ICSE Class 7 Maths Chapter 22 Indices

Chapter 22: Indices

22.1 Index Or Exponent

In 3 × 3 × 3 × 3 × 3, the factor 3 is being multiplied 5 times by itself and can also be written as 3^5, i.e., 3 × 3 × 3 × 3 × 3 = 3^5.

In 3^5, the repeated factor 3 is called the base and the number 5, written slightly raised at the right of the factor 3, is called the index or exponent.

Thus, an index or an exponent is a number which indicates how many times the base is used as a repeated factor.

The plural of index is indices.

1. If n is a whole number and a is any number, then:

a^n = a × a × a × a × ... × a (base = a and index = n.)

n factors

2. In 5 × 5 × 5 × 5 × 5 × 5 × 5 = 5^7, base = 5 and index = 7.

3. If base = -3 and index (exponent) = 8, the number = (-3)^8.

Teacher's Note

Understanding indices helps students write repeated multiplications in a compact form, similar to how shorthand notation is used in everyday writing and science to express large or small quantities efficiently.

22.2 Laws Of Indices

First Law (Product Law)

a^m × a^n = a^(m + n)

When numbers are in the exponent form with the same base, to get their product (multiplication), add their powers (indices) keeping the base same.

For example:

(i) a^3 × a^7 = a^(3 + 7) = a^10

(ii) x^2 y^3 × x^4 y^2 = (x^2 × x^4) × (y^3 × y^2) = x^(2 + 4) × y^(3 + 2) = x^6 y^5

(iii) 4a^2 b^3 c^2 × 8a^9 b^6 c × 3ab^10 c^5 = 4 × 8 × 3 × a^(2 + 9 + 1) × b^(3 + 6 + 10) × c^(2 + 1 + 5) = 96a^12 b^19 c^8

Second Law (Quotient Law)

a^m / a^n = a^(m - n); if m > n and a^m / a^n = 1 / a^(n - m); if m < n.

When a number in exponent form is divided by another number in the exponent form (both the numbers having the same base), the smaller index (power) is subtracted from the bigger index (power) and the base is kept the same.

For example:

(i) x^5 ÷ x^3 = x^5 / x^3 = x^(5 - 3) = x^2

(ii) 15a^2 ÷ 5a^10 = 15a^2 / 5a^10 = 3 / a^(10 - 2) = 3 / a^8

Teacher's Note

The product and quotient laws mirror how we combine or separate quantities in real life, such as combining ingredient portions in cooking or dividing materials into smaller portions while maintaining consistency.

Third Law (Power Law)

(a^m)^n = a^(mn)

When a number in the index form is raised to another index, the base is raised to the product of these two indices.

For example:

(i) (a^2)^6 = a^(3 × 6) = a^18

(ii) (x^6)^(3/2) = x^(6 × 3/2) = x^9

22.3 More About Indices

1. (ab)^m = a^m b^m And (a / b)^m = a^m / b^m

e.g. (i) (xy)^3 = x^3 y^3

(ii) (2a^3)^2 = 2^2 (a^3)^2 = 4a^6

(iii) (3x / 4y^2)^4 = 3^4 · x^4 / 4^4 · (y^2)^4 = 81x^4 / 256 y^8 and so on.

2. Any non-zero base raised to the power zero is equal to unity (i.e., 1).

i.e. a^0 = 1, if a ≠ 0

e.g. (i) 5^0 = 1

(ii) (-3)^0 = 1

(iii) (x^2 / 2y)^0 = 1 and so on.

3. Negative Index: If a ≠ 0 then: a^(-m) = 1 / a^m and 1 / a^(-m) = a^m

e.g. (i) a^(-4) = 1 / a^4

(ii) 1 / x^(-7) = x^7 and so on.

Also note that:

(i) √a = a^(1/2), e.g. √3 = 3^(1/2)

(ii) ∛a = a^(1/3), e.g. ∛3 = 3^(1/3)

(iii) √(a^5) = a^(5/2), e.g. √(3^5) = 3^(5/2)

(iv) ^n√a = a^(1/n), e.g. ^n√3 = 3^(1/n)

Teacher's Note

Understanding zero exponents and negative indices connects to real-world scenarios like comparing very small measurements or understanding inverse relationships in science and engineering.

Exercise 22

1. Fill in the blanks:

(i) In 5^2 = 25, base = ______________ and index = ________________

(ii) If index = 3x and base = 2y, the number = ________________

2. Evaluate:

(i) 2^8 ÷ 2^3

(ii) 2^3 ÷ 2^8

(iii) (2^6)^0

(iv) (3^0)^6

(v) 8^3 × 8^(-5) × 8^4

(vi) 5^4 × 5^3 ÷ 5^5

(vii) 5^4 ÷ 5^3 × 5^5

(viii) 4^4 ÷ 4^3 × 4^0

(ix) (3^5 × 4^7 × 5^8)^0

3. Simplify, giving answers with positive index:

(i) 2b^6 · b^3 · 5b^4

(ii) x^2 y^3 · 6x^6 y · 9x^3 y^4

(iii) (-a^2) (a^2)

(iv) (-y^2) (-y^3)

(v) (-3)^2 (3)^3

(vi) (-4x) (-5x^2)

(vii) (5a^2 b) (2ab^2) (a^3 b)

(viii) x^(2a + 7) · x^(2a - 8)

(ix) 3y · 3^2 · 3^(-4)

(x) 2^(4a) · 2^(3a) · 2^(-a)

(xi) 4x^2 y^2 ÷ 9x^3 y^3

(xii) (10^2)^3 (x^8)^12

(xiii) (a^10)^10 (1^6)^10

(xiv) (n^2)^2 (-n^2)^3

(xv) -(3ab)^2 (-5a^2 bc^4)^2

(xvi) (-2)^2 × (0)^3 × (3)^3

(xvii) (2a^3)^4 (4a^2)^2

(xviii) (4x^2 y^3)^3 ÷ (3x^2 y^3)^3

(xix) (1 / 2x)^3 × (6x)^2

(xx) (1 / 4ab^2 c)^2 ÷ (3 / 2a^2 bc^2)^4

(xxi) (5x^7)^3 · (10x^2)^2 / (2x^6)^7

(xxii) (7p^2 q^4 r^5)^2 (4pqn^3) / (14p^6 q^10 r^4)^2

4. Simplify and express the answer in the positive exponent form:

(i) (-3)^3 × 2^6 / 6 × 2^3

(ii) (2^3)^5 × 5^4 / 4^3 × 5^2

(iii) 36 × (-6)^2 × 3^6 / 12^3 × 3^5

(iv) -128 / 2187

(v) a^(-7) × b^(-7) × c^5 × d^4 / a^3 × b^(-5) × c^(-3) × d^8

(vi) (a^2 b^(-5))^(-2)

5. Evaluate:

(i) 6^(-2) ÷ (4^(-2) × 3^(-2))

(ii) [(5 / 6)^2 × 9 / 4] ÷ [(-3 / 2)^2 × 125 / 216]

(iii) 5^3 × 3^2 + (17)^0 × 7^3

(iv) 2^5 × 15^0 + (-3)^3 - (2 / 7)^(-2)

(v) (2^2)^0 + 2^(-4) ÷ 2^(-6) + (1 / 2)^(-3)

(vi) 5^n × 25^(n-1) ÷ (5^(n-1) × 25^(n-1))

6. If m = -2 and n = 2; find the value of:

(i) m^2 + n^2 - 2mn

(ii) m^n + n^m

(iii) 6m^(-3) + 4n^2

(iv) 2n^3 - 3m

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