Chapter 2: Exponents Of Real Numbers Exercise – 2.2

Question: 1

Assuming that x, y, z are positive real numbers, simplify each of the following

Solution:

= (243x¹⁰y⁵z¹⁰)^1/5

= (243)^1/5x^10/5y^5/5z^10/5

= (3⁵)^1/5x²yz²

= 3x²yz²

 

Question: 2

Simplify

Solution:

= (4^-1)

= 1/4

= [(2⁵)⁻³]1/5

= (2⁻¹⁵)^1/5

= 2⁻³

= 1/2³ = 1/8

= [(343)⁻²]^1/3

= (343)^−2×1/3

= (7³)^−2/3

= (7⁻²)

=(1/7²)

= (1/49)

(iv) (0.001)^1/3

= (1/1000)^1/3

= (1/10³)^1/3

= 7^21-20 × 5^{25/2 - 21/2}

= 7¹ × 5^4/2

= 7¹ × 5²

= 7 × 25

= 175

 

Question: 3

Prove that

(ii) 9^3/2 − 3 × 5⁰ − (1/81)^−1/2

Solution:

= ((3 × 5⁻³)^1/2 ÷ (3⁻¹)^1/3(5)^1/2) × (3 × 5⁶)^1/6

= ((3)^1/2(5⁻³)^1/2 ÷ (3⁻¹)^1/3(5)^1/2) × (3 × 5⁶)^1/6

= ((3)^1/2(5)^−3/2 ÷ (3)^−1/3(5)^1/2) × ((3)^1/6 × (5)^6/6)

= ((3)^{1/2 − (−1/3)} × (5)^−3/2−1/2) × ((3)^1/6 × (5))

= ((3)^5/6 × (5)⁻²) × ((3)^1/6 × (5))

= ((3)^5/6+1/6 × (5)⁻²⁺¹)

= ((3)^6/6 × (5)⁻¹)

= ((3)¹ × (5)⁻¹)

= ((3) × (5)⁻¹)

= ((3) × (1/5))

= (3/5)

(ii) 9^3/2 − 3 × 5⁰ − (1/81)^−1/2

= (3²)^3/2 − 3 − (1/9²)^−1/2

= 3³ − 3 − (9)^−2×−1/2

= 27 − 3 − 9

= 15

= 2⁴ − 3 × 2^3×2/3 + 4/3

= 16 − 3 × 2² + 4/3

= 16 − 3 × 4 + 4/3

= 16 − 12 + 4/3

= (12 + 4)/3

= 16/3

= 2 × 1 × 5

= 10

= 1/2 + 1/(0.1)¹ − (3)²

= 1/2 +1/(0.1) −9

= 1/2 + 10 − 9

= 1/2 + 1

= 3/2

= (5/4)² + 5/4 + 5/4

= 25/16 + 10/4

= 25/16 + 40/16

= (26 + 40)/16

= 65/16

 

Question: 4

Show that

(v) (x^a−b)^{a + b} (x^b−c)^{b + c}(x^{c − a})^{c + a} = 1

Solution:

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

Considering LHS,

Therefore, LHS = RHS

Hence proved

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

Considering LHS,

Therefore, LHS = RHS

Hence proved

(v) (x^a−b)^{a + b} (x^b−c)^{b + c}(x^{c − a})^{c + a} = 1

(x^a−b)^a+b(x^b−c)^b+c(x^c−a)^c+a

 

Question: 5

If 2^x = 3^y = 12^z, show that 1/z = 1/y + 2/x

Solution:

2^x = 3^y = (2 × 3 × 2)^z

2^x = 3^y = (2² × 3)^z

2^x = 3^y = (2^2z × 3^z)

2^x = 3^y = 12^z = k

2 = k^1/x

3 = k^1/y

12 = k^1/z

12 = 2 × 3 × 2

12 = k^1/z = k^1/y × k^1/x × k^1/x

k^1/z = k^2/x + ^1/y

1/z = 1/y + 2/x

 

