Indices — Competency Focused Questions

Which of the following is equal to x?

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

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

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

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

Answer

Solving Option 2,

Given,

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

Now simplifying:

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

Hence, option 2 is the correct option.

On simplifying $\Big[5\Big(8^\dfrac{1}{3} + 27^\dfrac{1}{3}\Big)^{3}\Big]^{\dfrac{1}{4}}$, we get:

1. 5

2. $\sqrt{5}$

3. 10

4. 1

Answer

Given,

$\Big[5\Big(8^\dfrac{1}{3} + 27^\dfrac{1}{3}\Big)^{3}\Big]^{\dfrac{1}{4}}$

Now simplifying:

$\Rightarrow \Big[5\Big((2^3)^{\dfrac{1}{3}} + (3^3)^{\dfrac{1}{3}}\Big)^{3}\Big]^{\dfrac{1}{4}} \\[1em] \Rightarrow \Big[5(2 + 3)^{3}\Big]^{\dfrac{1}{4}} \\[1em] \Rightarrow \Big[5(5)^{3}\Big]^{\dfrac{1}{4}} \\[1em] \Rightarrow \Big[(5)^{4}\Big]^{\dfrac{1}{4}} \\[1em] \Rightarrow 5^{4 \times \dfrac{1}{4}} \\[1em] \Rightarrow 5.$

Hence, option 1 is the correct option.

The value of 4.2 × 10^-15 + 42 × 10^-16 + 4.2 × 10^-14 is:

1. 5 × 10^-14

2. 5.4 × 10^-15

3. 5.04 × 10^-14

4. 5.04 × 10^-15

Answer

Given,

4.2 × 10^-15 + 42 × 10^-16 + 4.2 × 10^-14

Now simplifying:

$\Rightarrow 10^{-14} (4.2 × 10^{-1} + 42 × 10^{-2} + 4.2) \\[1em] \Rightarrow 10^{-14} (4.2 \times \dfrac{1}{10} + 42 \times \dfrac{1}{100} + 4.2) \\[1em] \Rightarrow 10^{-14} \Big(\dfrac{4.2}{10} + \dfrac{42}{100} + 4.2\Big) \\[1em] \Rightarrow 10^{-14} (0.42 + 0.42 + 4.2) \\[1em] \Rightarrow 5.04 × 10^{-14}.$

Hence, option 3 is the correct option.

The value of (2^-1 × 2² × 2^-3 × 2⁴ × 2^-5 × 2⁶ × 2^-7 × 2⁸ × 2^-9 × 2¹⁰)^-1 is:

1. 2

2. 16

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

4. 32

Answer

Given,

⇒ (2^-1 × 2² × 2^-3 × 2⁴ × 2^-5 × 2⁶ × 2^-7 × 2⁸ × 2^-9 × 2¹⁰)^-1

⇒ [2^{(-1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 - 9 + 10)}]^-1

⇒ [2^{(-1 - 3 - 5 - 7 - 9 + 2 + 4 + 6 + 8 + 10)}]^-1

⇒ [2^{(-25 + 30)}]^-1

⇒ (2⁵)^-1

⇒ 2^-5

⇒ $\dfrac{1}{2^5}$

⇒ $\dfrac{1}{32}$.

Hence, option 3 is the correct option.

The value of (-1)⁰ - (-1)¹ - (-1)² - (-1)³ - .... - (-1)¹⁰ is:

1. 0

2. 1

3. -1

4. 11

Answer

Given,

(-1)⁰ - (-1)¹ - (-1)² - (-1)³ - (-1)⁴ - (-1)⁵ - (-1)⁶ - (-1)⁷ - (-1)⁸ - (-1)⁹ - (-1)¹⁰

A negative number raised to an even power gives a positive result, also a real number raised to power 0 results in 1.

A negative number raised to an odd power gives a negative result.

Now simplifying:

⇒ 1 - (-1) - (1) - (-1) - (1) - (-1) - (1) - (-1) - (1) - (-1) - (1)

⇒ 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1

⇒ 1.

Hence, option 2 is the correct option.

Which of the following is equal to 1?

1. 15² + 15^-2

2. 15² - 15^-2

3. 15² × 15^-2

4. $\dfrac{15^2}{15^{-2}}$

Answer

Simplifying option 3,

⇒ 15² × 15^-2

⇒ 15^{2 - 2}

⇒ 15⁰

⇒ 1.

Hence, option 3 is the correct option.

If you raise a number to a negative power, can the resulting number be greater than the original number? Justify your answer by giving suitable examples.

Answer

Yes, if the base is a fraction between 0 and 1, raising it to a negative power results in a large number.

Example 1: Let the number be 0.2

Then, 0.2^-1

⇒ $\dfrac{1}{0.2}$

⇒ 5, which is greater than original number.

Hence, yes if the base is a fraction between 0 and 1.

If $(x - 1)^{\dfrac{1}{2}} + (y - 2)^{\dfrac{1}{2}} + (z - 3)^{\dfrac{1}{2}} = 0$, then find the values of x, y, z.

Answer

Given,

$\Rightarrow (x - 1)^{\dfrac{1}{2}} + (y - 2)^{\dfrac{1}{2}} + (z - 3)^{\dfrac{1}{2}} = 0$

We can write,

⇒ $(x - 1)^{\dfrac{1}{2}}$= 0 ....(1)

⇒ $(y - 2)^{\dfrac{1}{2}}$ = 0 ....(2)

⇒ $(z - 3)^{\dfrac{1}{2}}$ = 0 ....(3)

Solving equation 1,

⇒ $(x - 1)^{\dfrac{1}{2}}$= 0

Squaring both sides,

⇒ (x - 1) = 0

⇒ x = 1

Solving equation 2,

⇒ $(y - 2)^{\dfrac{1}{2}}$ = 0

Squaring both sides,

⇒ (y - 2) = 0

⇒ y = 2

Solving equation 3,

⇒ $(z - 3)^{\dfrac{1}{2}}$ = 0

Squaring both sides,

⇒ (z - 3) = 0

⇒ z = 3

Hence, the values of x, y, z are 1, 2, 3.