}{10a^2b^2}$

= $\frac{-20×a^{1+1}×b^{2+3}}{10a^2b^2}$ [a^m× aⁿ= a^m+n]

= -2×a²×b⁵×a^-2×b^-2

= -2×a^2+(-2)×b^5+(-2) [a^m× aⁿ= a^m+n]

= -2×a⁰×b³

= -2b³ [a⁰=1]

Thus, $\frac{4ab^2(-5ab^3)}{10a^2b^2}$ =-2b³

Question 1 (v). Simplify $(\frac{x^2y^2}{a^2b^3})^n$

Solution:

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

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

= $\frac{x^{2n}y^{2n}}{a^{2n}b^{3n}}$ [a^m× aⁿ= a^m+n]

Thus, $(\frac{x^2y^2}{a^2b^3})^n =\frac{x^{2n}y^{2n}}{a^{2n}b^{3n}}$

Question 1 (vi). Simplify $\frac{(a^{3n-9})^6}{a^{2n-4}}$

Solution:

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

= $\frac{a^{6(3n-9)}}{a^{2n-4}}$ [(a^m)ⁿ= a^mn]

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

= a^18n-54 × a^-(2n-4) [a^m× aⁿ= a^m+n]

= a^18n-54-2n+4

= a^16n-50

Thus, $\frac{(a^{3n-9})^6}{a^{2n-4}}$ = a^16n-50

Question 2 (i) If a = 3 and b = -2,find the value of a^a+ b^b

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in a^a+ b^b, we get

a^a+ b^b= 3³+ (-2)^-2

= 27 + 1/4

= (108 + 1)/4

= 109/4

Thus, a^a+ b^b= 109/4

Question 2 (ii). If a = 3 and b = -2,find the value of a^b+ b^a

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in a^b+ b^a, we get

a^b+ b^a= 3^-2+ (-2)³

= 1/9 + (-8)

= (1 – 72)/9

= -71/9

Thus, a^b+ b^a= -71/9

Question 2 (iii). If a = 3 and b = -2,find the value of (a + b)^ab.

Solution:

Given a = 3 and b = -2

On substituting the value of a and b in (a + b)^ab, we get

(a + b)^ab= (3 + (-2))^3×-2

= (1)^-6

= 1

Thus, (a + b)^ab= 1

Question 3 (i). Prove that $(\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}=1$

Solution:

Let us first solve left-hand side of the given equation

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

By using the formula (a^m)ⁿ= a^mn, we get

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

By using the formula a^m/aⁿ= a^m-n, we get

= $x^{a(a^2+ab+b^2)-b(a^2+ab+b^2)}×x^{b(b^2+bc+c^2)-c(b^2+bc+c^2)}×x^{c(c^2+ca+a^2)-a(c^2+ca+a^2)}$

= $x^{(a-b)(a^2+ab+b^2)}×x^{(b-c)(b^2+bc+c^2)}×x^{(c-a)(c^2+ca+a^2)}$

= $x^{(a^3-b^3)}×x^{(b^3-c^3)}×x^{(c^3-a^3)}$

By using the formula a^m× aⁿ= a^m+n , we get

= $x^{(a^3-b^3+b^3-c^3+c^3-a^3)}$

= x

= 1

= Right-hand side of the given equation

Thus, we proved that $(\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}=1$

Question 3 (ii). Prove that $(\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b=1$

Solution:

Let us consider the left-hand side of the given equation

$(\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b$

By using the formula, (a^m)ⁿ= a^mn, we get

= $\frac{x^{ac}}{x^{bc}}×\frac{x^{ba}}{x^{ca}}×\frac{x^{cb}}{x^{ab}}$

= $\frac{x^{ac}×x^{ba}×x^{cb}}{x^{bc}×x^{ca}×x^{ab}}$

= $\frac{x^{ac+ba+bc}}{x^{bc+ca+ab}}$ [a^m× aⁿ= a^m+n]

= 1

= Right-hand side of the given equation

Thus, we proved that $(\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b=1$

Question 3 (iii). Prove that $(\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}$

Solution:

Let us first solve left-hand side of the given equation

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

By using the formula (a^m)ⁿ= a^mn, we get

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

By using the formula a^m/aⁿ= a^m-n, we get

= $x^{a(a^2-ab+b^2)+b(a^2-ab+b^2)}×x^{b(b^2-bc+c^2)+c(b^2-bc+c^2)}×x^{c(c^2-ca+a^2)+a(c^2-ca+a^2)}$