Question: 6

If 2^x = 3^y = 6^−z, show that 1/x + 1/y + 1/z = 0

Solution:

 2^x = 3^y = 6^−z

2^x = k

2 = k^1/x

3^y = k

3 = k^1/y

6^−z = k

k = 1/6^z

6 = k^−1/z

2 × 3 = 6

k^1/x × k^1/y = k^−1/z

1/x + 1/y = −1/z [by equating exponents]

1/x + 1/y + 1/z = 0

 

Question: 7

If a^x = b^y = c^z and b² = ac, then show that

Solution:

Let a^x = b^y = c^z = k

a = k^1/x, b = k^1/y, c = k^1/z

Now,

b² = ac

(k^1/y)² = k^1/x × k^1/z

k^2/y = k^{1/x + 1/z}

2/y = 1/x + 1/z

 

Question: 8

If 3^x = 5^y = (75)^z, Show that

Solution:

3^x = k

3 = k^1/x

5^y = k

5 = k^1/y

75^z = k

75 = k^1/z

3¹ × 5² = 75¹

k^1/x × k^2/y = k^1/z

1/x + 2/y = 1/z

 

Question: 9

If (27)^x = 9/3^x, find x

Solution:

We have,

(27)^x = 9/3^x

(3³)^x = 9/3^x

3^3x = 9/3^x

3^3x = 3²/3^x

3^3x = 3^2−x

3x = 2 − x [On equating exponents]

3x + x = 2

4x = 2

x = 2/4

x = 1/2

Here the value of x is ½

 

Question: 10

Find the values of x in each of the following

(ii) (2³)⁴ = (2²)^x

(iii) (3/5)^x(5/3)^2x = 125/27

(iv) 5^x−2 × 3^2x−3 = 135

(v) 2^x−7 × 5^x−4 = 1250

(vii) 5^2x+3 = 1

Solution:

We have

= 4x = 4 [On equating exponent]

x = 1

Hence the value of x is 1

(ii) (2³)⁴ = (2²)^x

We have

(2³)⁴ = (2²)^x

= 2^3×4 = 2^2×x

12 = 2x

2x = 12 [On equating exponents]

x = 6

Hence the value of x is 6

(iii) (3/5)^x(5/3)^2x = 125/27

We have

(3/5)^x(5/3)^2x = 125/27

⇒ 5^2x−x/3^2x−x  = 5³/3³

⇒ 5^x/3^x = 5³/3³

⇒ (5/3)^x = (5/3)³

x = 3 [on equating exponents]

Hence the value of x is 3

(iv) 5^x−2 × 3^2x−3 = 135

We have,

5^x−2 × 3^2x−3 = 135

⇒ 5^x−2 × 3^2x−3 = 5 × 27

⇒ 5^x−2 × 3^2x−3 = 5¹ × 3³

⇒ x − 2 = 1, 2x − 3 = 3 [On equating exponents]

⇒ x = 2 + 1, 2x = 3 + 3

⇒ x = 3, 2x = 6

⇒ x = 3

Hence the value of x is 3

(v) 2^x−7 × 5^x−4 = 1250

We have

2^x−7 × 5^x−4 = 1250

⇒ 2^x−7 × 5^x−4 = 2 × 625

⇒ 2^x−7 × 5^x−4 = 2 × 5⁴

⇒ x − 7 = 1

⇒ x = 8, x − 4 = 4

⇒ x = 8

Hence the value of x is 8

4x + 1 = -15

4x = -15 - 1

4x = -16

x = (-16)/4

x = - 4

Hence the value of x is 4

(vii) 5^2x+3 = 1

5^2x+3 = 1 × 5⁰

2x + 3 = 0 [By equating exponents]

2x = −3

x = −3/2

Hence the value of x is −3/2

√x = 2 [By equating exponents]

(√x)² = (2)²

x = 4

Hence the value of x is 4

x + 1 = - 6

x = - 6 - 1

x = -7

Hence the value of x is 7

 

