Exponents and Powers — Multiple Choice Questions

$\Big(\dfrac{1}{2}\Big)^{-5}$ is equal to

1. 32

2. $\dfrac{1}{32}$

3. -32

4. $-\dfrac{1}{32}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{1}{2}\Big)^{-5}\\[1em] \Rightarrow \Big(\dfrac{2}{1}\Big)^{5}\\[1em] \Rightarrow 2^5\\[1em] \Rightarrow 2 \times 2 \times 2 \times 2 \times 2\\[1em] \Rightarrow 32$

Hence, option 1 is the correct option.

The value of $\Big[\Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{3}\Big)^{-2}\Big] \div \Big(\dfrac{1}{5}\Big)^{-2}$ is

1. 0

2. -1

3. 1

4. $\dfrac{7}{5}$

Answer

Solving,

$\Rightarrow \Big[\Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{3}\Big)^{-2}\Big] \div \Big(\dfrac{1}{5}\Big)^{-2}$

⇒ [4² + 3²] ÷ 5²

⇒ [16 + 9] ÷ 25

⇒ 25 ÷ 25

⇒ 1

Hence, option 3 is the correct option.

If 2⁴ × 4³ = 4^x, then value of x is

1. 5

2. 7

3. 4

4. -5

Answer

We know that 4 = 2², so 2⁴ = (2²)² = 4²

$\Rightarrow 2^4 \times 4^3 = 4^x\\[1em] \Rightarrow 4^2 \times 4^3 = 4^x\\[1em] \Rightarrow 4^{2 + 3} = 4^x\\[1em] \Rightarrow 4^5 = 4^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 5$

Hence, option 1 is the correct option.

$\Big(\dfrac{-4}{3}\Big)^7 \times \Big(\dfrac{-4}{3}\Big)^8 \div \Big(\dfrac{-3}{4}\Big)^{-5}$ is equal to

1. $\Big(\dfrac{-4}{3}\Big)^{20}$

2. $\Big(\dfrac{-4}{3}\Big)^{10}$

3. $\Big(\dfrac{-3}{4}\Big)^{10}$

4. $\Big(\dfrac{-3}{4}\Big)^{20}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{-4}{3}\Big)^7 \times \Big(\dfrac{-4}{3}\Big)^8 \div \Big(\dfrac{-3}{4}\Big)^{-5}\\[1em] \Rightarrow \Big(\dfrac{-4}{3}\Big)^7 \times \Big(\dfrac{-4}{3}\Big)^8 \div \Big(\dfrac{-4}{3}\Big)^{5}\\[1em] \Rightarrow \Big(\dfrac{-4}{3}\Big)^{7 + 8 - 5}\\[1em] \Rightarrow \Big(\dfrac{-4}{3}\Big)^{10}$

Hence, option 2 is the correct option.

Multiplicative inverse of $\Big(\dfrac{-2}{3}\Big)^4$ is

1. $\dfrac{16}{81}$

2. $\dfrac{-16}{81}$

3. $\dfrac{81}{16}$

4. $\dfrac{-81}{16}$

Answer

Solving,

$\Rightarrow \Big(\dfrac{-2}{3}\Big)^4\\[1em] \Rightarrow \dfrac{(-2)^4}{3^4}\\[1em] \Rightarrow \dfrac{16}{81}$

The multiplicative inverse of $\dfrac{16}{81}$ is $\dfrac{81}{16}$.

Hence, option 3 is the correct option.

$2^0 + 3^0 + \Big(\dfrac{1}{4}\Big)^0$ is equal to

1. $\dfrac{21}{4}$

2. 0

3. 9

4. 3

Answer

We know that $a^0 = 1$

$\Rightarrow 2^0 + 3^0 + \Big(\dfrac{1}{4}\Big)^0$

⇒ 1 + 1 + 1

⇒ 3

Hence, option 4 is the correct option.

If $\Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{13}{12}\Big)^{-8} = \Big(\dfrac{12}{13}\Big)^{2x}$, then x is

1. -2

2. 6

3. 2

4. -6

Answer

Solving,

$\Rightarrow \Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{13}{12}\Big)^{-8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{12}{13}\Big)^{8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^{4 + 8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^{12} = \Big(\dfrac{12}{13}\Big)^{2x}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 12 = 2x\\[1em] \Rightarrow x = \dfrac{12}{2}\\[1em] \Rightarrow x = 6$

Hence, option 2 is the correct option.