ICSE Class 9 Maths Chapter 08 Indices

Chapter 8: Indices

Points To Remember

Indices: For any real number 'a' and positive integer 'n', we define a × a × a ... to n factors = aⁿ.

Here 'a' is called base and n is called index or exponent.

Some Laws of Indices

(i) a⁰ = 1

(ii) a^-n = \(\frac{1}{a^n}\)

(iii) a^m × aⁿ = a^m+n

(iv) a^m ÷ aⁿ = a^m-n

(v) (a^m)ⁿ = a^mn

(vi) a^-m = \(\frac{1}{a^m}\)

(vii) (ab)^m = a^m . b^m

(viii) \(\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\)

(ix) \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\)

(x) If a^m = aⁿ, then m = n where a > 0 and a ≠ 1.

(xi) If p^m × qⁿ × r^o = p^a q^b r^c, then m = a, n = b and o = c where p, q and r are different primes.

(xii) If a^m = b^m, then a = b where a and b are positive.

(xiii) \(\sqrt{a} = a^{\frac{1}{2}}\), \(\sqrt[3]{a} = a^{\frac{1}{3}}\), \(\sqrt[4]{a} = a^{\frac{1}{4}}\)

Some Formulae Which Are Used

(i) (a + b) (a - b) = a² - b²

(ii) (a - b) (a² + ab + b²) = a³ - b³

Teacher's Note

Understanding indices helps us express very large or very small numbers in a compact form, such as scientific notation used in astronomy and microscopy.

Exercise 8

Evaluate:

Q. 1. (i) (125)^1/3

(ii) (8)^2/3

(iii) \(\left(\frac{1}{5}\right)^{-2}\)

(iv) (16)^-3/4

(v) (32)^-4/5

(vi) \(\left(\frac{8}{125}\right)^{-1/3}\)

(vii) (-27)^2/3

(viii) (0.001)^-1/3

(ix) (0.027)^-2/3

Solutions:

(i) (125)^1/3 = (5 × 5 × 5)^1/3 = (5³)^1/3 = 5^{3 × 1/3} = 5 Ans.

(ii) (8)^2/3 = (2 × 2 × 2)^2/3 = (2³)^2/3 = 2^{3 × 2/3} = 2² = 2 × 2 = 4 Ans.

(iii) \(\left(\frac{1}{5}\right)^{-2} = \left(\frac{5}{1}\right)^2 = 5 × 5 = 25\) Ans.

(iv) (16)^-3/4 = (2 × 2 × 2 × 2)^-3/4 = (2⁴)^-3/4

= 2^{-3/4 × 4} = 2^-3 = \(\frac{1}{2^3}\)

= 2^{-3 × 4} = 2^-3 = \(\frac{1}{2^3}\)

= \(\frac{1}{2 × 2 × 2} = \frac{1}{8}\) Ans.

(v) (32)^-4/5 = (2 × 2 × 2 × 2 × 2)^-4/5

= (2⁵)^-4/5 = 2^{-4/5 × 5} = 2^-4 = \(\frac{1}{2^4}\)

= \(\frac{1}{2 × 2 × 2 × 2} = \frac{1}{16}\) Ans.

(vi) \(\left(\frac{8}{125}\right)^{-1/3} = \left(\frac{2 × 2 × 2}{5 × 5 × 5}\right)^{-1/3}\)

= \(\left[\left(\frac{2}{5}\right)^3\right]^{-1/3}\)

= \(\left(\frac{2}{5}\right)^{-1/3 × 3} = \left(\frac{2}{5}\right)^{-1} = \frac{5}{2}\) Ans.

(vii) (-27)^2/3 = [(-3) × (-3) × (-3)]^2/3

= (-3)^{3 × 2/3} = (-3)² = -3 × -3 = 9 Ans.

(viii) (0.001)^-1/3 = (0.1 × 0.1 × 0.1)^-1/3

= (0.1)^{3 × (-1/3)}

= (0.1)^-1 = \(\frac{1}{0.1} = \frac{1}{1/10} = \frac{1 × 10}{1}\) = 10 Ans.

(ix) (0.027)^-2/3 = (0.3 × 0.3 × 0.3)^-2/3

= (0.3)^{-2/3 × 3}

= (0.3)^-2 = \(\frac{1}{(0.3)^2} = \frac{1}{0.3 × 0.3}\)

= \(\frac{1}{0.09} = \frac{1}{9/100} = \frac{100}{9}\) Ans.

Teacher's Note

Fractional exponents like 1/3 and 2/3 represent roots - understanding them helps in calculating compound interest rates and population growth models.