Question: 11

If x = 2^1/3 + 2^2/3, show that x³ − 6x = 6

Solution:

x³ − 6x = 6

x = 2^1/3 + 2^2/3

Putting cube on both the sides, we get

x³  = (2^1/3 + 2^2/3)³

As we know, (a + b)³ = a³ + b³ + 3ab(a + b)

x³ = (2^1/3)³ + (2^2/3)³ + 3(2^1/3)(2^2/3)(2^1/3 + 2^2/3)

x³ = (2^1/3)³ + (2^2/3)³ + 3(2^1/3+2/3)(x)

x³ = (2^1/3)³ + (2^2/3)³ + 3(2)(x)

x³ = 6 + 6x

x³ - 6x = 6

Hence proved

 

Question: 12

Determine (8x)^x, if 9^x+2 = 240 + 9^x.

Solution:

 9^x+2 = 240 + 9^x

9^x .9² = 240 + 9^x

Let 9^x be y

81y = 240 + y

81y - y = 240

80y = 240

y = 3

Since, y = 3

Then,

9^x = 3

3^2x = 3

Therefore, x = ½

(8x)^x = (8 × 1/2)^1/2

= (4)^1/2

= 2

Therefore (8x)^x = 2

 

Question: 13

If 3^x+1 = 9^x-2, find the value of 2^1+x

Solution:

3^x+1 = 9^x-2

3^x+1 = 3^2x-4

x + 1 = 2x - 4

x = 5

Therefore the value of 2^1+x = 2¹⁺⁵ = 2⁶ = 64

 

Question: 14

If 3^4x = (81)^-1 and (10)^1/y = 0.0001, find the value of 2^-x+4y.

Solution:

3^4x = (81)^-1 and (10)^1/y = 0.0001

3^4x = (3)^-4

x = -1

And, (10)^1/y = 0.0001

(10)^1/y = (10)⁻⁴

1/y = -4

y = 1/−4

To find the value of 2^-x+4y, we need to substitute the value of x and y

2^-x+4y = 2^1+4(1/−4) = 2^1-1 = 2⁰ = 1

 

Question: 15

If 5^3x = 125 and 10^y = 0.001. Find x and y.

Solution:

5^3x = 125 and 10^y = 0.001

5^3x = 5³

x = 1

Now,

10^y = 0.001

10^y = 10^-3

y = -3

Therefore, the value of x = 1 and the value of y = – 3

 

Question: 16

Solve the following equations

(i) 3^x+1 = 27 × 3⁴

(iii) 3^x−1 × 5^2y−3 = 225

(iv) 8^x+1 = 16^y+2 and (1/2)^3+x = (1/4)^3y

Solution:

(i) 3^x+1 = 27 × 3⁴

3^x+1 = 3³ × 3⁴

3^x+1 = 3³⁺⁴

x + 1 = 3 + 4 [By equating exponents]

x + 1 = 7

x = 7 − 1

x = 6

4x = 3 (By equating exponents)

(iii) 3^x−1 × 5^2y−3 = 225

3^x−1 × 5^2y−3 = 3² × 5²

x − 1 = 2 [By equating exponents]

x = 3

3^x−1 × 5^2y−3 = 3² × 5²

2y − 3 = 2 [By equating exponents]

2y = 5

y = 5/2

(iv) 8^x+1 = 16^y+2 and (1/2)^3+x = (1/4)^3y

(2³)^x+1 and (2⁻¹)^3+x = (2⁻²)^3y

3x + 3 = 4y + 8 and − 3 − x = −6y

3x + 3 = 4y + 8 and 3 + x = 6y

3x + 3 = 4y + 8 and y = (3+x)/6

3x + 3 = 4y + 8... eq1

Substitute eq2 in eq1

3(3x + 3) = 6 + 2x + 24

9x + 9 = 30 + 2x

7x = 21

x = 21/7

x = 3

Putting value of x in eq2

y = 1

(v) 4^x−1 × (0.5)^3−2x = (1/8)^x

2^2x−2 × (5/10)^3−2x = (1/2³)^x

2^2x−2 × (1/2)^3−2x = 2^−3x

2^2x−2 × 2^−3+2x = 2^−3x

2x − 2 − 3 + 2x = −3x [By equating exponents]

