Indices — Multiple Choice Questions

$128 \times 32^{\Big(-\dfrac{4}{3}\Big)}$ =

1. $\sqrt[3]{4}$

2. $\sqrt[3]{2}$

3. 8

4. 2

Answer

Given,

$\Rightarrow 128 \times 32^{\Big(\dfrac{-4}{3}\Big)} \\[1em] \Rightarrow (2)^7 \times [(2)^5]^{\dfrac{-4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\dfrac{5 \times -4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\dfrac{-20}{3}} \\[1em] \Rightarrow (2)^{7 - \dfrac{20}{3}} \\[1em] \Rightarrow (2)^{\dfrac{21 - 20}{3}} \\[1em] \Rightarrow (2)^{\dfrac{1}{3}} \\[1em] \Rightarrow \sqrt[3]{2}.$

Hence, option 2 is the correct option.

$\Big[\Big(\sqrt[3]{a^4}\Big)^{\dfrac{-3}{2}}\Big]^{\dfrac{-1}{2}}$ =

1. a

2. a²

3. $\dfrac{1}{a}$

4. $\dfrac{1}{a^2}$

Answer

Given,

$\Big[\Big(\sqrt[3]{a^4}\Big)^{\dfrac{-3}{2}}\Big]^{\dfrac{-1}{2}}$

Simplifying the expression:

$\Rightarrow \Big[\Big(a^4\Big)^{\dfrac{1}{3}}\Big]^{\dfrac{-3}{2} \times \dfrac{-1}{2}} \\[1em] \Rightarrow \Big[\Big(a^4\Big)^{\dfrac{1}{3}}\Big]^{\dfrac{3}{4}} \\[1em] \Rightarrow a^{4 \times \dfrac{1}{3} \times \dfrac{3}{4}} \\[1em] \Rightarrow a^1 \\[1em] \Rightarrow a.$

Hence, option 1 is the correct option.

$\dfrac{\sqrt{5 \times 3^{-3}} \times \sqrt[6]{5 \times 3^6}}{\sqrt[6]{5} \times \sqrt{3}}$ =

1. $\sqrt{\dfrac{3}{5}}$

2. $\sqrt{\dfrac{5}{3}}$

3. $\dfrac{3}{5}$

4. $\dfrac{\sqrt{5}}{3}$

Answer

Given,

$\dfrac{\sqrt{5 \times 3^{-3}} \times \sqrt[6]{5 \times 3^6}}{\sqrt[6]{5} \times \sqrt{3}}$

Simplifying the expression:

$\Rightarrow \dfrac{\sqrt{5} \times \sqrt{3^{-3}} \times \sqrt[6]{5} \times \sqrt[6]{3^6}}{\sqrt[6]{5} \times \sqrt{3}} \\[1em] \Rightarrow \dfrac{\sqrt{5} \times {(3^{-3})}^{\dfrac{1}{2}} \times {3}^{6 \times \dfrac{1}{6}}}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{\sqrt{5} \times {3}^{\dfrac{-3}{2}} \times {3^1}}{{3}^{\dfrac{1}{2}}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\dfrac{-3}{2} + 1 - \dfrac{1}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\dfrac{-3 + 2 - 1}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{\dfrac{-2}{2}} \\[1em] \Rightarrow \sqrt{5} \times {3}^{-1} \\[1em] \Rightarrow \dfrac {\sqrt{5}}{3}.$

Hence, option 4 is the correct option.

(81)^0.13 × (81)^0.12 =

1. 1

2. 3

3. $\sqrt{3}$

4. $\dfrac{1}{\sqrt{3}}$

Answer

Given,

⇒ (81)^0.13 × (81)^0.12

Simplifying the expression:

⇒ (81)^{(0.13 + 0.12)}

⇒ (81)^0.25

⇒ [(3)⁴]^0.25

⇒ 3^{4 × 0.25}

⇒ 3¹

⇒ 3.

Hence, option 2 is the correct option.

If 3^x = 3^-x, then (1.2)^x =

1. 0

2. 1

3. 1.2

4. 1.44

Answer

Given,

⇒ 3^x = 3^(-x)

$\Rightarrow 3^x = \dfrac{1}{3^x} \\[1em] \Rightarrow 3^x \times 3^x = 1 \\[1em] \Rightarrow 3^{2x} = 1 \\[1em] \Rightarrow 3^{2x} = 3^0 \\[1em]$

Equating the exponents,

$\Rightarrow 2x = 0 \\[1em] \Rightarrow x = 0$

Substituting value of x in (1.2)^x, we get :

⇒ (1.2)⁰

⇒ 1.

