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

### Question 1 (i). Simplify 3(a^{4}b^{3})^{10 }× 5(a^{2}b^{2})^{3}

**Solution:**

Given 3(a

^{4}b^{3})^{10 }× 5(a^{2}b^{2})^{3}Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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^{40 }× b^{30 }× 5 × a^{6 }× b^{6}= 3 × a

^{46 }× b^{36 }× 5 [a^{m }× a^{n }= a^{m+n}]= 15 × a

^{46 }× b^{36}= 15a

^{46}b^{36}Thus, 3(a

^{4}b^{3})^{10 }× 5(a^{2}b^{2})^{3 }= 15a^{46}b^{36}

### Question 1 (ii). Simplify (2x^{-2}y^{3})^{3}

**Solution:**

Given (2x

^{-2}y^{3})^{3}= 2

^{3 }× x^{-6 }× y^{9}= 8 × x

^{-6 }× y^{9}[a^{m }× a^{n }=^{ }a^{m+n}]= 8x

^{-6}y^{9}Thus, (2x

^{-2}y^{3})^{3}= 8x^{-6}y^{9}

### Question 1 (iii). Simplify

**Solution:**

Given

=

= [a

^{m }× a^{n }= a^{m+n}]=

= 3/10

^{2}= 3/100

Thus,

### Question 1 (iv). Simplify

**Solution:**

Given

=

=

= [a

^{m }× a^{n }= a^{m+n}]= -2×a

^{2}×b^{5}×a^{-2}×b^{-2}= -2×a

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

^{0}×b^{3}= -2b

^{3}[a^{0}=1]Thus, =-2b

^{3}

### Question 1 (v). Simplify

**Solution:**

Given

=

=

[a^{m }× a^{n }= a^{m+n}]Thus,

### Question 1 (vi). Simplify

**Solution:**

Given

= [(a

^{m})^{n }= a^{mn}]=

= a

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

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

^{16n-50}Thus, = 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 geta

^{a }+ b^{b }= 3^{3 }+ (-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 geta

^{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

**Solution:**

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

By using the formula (a

^{m})^{n }= a^{mn}, we get=

By using the formula a

^{m}/a^{n }= a^{m-n}, we get=

=

=

By using the formula a

^{m }× a^{n }= a^{m+n}, we get=

= x

= 1

= Right-hand side of the given equation

Thus, we proved that

### Question 3 (ii). Prove that

**Solution:**

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

By using the formula, (a

^{m})^{n }= a^{mn}, we get=

=

= [a

^{m }× a^{n }= a^{m+n}]= 1

= Right-hand side of the given equation

Thus, we proved that

### Question 3 (iii). Prove that

**Solution:**

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

By using the formula (a

^{m})^{n }= a^{mn}, we get=

By using the formula a

^{m}/a^{n }= a^{m-n}, we get=

=

=

By using the formula a

^{m }× a^{n }= a^{m+n}, we get=

=

= Right-hand side of the given equation

Thus, we proved that

### Question 4 (i). Prove that

**Solution:**

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

=

=

=

=

= 1

= Right-hand side of the given equation

Thus, we proved that

### Question 4 (ii). Prove that

**Solution:**

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

=

=

=

=

= 1

= Right-hand side of the given equation

Thus, we proved that

### Question 5 (i). Prove that

**Solution:**

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

=

=

= abc

= Right hand side of the given equation

Thus, we proved that

### Question 5 (ii). Prove that

**Solution:**

Let us first consider the left hand side of given equation

=

=

=

=

= Right hand side of the given equation

Thus, we proved that

### Question 6. If abc = 1, show that

**Solution:**

Given abc = 1

⇒ c = 1/ab

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

=

=

=

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

=

=

=

=

=

= 1

= Right hand side of the given equation

Thus, we have shown that if abc = 1,

### Question 7 (i). Simplify

**Solution:**

Given

=

=

= [a

^{m }× a^{n }= a^{m+n}]=

= 3

^{3n+2-(3n-3)}[a^{m}/a^{n }= a^{m-n}]= 3

^{5}= 243

Thus, = 243

### Question 7 (ii). Simplify

**Solution:**

Given

=

=

= [a

^{m }× a^{n }= a^{m+n}]=

= 4/24

= 1/6

Thus, = 1/6

### Question 7 (iii). Simplify

**Solution:**

Given,

=

=

= (19 × 5)/5

= 19

Thus,

### Question 7 (iv). Simplify

**Solution:**

Given

=

=

=

=

=

= (48 + 4)/13

= 52/13

= 4

Thus,

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

**Solution:**

Given equation 7

^{2x+3 }= 1We know that, for any a∈ Real numbers, a

^{0 }= 1Let a = 7

⇒ 7

^{2x+3 }= 7^{0}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 = 2

^{2}⇒ 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 = 2

^{3}⇒ 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 }× 2^{5 }= 1⇒ 2

^{4x+5 }= 2^{0}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) }× 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 }= 2^{8}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 }+ 2^{4 }= 0 for x.

**Solution:**

Given 2

^{2x }– 2^{x+3 }+ 2^{4 }= 0⇒ (2

^{x})^{2 }– 2 × 2^{x }× 2^{2 }+ (2^{2})^{2 }= 0⇒ (2

^{x }– 2^{2})^{2 }= 0⇒ 2

^{x }– 2^{2 }= 0⇒ 2

^{x }= 2^{2}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 }– 1)^{2 }= 0⇒ 3

^{x+2 }– 1 = 0⇒ 3

^{x+2 }= 3^{0}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 = a^{4}b^{2}c^{3}, 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 49392Thus, 49392 = 2

^{4 }× 3^{2 }× 7^{3}Where 2, 3 and 7 are positive primes

49392 = 2

^{4}3^{2}7^{3 }= a^{4}b^{2}c^{3}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^{a}3^{b}7^{c}, find a, b and c.

**Solution:**

Given 1176 = 2

^{a}3^{b}7^{c}Let us first find out prime factorization of 1176

Thus, 1176 = 2

^{3 }× 3^{1 }× 7^{2}1176 = 2

^{3}3^{1}7^{2 }= 2^{a}3^{b}7^{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^{a}5^{b}7^{c}, find

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

### (ii) the value of 2^{-a}3^{b}7^{c}

**Solution:**

Given 4725 = 3

^{a}5^{b}7^{c}

(i)Let us first find out prime factorization of 4725Thus, 4725 = 3

^{3 }× 5^{2 }× 7^{1}4725 = 3

^{3}5^{2}7^{1 }= 3^{a}5^{b}7^{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 = 1On substituting these values in 2

^{-a}3^{b}7^{c}2

^{-a}3^{b}7^{c}= 2^{-3}×3^{2}×7^{1}= 1/8 × 9 × 7 = 63/8

Thus, the value of 2

^{-a}3^{b}7^{c}is 63/8

### Question 13. If a = xy^{p-1}, b = xy^{q-1}, 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}, c = xy^{r-1}a

^{q-r}b^{r-p}c^{p-q}==

= x

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

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

^{0}y^{0}= 1

Thus, we proved that a

^{q-r}b^{r-p}c^{p-q }= 1