Indices — Multiple Choice Questions

The value of $\Big(5\dfrac{1}{16}\Big)^{-\dfrac{3}{4}}$ is

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

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

3. $\dfrac{27}{8}$

4. $\dfrac{8}{27}$

Answer

Given,

$\Rightarrow \Big(5\dfrac{1}{16}\Big)^{-\dfrac{3}{4}} = \Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}} \\[1em] = \Big(\dfrac{16}{81}\Big)^{\dfrac{3}{4}} \\[1em] = \Big(\dfrac{2^4}{3^4}\Big)^{\dfrac{3}{4}} = \dfrac{2^{4 \times \frac{3}{4}}}{3^{4 \times \dfrac{3}{4}}} \\[1em] = \dfrac{2^3}{3^3} = \dfrac{8}{27}.$

Hence, Option 4 is the correct option.

$\sqrt[4]{\sqrt[3]{2^2}}$ is equal to

1. $2^{-\dfrac{1}{6}}$

2. 2^-6

3. $2^{\dfrac{1}{6}}$

4. 2⁶

Answer

Given,

$\Rightarrow \sqrt[4]{\sqrt[3]{2^2}} = ((2^2)^{\dfrac{1}{3}})^{\dfrac{1}{4}} \\[1em] = 2^{2 \times \dfrac{1}{3} \times \dfrac{1}{4}} \\[1em] = 2^{\dfrac{1}{6}}.$

Hence, Option 3 is the correct option.

The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ equals

1. $\sqrt{2}$

2. 2

3. $\sqrt[12]{2}$

4. $\sqrt[12]{32}$

Answer

Given,

$\Rightarrow \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32} = (2)^{\dfrac{1}{3}}.(2)^{\dfrac{1}{4}}.(2^5)^{\dfrac{1}{12}} \\[1em] = (2)^{\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{5}{12}} \\[1em] = (2)^{\dfrac{4}{12} + \dfrac{3}{12} + \dfrac{5}{12}} \\[1em] = (2)^{\dfrac{12}{12}} \\[1em] = 2^1 = 2.$

Hence, Option 2 is the correct option.

The value of $\sqrt[4]{(81)^{-2}}$ is

1. $\dfrac{1}{9}$

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

3. 9

4. $\dfrac{1}{81}$

Answer

Given,

$\Rightarrow \sqrt[4]{(81)^{-2}} = \Big[\Big(\dfrac{1}{81}\Big)^{2}\Big]^{\dfrac{1}{4}} \\[1em] = \Big(\dfrac{1}{81}\Big)^{\dfrac{1}{2}} = \dfrac{1}{9}.$

Hence, Option 1 is the correct option.

Value of (256)^0.16 × (256)^0.09 is

1. 4

2. 16

3. 64

4. 256.25

Answer

Given,

$\Rightarrow (256)^{0.16} \times (256)^{0.09} = (256)^{\dfrac{16}{100}} \times (256)^{\dfrac{9}{100}} \\[1em] = (256)^{\dfrac{16 + 9}{100}} = (256)^{\dfrac{25}{100}}\\[1em] = (256)^{\dfrac{1}{4}} = (4^4)^{\dfrac{1}{4}} \\[1em] = 4.$

Hence, Option 1 is the correct option.

Which of the following is equal to x?

Answer

1. $x^{\dfrac{12}{7}} - x^{\dfrac{5}{7}}$

2. $\sqrt[12]{(x^4)^{\dfrac{1}{3}}}$

3. $(\sqrt{x^3})^{\dfrac{2}{3}}$

4. $x^{\dfrac{12}{7}} \times x^{\dfrac{7}{12}}$

Answer

On solving $(\sqrt{x^3})^{\frac{2}{3}}$ we get,

$\Rightarrow (\sqrt{x^3})^{\frac{2}{3}} = ((x^3)^{\frac{1}{2}})^{\frac{2}{3}} \\[1em] = x^{3 \times \frac{1}{2} \times \frac{2}{3}} \\[1em] = x.$

Hence, Option 3 is the correct option.

Consider the following two statements.

Statement 1: 3^m + 2ⁿ = 5^{m + n}

Statement 2: a^m + bⁿ = (a + b)^{m + n}, where a, b, m, n are positive integers.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

According to statement 1 :

3^m + 2ⁿ = 5^{m + n}

Lets take some examples

Let m = 1, n = 1

Taking L.H.S.: 3^m + 2ⁿ

= 3¹ + 2¹

= 3 + 2

= 5

Taking R.H.S.: 5^{m + n}

= 5^{1 + 1}

= 5²

= 25

As L.H.S. ≠ R.H.S.

We can conclude that 3^m + 2ⁿ ≠ 5^{m + n}

∴ Statement 1 is false.

According to statement 2; a^m + bⁿ = (a + b)^{m + n}

Lets take some examples

Let a = 1, b = 1, m = 1, n = 1

Taking L.H.S.: a^m + bⁿ

= 1¹ + 1¹

= 1 + 1

= 2.

Taking R.H.S.: (a + b)^{m + n}

= (1 + 1)^{1 + 1}

= 2²

= 4.

As L.H.S. ≠ R.H.S.

We can conclude that a^m + bⁿ ≠ (a + b)^{m + n}

∴ Statement 2 is false.

∴ Both the statements are false.

Hence, option 2 is the correct option.