Hence, option 2 is the correct option.

If $9 \times 81^x = \dfrac{1}{27^{(x - 3)}}$, then x =

1. 0

2. -1

3. 1

4. 3

Answer

Given,

$\Rightarrow 9 \times 81^x = \dfrac{1}{27^{x - 3}} \\[1em] \Rightarrow 3^2 \times (3^4)^{x} = \dfrac{1}{3^{3(x - 3)}} \\[1em] \Rightarrow 3^2 \times 3^{4x} = 3^{-3(x - 3)} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3 \times x - 3 \times - 3} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3x + 9}$

Equating the exponents:

$\Rightarrow 2 + 4x = -3x + 9 \\[1em] \Rightarrow 4x + 3x = 9 - 2 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \dfrac{7}{7} \\[1em] \Rightarrow x = 1.$

Hence, option 3 is the correct option.

If 4 × 2^{x + 3} = 8^{x + 1}, then 2^x =

1. 1

2. 2

3. 4

4. 8

Answer

Given,

⇒ 4 × 2^{x + 3} = 8^{x + 1}

Simplifying the expression,

$\Rightarrow 4 \times 2^{x + 3} = (2^3)^{x + 1} \\[1em] \Rightarrow 4 \times 2^{x + 3} = 2^{3x + 3} \\[1em] \Rightarrow 4 = \dfrac{2^{3x + 3}}{2^{x + 3} } \\[1em] \Rightarrow 2^2 = 2^{3x + 3} \times 2^{-(x + 3)} \\[1em] \Rightarrow 2^2 = 2^{3x + 3 - (x + 3)} \\[1em] \Rightarrow 2^2 = 2^{3x - x + 3 - 3} \\[1em] \Rightarrow 2^2 = 2^{2x}$

Equating the exponents:

⇒ 2 = 2x

⇒ x = $\dfrac{2}{2}$

⇒ x = 1.

⇒ 2^x = 2¹ = 2.

Hence, option 2 is the correct option.

If 2^{x + 3} + 2^{x + 1} = 320, then x =

1. 2

2. 3

3. 4

4. 5

Answer

Given,

2^{(x + 3)} + 2^{(x + 1)} = 320

Simplifying the expression:

$\Rightarrow 2^{x + 1 + 2} + 2^{x + 1} = 320 \\[1em] \Rightarrow 2^{x + 1} \times 2^2 + 2^{x + 1} = 320 \\[1em] \Rightarrow 2^{x + 1} \Big(2^2 + 1\Big) = 320 \\[1em] \Rightarrow 2^{x + 1} \times 5 = 320 \\[1em] \Rightarrow 2^{x + 1} = \dfrac{320}{5} \\[1em] \Rightarrow 2^{x + 1} = 64 \\[1em] \Rightarrow 2^{x + 1} = 2^6$

Equating the exponents:

⇒ x + 1 = 6

⇒ x = 6 - 1

⇒ x = 5.

Hence, option 4 is the correct option.

If 4^x = 8^y, then x : y =

1. 2 : 3

2. 3 : 2

3. 3 : 4

4. 4 : 3

Answer

Given,

⇒ 4^x = 8^y

Simplifying the expression:

⇒ (2²)^x = (2³)^y

⇒ 2^2x = 2^3y

Equating the exponents:

$\Rightarrow 2x = 3y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{3}{2} \\[1em] \Rightarrow x : y = 3 : 2.$

Hence, option 2 is the correct option.

If (2⁵ + 0.125)² - (2⁵ - 0.125)² = 2^x, then the value of x is :

1. 2

2. 3

3. 4

4. 5

Answer

Given,

(2⁵ + 0.125)² - (2⁵ - 0.125)² = 2^x

By using the identity:

(a + b)² - (a - b)² = 4ab

Let a = 2⁵, b = 0.125

Using identity in L.H.S. of the given equation,

$\Rightarrow (2^5 + 0.125)^2 - (2^5 - 0.125)^2 = 4 \times 2^5 \times 0.125 \\[1em] = 4 \times 32 \times \dfrac{125}{1000} \\[1em] = 4 \times 32 \times \dfrac{1}{8} \\[1em] = 16 \\[1em] = 2^4.$

Equation L.H.S. and R.H.S.,

⇒ 2^x = 2⁴

⇒ x = 4.