= $x^{(a+b)(a^2-ab+b^2)}×x^{(b+c)(b^2-bc+c^2)}×x^{(c+a)(c^2-ca+a^2)}$

= $x^{(a^3+b^3)}×x^{(b^3+c^3)}×x^{(c^3+a^3)}$

By using the formula a^m× aⁿ= a^m+n , we get

= $x^{(a^3+b^3+b^3+c^3+c^3+a^3)}$

= $x^{2(a^3+b^3+c^3)}$

= Right-hand side of the given equation

Thus, we proved that $(\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}$

Question 4 (i). Prove that $\frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}=1$

Solution:

Let us first consider the left-hand side of given equation

$\frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}$

= $\frac1{1+\frac{x^a}{x^b}}+\frac1{1+\frac{x^b}{x^a}}$

= $\frac1{\frac{x^b+x^a}{x^b}}+\frac1{\frac{x^a+x^b}{x^a}}$

= $\frac{x^b}{x^a+x^b}+\frac{x^a}{x^a+x^b}$

= $\frac{x^a+x^b}{x^b+x^a}$

= 1

= Right-hand side of the given equation

Thus, we proved that $\frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}=1$

Question 4 (ii). Prove that $\frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}=1$

Solution:

Let us first consider the left-hand side of given equation

$\frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}$

= $\frac1{1+\frac{x^b}{x^a}+\frac{x^c}{x^a}}+\frac1{1+\frac{x^a}{x^b}+\frac{x^c}{x^b}}+\frac1{1+\frac{x^b}{x^c}+\frac{x^a}{x^c}}$

= $\frac1{\frac{x^a+x^b+x^c}{x^a}}+\frac1{\frac{x^a+x^b+x^c}{x^b}}+\frac1{\frac{x^a+x^b+x^c}{x^c}}$

= $\frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^a+x^b+x^c}+\frac{x^c}{x^a+x^b+x^c}$

= $\frac{x^a+x^b+x^c}{x^a+x^b+x^c}$

= 1

= Right-hand side of the given equation

Thus, we proved that $\frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}=1$

Question 5 (i). Prove that $\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}=abc$

Solution:

Let us first consider the left-hand side of given equation

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

= $\frac{a+b+c}{\frac1{ab}+\frac1{bc}+\frac1{ca}}$

= $\frac{a+b+c}{\frac{a+b+c}{abc}}$

= abc

= Right hand side of the given equation

Thus, we proved that $\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}=abc$

Question 5 (ii). Prove that $(a^{-1}+b^{-1})^{-1}=\frac{ab}{a+b}$

Solution:

Let us first consider the left hand side of given equation

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

= $\frac1{a^{-1}+b^{-1}}$

= $\frac1{\frac1a+\frac1b}$

= $\frac1{\frac{b+a}{ab}}$

= $\frac{ab}{a+b}$

= Right hand side of the given equation

Thus, we proved that $(a^{-1}+b^{-1})^{-1}=\frac{ab}{a+b}$

Question 6. If abc = 1, show that $\frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}=1$

Solution:

Given abc = 1

⇒ c = 1/ab

Let us first consider the left-hand side of given equation

$\frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}$

= $\frac1{1+a+\frac1b}+\frac1{1+b+\frac1c}+\frac1{1+c+\frac1a}$

= $\frac1{\frac{b+ab+1}{b}}+\frac1{\frac{c+bc+1}c}+\frac1{\frac{a+ac+1}a}$

= $\frac{b}{b+ab+1}+\frac{c}{c+bc+1}+\frac{a}{a+ac+1}$

By substituting the value of c in above equation, we get

= $\frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac1{ab}+b(\frac1{ab})+1}+\frac{a}{a+a(\frac1{ab})+1}$

= $\frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac1{ab}+\frac{b}{ab}+\frac{ab}{ab}}+\frac{a}{\frac{ab}{b}+\frac1{b}+\frac{b}{b}}$

= $\frac{b}{b+ab+1}+\frac{\frac1{ab}}{\frac{1+b+ab}{ab}}+\frac{a}{\frac{ab+1+b}{b}}$

= $\frac{b}{b+ab+1}+\frac{\frac{ab}{ab}}{1+b+ab}+\frac{ab}{ab+b+1}$

= $\frac{b+1+ab}{b+ab+1}$

= 1

= Right hand side of the given equation

Thus, we have shown that if abc = 1, $\frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}=1$