4x + 3x = 5

7x = 5x = 5/7

(a/b)^1/2 = (a/b)^−(1−2x)1/2 = −1 + 2x [By equating exponents]

1/2 + 1 = 2x

2x = 3/2

x = ¾

 

Question: 17

If a and b are distinct positive primes such that

, find x and y

Solution:

(a⁶b⁻⁴)^1/3 = a^xb^2y

a^6/3b^−4/3 = a^xb^2y

a²b^−4/3 = a^xb^2y

x = 2, 2y = −4/3

 

Question: 18

If a and b are different positive primes such that

(ii) (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y, find x and y

Solution:

(a⁻¹⁻²b²⁺⁴)⁷ ÷ (a³⁺²b⁻⁵⁻³) = a^xb^y

(a⁻³b⁶)⁷ ÷ (a⁵b⁻⁸) = a^xb^y

(a⁻²¹b⁴²) ÷ (a⁵b⁻⁸) = a^xb^y

(a⁻²¹⁻⁵b⁴²⁺⁸) = a^xb^y

(a⁻²⁶b⁵⁰) = a^xb^y

x = −26, y = 50

(ii) (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y, find x and y

(a + b)⁻¹(a⁻¹ + b⁻¹)

= 1/ab

= (ab)⁻¹ = a⁻¹b⁻¹

By equating exponents

x = −1, y = −1

Therefore x + y + 2 = −1 − 1 + 2 = 0

 

Question: 19

If 2^x × 3^y × 5^z = 2160, find x, y and z. Hence compute the value of 3^x × 2^−y × 5^−z

Solution:

2^x × 3^y × 5^z = 2160

2^x × 3^y × 5^z = 2⁴ × 3³ × 5¹

x = 4, y = 3, z = 1

3^x × 2^−y × 5^−z = 3⁴ × 2⁻³ × 5⁻¹

= 81/40

 

Question: 20

If 1176 = 2^a × 3^b × 7^c, find the values of a, b and c.

Solution:

Hence compute the value of 2^a × 3^b × 7^−c as a fraction

1176 = 2^a × 3^b × 7^c

2³ × 3¹ × 7² = 2^a × 3^b × 7^c

a = 3, b = 1, c = 2

We have to find the value of 2^a × 3^b × 7^−c

2^a × 3^b × 7^−c = 2³ × 3¹ × 7⁻²

= 24/49

 

Question: 21

Simplify

Solution:

(x^a+b−c)^a−b(x^b+c−a)^b−c(x^c+a−b)^c−a

 

Question: 22

Show that

Solution:

Hence, LHS = RHS

 

Question: 23

(i) If a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ, prove that a^m−nb^n−lc^l−m = 1

(ii) If x = a^m+n, y = a^n+l and z = a^l+m, prove that x^myⁿz^l = xⁿy^lz^m

Solution:

(i) If a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ, prove that a^m−nb^n−lc^l−m = 1

(x^m+ny^l)^m−n(x^n+ly^m)^n−l(x^l+myⁿ)^l−m

= (x^(m+n)(m−n)y^l(m−n))(x^(n+l)(n−l)y^m(n−l))(x^(l+m)(l−m)y^n(l−m))

= x⁰y⁰ = 1

(ii) If x = a^m+n, y = a^n+l and z = a^l+m, prove that x^myⁿz^l = xⁿy^lz^m

LHS = x^myⁿz^l

(a^m+n)^m(a^n+l)ⁿ(a^l+m)^l

=a^(m+n)na^(n+l)la^(l+m)m

= xⁿy^lz^m