Hence, option 3 is the correct option.

If 2^{x + 1} + 2^x = 3, then 3^x + 3^-x =

1. 0

2. 1

3. 2

4. $\dfrac{4}{3}$

Answer

Given,

2^{x + 1} + 2^x = 3

Now simplifying:

⇒ 2 × 2^x + 2^x = 3

⇒ 2^x(2 + 1) = 3

⇒ 2^x × 3 = 3

⇒ 2^x = 1

⇒ 2^x = 2⁰

Equating the exponents:

⇒ x = 0

Substituting value of x in 3^x + 3^-x, we get :

⇒ 3⁰ + 3⁰

⇒ 1 + 1

⇒ 2.

Hence, option 3 is the correct option.

If l^x = m^y = n^z and lmn = 1, then yz + zx + xy =

1. 0

2. 1

3. -1

4. $\dfrac{1}{2}$

Answer

Let us consider l^x = m^y = n^z = k, and lmn = 1

From, l^x = k, we get $l = k^{\dfrac{1}{x}}$

From, m^y = k, we get $m = k^{\dfrac{1}{y}}$

From, n^z = k, we get $n = k^{\dfrac{1}{z}}$

Substituting in lmn = 1:

$\Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} \times k^{\dfrac{1}{z}} = 1 \\[1em] \Rightarrow k^{\left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\right)} = 1 \\[1em] \Rightarrow k^{\left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\right)} = k^0$

Equating the exponents:

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

Multiply both sides by xyz:

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0 \\[1em] \Rightarrow xyz \Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big) = 0 \times xyz \\[1em] \Rightarrow \Big(\dfrac{xyz}{x} + \dfrac{xyz}{y} + \dfrac{xyz}{z}\Big) = 0 \\[1em] \Rightarrow yz + zx + xy = 0.$

Hence, option 1 is the correct option.

$\dfrac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x} =$

1. 0

2. 1

3. $\dfrac{14}{9}$

4. $\dfrac{9}{14}$

Answer

Given,

$\dfrac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x}$

Solving Numerator:

⇒ 9 × (4^x)²

⇒ 9 × [(2²)^x]²

⇒ 9 × 2^4x

Solving Denominator:

⇒ 16^{x + 1} - 2^{x + 1} × 8^x

⇒ (2⁴)^{x + 1} - 2^{x + 1} × (2³)^x

⇒ 2^{4x + 4} - 2^{x + 1} × 2^3x

⇒ 2^{4x + 4} - 2^{x + 1 + 3x}

⇒ 2^{4x + 4} - 2^{4x + 1}

Substituting the simplified numerator and denominator in the original expression:

$\Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x + 4} - 2^{4x + 1}} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x}.2^4 - 2^{4x}.2^1} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x}(2^4 - 2^1)} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x} \times (16 - 2)} \\[1em] \Rightarrow \dfrac{9}{14}.$

Hence, option 4 is the correct option.

If x = 0.1, then the value of $[1 - ({1 - [1 - x^3]^{(-1)}}) ^{(-1)}]^{\Big(\dfrac{-1}{3}\Big)}$ is:

1. 0

2. 1

3. 0.1

4. -1.1

Answer

Simplifying the expression :

$\Rightarrow [1 - (1 - [1 - x^3]^{(-1)})^{(-1)}]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(1 - \dfrac{1}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{1 - x^3 - 1}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{-x^3}{1 - x^3}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \Big(\dfrac{x^3}{x^3 - 1}\Big)^{-1}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[1 - \dfrac{x^3 - 1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\dfrac{x^3 - x^3 + 1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\dfrac{1}{x^3}\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow (x^3)^{\dfrac{1}{3}} \\[1em] \Rightarrow x \\[1em] \Rightarrow 0.1$

Hence, option 3 is the correct option.

(x^a)^{(b - c)} (x^b)^{(c - a)}(x^c)^{(a - b)} =

1. 0

2. 1

3. 2

4. 3

Answer

Given,

⇒ (x^a)^{(b - c)}(x^b)^{(c - a)}(x^c)^{(a - b)}

⇒ x^{(ab - ac)}x^{(bc - ba)}x^{(ca - cb)}

⇒ x^{(ab - ac) + (bc - ba) + (ca - cb)}

⇒ x^{(ab - ac + bc - ba + ca - cb)}

⇒ x^{(ab - ab + bc -bc - ac + ac)}

⇒ x⁰

⇒ 1.

Hence, option 2 is the correct option.