Question 7 (i). Simplify $\frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}$

Solution:

Given $\frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}$

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

= $\frac{3^n×3^{2n+2}}{3^{n-1}×3^{2n-2}}$

= $\frac{3^{n+2n+2}}{3^{n-1+2n-2}}$ [a^m× aⁿ= a^m+n]

= $\frac{3^{3n+2}}{3^{3n-3}}$

= 3^3n+2-(3n-3) [a^m/aⁿ= a^m-n]

= 3⁵

= 243

Thus, $\frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}$ = 243

Question 7 (ii). Simplify $\frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}$

Solution:

Given $\frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}$

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

= $\frac{5×(5^{2n+2})-(5^2)×5^{2n}}{5×5^{2n+3}-5^{2n+2}}$

= $\frac{5^{1+2n+2}-5^{2+2n}}{5^{1+2n+3}-5^{2n+2}}$ [a^m× aⁿ= a^m+n]

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

= 4/24

= 1/6

Thus, $\frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}$ = 1/6

Question 7 (iii). Simplify $\frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}$

Solution:

Given, $\frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}$

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

= $\frac{5^n×5×(25-6)}{5^n(9-4)}$

= (19 × 5)/5

= 19

Thus, $\frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}=19$

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

Solution:

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

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

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

= $\frac{6×2^{3n}(2^3)+16(2^{3n})2^{-2}}{10(2)^{3n}(2^1)-7(2^{3n})}$

= $\frac{2^{3n}((6×2^3)+(16×\frac1{2^2}))}{2^{3n}((10×2)-7)}$

= $\frac{6×8+(16×\frac14)}{20-7}$

= (48 + 4)/13

= 52/13

= 4

Thus, $\frac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^n}=4$

Question 8 (i). Solve the equation 7^2x+3= 1 for x.

Solution:

Given equation 7^2x+3= 1

We know that, for any a∈ Real numbers, a⁰= 1

Let a = 7

⇒ 7^2x+3= 7⁰

Since the bases are equal, let us equate the exponents

⇒ 2x + 3 = 0

⇒ x = -3/2

Thus, the value of x is -3/2

Question 8 (ii). Solve the equation 2^x+1= 4^x-3 for x.

Solution:

Given 2^x+1= 4^x-3

We can write 4 = 2²

⇒ 2^x+1=2^2(x-3)

⇒ 2^x+1= 2^2x-6

Since the bases are equal, let us equate the exponents

⇒ x + 1 = 2x – 6

⇒ x = 7

Thus, the value of x is 7

Question 8 (iii). Solve the equation 2^5x+3= 8^x+3 for x.

Solution:

Given 2^5x+3= 8^x+3

We know that 8 = 2³

⇒ 2^5x+3= 2^3(x+3)

⇒ 2^5x+3= 2^3x+9

Since the bases are equal, let us equate the exponents

⇒ 5x + 3 = 3x + 9

⇒ 5x – 3x = 9 – 3

⇒ 2x = 6

⇒ x = 3

Thus, the value of x is 3

Question 8 (iv). Solve the equation 4^2x= 1/32 for x.

Solution:

Given 4^2x= 1/32

⇒ 2^2(2x)= 1/32

⇒ 2^2(2x)× 32 = 1

⇒ 2^4x× 2⁵= 1

⇒ 2^4x+5= 2⁰

Since the bases are equal, let us equate the exponents

⇒ 4x + 5 = 0

⇒ x = -5/4

Thus, the value of x is -5/4

Question 8 (v). Solve the equation 4^{x – 1}× (0.5)^3-2x = (1/8)^x for x.

Solution:

Given 4^{x – 1}× (0.5)^3-2x = (1/8)^x

⇒ $(2^2)^{x-1}×(\frac12)^{3-2x}=(\frac1{2^3})^x$

⇒ $(2^2)^{x-1}×(2^{-1})^{3-2x}=(2^{-3})^x$

⇒ 2^2(x-1)× 2^-(3-2x)= 2^-3x

⇒ 2^2x-2-3+2x= 2^-3x

⇒ 2^4x-5= 2^-3x

Since the bases are equal, let us equate the exponents

⇒ 4x – 5 = -3x

⇒ 7x = 5

⇒ x = 5/7

Thus, the value of x is 5/7

Question 8 (vi). Solve the equation 2^3x-7= 256 for x.