Q. 2. (i) \(\left(\frac{1}{4}\right)^{-2} - 3 × 8^{2/3} × 5^0 + \left(\frac{9}{16}\right)^{-1/2}\)

(ii) \(\sqrt{\frac{1}{4} + (0.01)^{-1/2} - (27)^{2/3} × 3^0}\)

Solutions:

(i) \(\left(\frac{1}{4}\right)^{-2} - 3 × 8^{2/3} × 5^0 + \left(\frac{9}{16}\right)^{-1/2}\)

= \(\left(\frac{1}{2} × \frac{1}{2}\right)^{-2} - 3 × (2 × 2 × 2)^{2/3} × 5^0 + \left(\frac{3}{4} × \frac{3}{4}\right)^{-1/2}\)

= \(\left(\frac{1}{2}\right)^{2 × (-2)} - 3 × (2^3)^{2/3} × 5^0 + \left(\frac{3}{4}\right)^{2 × (-1/2)}\)

= \(\left(\frac{1}{2}\right)^{-4} - 3 × 2^2 × 1 + \left(\frac{3}{4}\right)^{-1}\)

= (2)⁴ - 3 × 4 × 1 + \(\frac{4}{3}\)

= (2 × 2 × 2 × 2) - 12 + \(\frac{4}{3}\)

= 16 - 12 + \(\frac{4}{3}\) = 4 + \(\frac{4}{3}\)

= \(\frac{12 + 4}{3} = \frac{16}{3} = 5\frac{1}{3}\) Ans.

(ii) \(\sqrt{\frac{1}{4} + (0.01)^{-1/2} - (27)^{2/3} × 3^0}\)

= \(\sqrt{\left(\frac{1}{2}\right)^{2 × 1/2} + (0.1 × 0.1)^{-1/2} - (3 × 3 × 3)^{2/3} × 3^0}\)

= \(\sqrt{\left(\frac{1}{2}\right)^{2 × 1/2} + (0.1)^{-2 × (-1/2)} - (3^3)^{2/3} × 3^0}\)

= \(\sqrt{\left(\frac{1}{2}\right)^1 + (0.1)^{-1} - (3)^2 × 3^0}\)

= \(\sqrt{\frac{1}{2} + \frac{1}{0.1} - 9 × 1}\)

= \(\sqrt{\frac{1}{2} + 10 - 9}\) = \(\sqrt{1\frac{1}{2}}\) Ans.

Teacher's Note

Evaluating complex expressions with multiple indices strengthens problem-solving skills needed in advanced mathematics and engineering calculations.

Q. 3. (i) \(\left(\frac{81}{16}\right)^{-3/4} × \left[\left(\frac{25}{9}\right)^{-3/2} + \left(\frac{5}{2}\right)^{-3}\right]\)

(ii) \(\left[(64)^{2/3} × 2^{-2} ÷ 7^0\right]^{-1/2}\)

Solutions:

(i) \(\left(\frac{81}{16}\right)^{-3/4} × \left[\left(\frac{25}{9}\right)^{-3/2} + \left(\frac{5}{2}\right)^{-3}\right]\)

= \(\left(\frac{3 × 3 × 3 × 3}{2 × 2 × 2 × 2}\right)^{-3/4} × \left[\left(\frac{5 × 5}{3 × 3}\right)^{-3/2} + \left(\frac{5}{2}\right)^{-3}\right]\)

= \(\left[\left(\frac{3}{2}\right)^4\right]^{-3/4} × \left[\left(\frac{5}{3}\right)^{2 × (-3/2)} + \left(\frac{5}{2}\right)^{-3}\right]\)

= \(\left(\frac{3}{2}\right)^{4 × (-3/4)} × \left[\left(\frac{5}{3}\right)^{-3} + \left(\frac{2}{5}\right)^3\right]\)

= \(\left(\frac{2}{3}\right)^3 × \left[\left(\frac{3}{5}\right)^3 + \left(\frac{2}{5}\right)^3\right]\)

= \(\frac{2 × 2 × 2}{3 × 3 × 3} × \left[\frac{3 × 3 × 3}{5 × 5 × 5} + \frac{2 × 2 × 2}{5 × 5 × 5}\right]\)

= \(\frac{8}{27} × \left[\frac{27}{125} + \frac{8}{125}\right]\)

= \(\frac{8}{27} × \frac{27 × 125}{8} = \frac{27}{27} × \frac{27}{8} = 1\) Ans.

(ii) \(\left[(64)^{2/3} × 2^{-2} ÷ 7^0\right]^{-1/2}\)

= \(\left[(4 × 4 × 4)^{2/3} × \frac{1}{2^2} ÷ 1\right]^{-1/2}\)

= \(\left[(4^3)^{2/3} × \frac{1}{4} ÷ 1\right]^{-1/2}\)

= \(\left[4^2 × \frac{1}{4} ÷ 1\right]^{-1/2}\)

= \(\left[16 × \frac{1}{4} × 1\right]^{-1/2}\)

= \((4)^{-1/2} = \frac{1}{\sqrt{4}} = \frac{1}{\sqrt{(2)^2}} = \frac{1}{2}\) Ans.

Teacher's Note

Simplifying complex exponential expressions with nested brackets is fundamental for higher mathematics and physics problem-solving.

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