Index Questions

Index or Indices refers to the powers of a number or variable. Index questions are provided here to help students understand how to simplify expressions using simple formulas and techniques. In this article, you can practise various problems on indices of numbers, variables and some special expressions.

What is an Index?

In mathematics, an index is a number that tells us how many times a particular number is multiplied by itself. For instance, 35, where 5 is the index, means 3 is multiplied by itself five times.

Also, read: Index

Index Questions and Answers

1. Evaluate: 20⁷ ÷ 20⁵

Solution:

20⁷ ÷ 20⁵

Using the formula a^m/aⁿ = a^m-n, we have;

20⁷ ÷ 20⁵ = 20⁷/ 20⁵

= 20^(7-5)

= 20²

= 20 × 20

= 400

Therefore, 20⁷ ÷ 20⁵ = 400.

2. If 3^(x-y) = 27 and 3^(x+y) = 243, find the value of x and y.

Solution:

Given,

3^(x-y) = 27

3^(x-y) = 3³

Bases are equal hence, equate the powers (indices).

That means, x – y = 3….(i)

and

3^(x+y) = 243

3^(x+y) = 3⁵

Bases are equal hence, equate the powers (indices).

That means, x + y = 5….(ii)

Adding equations (i) and (ii), we get;

2x = 8

x = 4

Substituting x = 4 in equation (ii), we get;

4 + y = 5

y = 5 – 4 = 1

Therefore, x = 4 and y = 1.

3. Simplify the expression: (p³)⁰ × (p^1/2)^4.

Solution:

(p³)⁰ × (p^1/2)⁴

We know that a⁰ = 1 and (a^m)ⁿ = a^mn.

So, (p³)⁰ × (p^1/2)⁴ = p^(3×0) × p^{(½ × 4)}

= p⁰ × p²

= 1 × p²

= p²

Therefore, (p³)⁰ × (p^1/2)⁴ = p².

4. Write 8^2x+3 in the form of 2^y and express the relation between x and y.

Solution:

8^2x+3

As we know, 8 = 2³.

So, 8^2x+3 = (2³)^2x+3

Using the formula (a^m)ⁿ = a^mn, we get;

8^2x+3 = 2^3(2x+3)

= 2^6x+9

= 2^y

Thus, y = 6x + 9.

5. Evaluate the expression: 125^2/3 × 625^3/4.

Solution:

125^2/3 × 625^3/4

As we know, 125 = 5³ and 625 = 5⁴

So, 125^2/3 × 625^3/4 = (5³)^⅔ × (5⁴)^¾

= 5² × 5³

= 25 + 125

= 150

Therefore, 125^2/3 × 625^3/4 = 150.

6. Simplify: [5 a⁵ b² × 3(a b³)²]/ (15 a² b).

Solution:

[5 a⁵ b² × 3(a b³)²]/ (15 a² b)

Using the formula (a^m)ⁿ = a^mn, we have;

= [(5 × 3) × (a⁵ b²) × a² b⁶]/ (15 a² b)

= (15 a⁵⁺² b²⁺⁶)/ (15 a² b)

= (a⁷ b⁸)/(a² b)

= a^7-2 b^8-1

= a⁵ b⁷

Therefore, [5 a⁵ b² × 3(a b³)²]/ (15 a² b) = a⁵ b⁷.

7. If a and b are whole numbers such that a^b = 121, find the value of (a – 1)^{b + 1}.

Solution:

Given,

a^b = 121

We know that 11² = 121

Thus, ab = 11²

⇒ a = 11 and b = 2

Now, (a – 1)^{b + 1}

= (11 – 1)^{2 + 1}

= 10³

= 10 × 10 × 10

= 1000

Hence, (a – 1)^{b + 1} = 1000.

8. If 5^m = 3125, find the value of 4^{(m – 2)}.

Solution:

Given,

5^m = 3125

We know that 5⁵ = 3125

⇒ 5^m = 5⁵

⇒ m = 5

Now, 4^{(m – 2)} = 4^{(5 – 2)} = 4³ = 4 × 4 × 4 = 64

Therefore, 4^{(m – 2)} = 64.

9. Evaluate: (1/216)^-2/3 ÷ (1/27)^-4/3.

Solution:

(1/216)^-2/3 ÷ (1/27)^-4/3

Using the formula a^-m = 1/a^m or 1/a^-m = a^m, we get;

= 216^2/3 ÷ 27^4/3

= (6³)^2/3 ÷ (3³)^4/3

Using the formula (a^m)ⁿ = a^mn, we get;

= 6² ÷ 3⁴

= 36/81

= 4/9

Thus, (1/216)^-2/3 ÷ (1/27)^-4/3 = 4/9.

10. Simplify: 1 – {1 + (x² – 1)^-1}^-1.

Solution:

1 – {1 + (x² – 1)^-1}^-1

= 1 – {1 + [1/(x² – 1)] }^-1

= 1 – [(x² – 1 + 1)/ (x² – 1)]^-1

= 1 – [x²/(x² – 1)]^-1

= 1 – [(x² – 1)/x²]

= [x² – (x² – 1)]/ x²

= (x² – x² + 1)/x²

= 1/x²

Therefore, 1 – {1 + (x² – 1)^-1}^-1 = 1/x².

Practice Questions on Index (Indices)

1. If (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^m, find the value of m.
2. Simplify the expression: a³ b⁻² × (a² b²)⁴
3. If (a/b)^(x-1) = (b/a)^(x-3), then what is the value of x?
4. Simplify: (2x)^-4/x⁴
5. Evaluate: (16)^0.16 × (16)^0.04 × (2)^0.2