Solution:

Given 2^3x-7= 256

⇒ 2^3x-7= 2⁸

Since the bases are equal, let us equate the exponents

⇒ 3x – 7 = 8

⇒ x = 15/3

⇒ x = 5

Thus, the value of x is 5

Question 9 (i). Solve the equation 2^2x– 2^x+3+ 2⁴= 0 for x.

Solution:

Given 2^2x– 2^x+3+ 2⁴= 0

⇒ (2^x)²– 2 × 2^x× 2²+ (2²)²= 0

⇒ (2^x– 2²)²= 0

⇒ 2^x– 2²= 0

⇒ 2^x= 2²

Since the bases are equal, let us equate the exponents

⇒ x = 2

Thus, the value of x is 2

Question 9 (ii). Solve the equation 3^2x+4 + 1 = 2.3^x+2 for x.

Solution:

Given 3^2x+4 + 1 = 2.3^x+2

⇒ $(3^{x+2})^2+1=2.3^{x+2}$

⇒ $(3^{x+2})^2-2.3^{x+2}+1=0$

⇒ (3^x+2– 1)²= 0

⇒ 3^x+2– 1 = 0

⇒ 3^x+2= 3⁰

Since the bases are equal, let us equate the exponents

⇒ x + 2 = 0

⇒ x = -2

Thus, the value of x is -2

Question 10. If 49392 = a⁴b²c³, find the values of a, b, and c where a, b and c are different positive primes.

Solution:

Let us first find out prime factorization of 49392

Thus, 49392 = 2⁴× 3²× 7³

Where 2, 3 and 7 are positive primes

49392 = 2⁴3²7³= a⁴b²c³

Thus, on comparing, we get

a = 2,b = 3 and c = 7

Thus, the values of a, b and c are 2, 3, 7 respectively.

Question 11. If 1176 = 2^a3^b7^c, find a, b and c.

Solution:

Given 1176 = 2^a3^b7^c

Let us first find out prime factorization of 1176

Thus, 1176 = 2³× 3¹× 7²

1176 = 2³3¹7²= 2^a3^b7^c

Thus, on comparing, we get

a = 3, b = 1, c = 2

Thus, the values of a, b and c are 3, 1, 2 respectively.

Question 12. Given 4725 = 3^a5^b7^c, find

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

(ii) the value of 2^-a3^b7^c

Solution:

Given 4725 = 3^a5^b7^c

(i) Let us first find out prime factorization of 4725

Thus, 4725 = 3³× 5²× 7¹

4725 = 3³5²7¹= 3^a5^b7^c

Thus, on comparing, we get

a = 3,b = 2,c = 1

Thus, the values of a, b and c are 3,2,1 respectively.

(ii) Here a = 3, b = 2, c = 1

On substituting these values in 2^-a3^b7^c

2^-a3^b7^c= 2^-3×3²×7¹

= 1/8 × 9 × 7 = 63/8

Thus, the value of 2^-a3^b7^c is 63/8

Question 13. If a = xy^p-1, b = xy^q-1, c = xy^r-1, prove that a^q-rb^r-pc^p-q= 1.

Solution:

Given a = xy^p-1, b = xy^q-1, c = xy^r-1

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

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

= x^q-r+r-p+p-qy^{(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)}

= x^q-r+r-p+p-qy^{pq-q-pr+r+rq-r-pq+p+pr-p-qr+q}

= x⁰y⁰

= 1

Thus, we proved that a^q-rb^r-pc^p-q= 1

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Last Updated : 04 May, 2021

Class 9 RD Sharma Solutions - Chapter 1 Number System - Exercise 1.4

Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 1

Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.1

Question 1 (i). Simplify 3(a4b3)10 × 5(a2b2)3

Question 1 (ii). Simplify (2x-2y3)3

Question 1 (iii). Simplify

Question 1 (iv). Simplify

Question 1 (v). Simplify

Question 1 (vi). Simplify

Question 2 (i) If a = 3 and b = -2,find the value of aa + bb

Question 2 (ii). If a = 3 and b = -2,find the value of ab + ba

Question 2 (iii). If a = 3 and b = -2,find the value of (a + b)ab.

Question 3 (i). Prove that

Question 3 (ii). Prove that

Question 3 (iii). Prove that

Question 4 (i). Prove that

Question 4 (ii). Prove that

Question 5 (i). Prove that

Question 5 (ii). Prove that

Question 6. If abc = 1, show that

Question 7 (i). Simplify

Question 7 (ii). Simplify

Question 7 (iii). Simplify

Question 7 (iv). Simplify

Question 8 (i). Solve the equation 72x+3 = 1 for x.

Question 8 (ii). Solve the equation 2x+1 = 4x-3 for x.

Question 8 (iii). Solve the equation 25x+3 = 8x+3 for x.

Question 8 (iv). Solve the equation 42x = 1/32 for x.

Question 8 (v). Solve the equation 4x – 1 × (0.5)3-2x = (1/8)x for x.

Question 8 (vi). Solve the equation 23x-7 = 256 for x.

Question 9 (i). Solve the equation 22x – 2x+3 + 24 = 0 for x.

Question 9 (ii). Solve the equation 32x+4 + 1 = 2.3x+2 for x.

Question 1 (i). Simplify 3(a⁴b³)¹⁰× 5(a²b²)³

Question 1 (ii). Simplify (2x^-2y³)³

Question 1 (iii). Simplify $\frac{(4×10^7)(6×10^{-5})}{8×10^4}$

Question 1 (iv). Simplify $\frac{4ab^2(-5ab^3)}{10a^2b^2}$

Question 1 (v). Simplify $(\frac{x^2y^2}{a^2b^3})^n$

Question 1 (vi). Simplify $\frac{(a^{3n-9})^6}{a^{2n-4}}$

Question 2 (i) If a = 3 and b = -2,find the value of a^a+ b^b

Question 2 (ii). If a = 3 and b = -2,find the value of a^b+ b^a

Question 2 (iii). If a = 3 and b = -2,find the value of (a + b)^ab.

Question 3 (i). Prove that $(\frac{x^a}{x^b})^{a^2+ab+b^2}×(\frac{x^b}{x^c})^{b^2+bc+c^2}×(\frac{x^c}{x^a})^{c^2+ca+a^2}=1$

Question 3 (ii). Prove that $(\frac{x^a}{x^b})^c×(\frac{x^b}{x^c})^a×(\frac{x^c}{x^a})^b=1$

Question 3 (iii). Prove that $(\frac{x^a}{x^{-b}})^{a^2-ab+b^2}×(\frac{x^b}{x^{-c}})^{b^2-bc+c^2}×(\frac{x^c}{x^{-a}})^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}$

Question 4 (i). Prove that $\frac1{1+x^{a-b}}+\frac1{1+x^{b-a}}=1$

Question 4 (ii). Prove that $\frac1{1+x^{b-a}+x^{c-a}}+\frac1{1+x^{a-b}+x^{c-b}}+\frac1{1+x^{b-c}+x^{c-a}}=1$

Question 5 (i). Prove that $\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+a^{-1}c^{-1}}=abc$

Question 5 (ii). Prove that $(a^{-1}+b^{-1})^{-1}=\frac{ab}{a+b}$

Question 6. If abc = 1, show that $\frac1{1+a+b^{-1}}+\frac1{1+b+c^{-1}}+\frac1{1+c+a^{-1}}=1$

Question 7 (i). Simplify $\frac{3^n×9^{n+1}}{3^{n-1}×9^{n-1}}$

Question 7 (ii). Simplify $\frac{5×25^{n+1}-25×5^{2n}}{5×5^{2n+3}-25^{n+1}}$

Question 7 (iii). Simplify $\frac{5^{n+3}-6×5^{n+1}}{9×5^n-2^2×5^n}$

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

Question 8 (i). Solve the equation 7^2x+3= 1 for x.

Question 8 (ii). Solve the equation 2^x+1= 4^x-3 for x.

Question 8 (iii). Solve the equation 2^5x+3= 8^x+3 for x.

Question 8 (iv). Solve the equation 4^2x= 1/32 for x.

Question 8 (v). Solve the equation 4^{x – 1}× (0.5)^3-2x = (1/8)^x for x.

Question 8 (vi). Solve the equation 2^3x-7= 256 for x.

Question 9 (i). Solve the equation 2^2x– 2^x+3+ 2⁴= 0 for x.

Question 9 (ii). Solve the equation 3^2x+4 + 1 = 2.3^x+2 for x.

Question 10. If 49392 = a⁴b²c³, find the values of a, b, and c where a, b and c are different positive primes.

Question 11. If 1176 = 2^a3^b7^c, find a, b and c.

Question 12. Given 4725 = 3^a5^b7^c, find

(ii) the value of 2^-a3^b7^c

Question 13. If a = xy^p-1, b = xy^q-1, c = xy^r-1, prove that a^q-rb^r-pc^p-q= 1.