ICSE Class 9 Maths Chapter 07 Indices Exponents

Indices [Exponents]

Introduction

If \(m\) is a positive integer, then \(a \times a \times a \times a\) - upto \(m\) terms, is written as \(a^m\); where \(a\) is called the base and \(m\) is called the power (or exponent or index).

\(a^m\) is read as \(a\) power \(m\) or \(a\) raised to the power \(m\).

Thus: (i) \(a \times a \times a \times\) - upto 10 terms \(= a^{10}\) [a raised to the power 10]

(ii) \(2 \times 2 \times 2 \times\) - upto 7 terms \(= 2^7\) [2 raised to the power 7] and so on.

Laws Of Indices

1st Law (Product Law): \(a^m \times a^n = a^{m+n}\)

e.g. (i) \(a^7 \times a^4 = a^{7+4} = a^{11}\) (ii) \(a^3 \times a^{-6} = a^{3-6} = a^{-3}\) and so on.

2nd Law (Quotient Law): \(\frac{a^m}{a^n} = a^{m-n}\)

e.g. (i) \(\frac{a^7}{a^4} = a^{7-4} = a^3\) (ii) \(\frac{a^3}{a^6} = a^{3-6} = a^{-3}\) and so on.

3rd Law (Power Law): \((a^m)^n = a^{mn}\)

e.g. (i) \((a^3)^4 = a^{12}\) (ii) \((a^{-2})^5 = a^{-10}\) and so on.

Handling Positive, Fractional, Negative And Zero Indices

1. \((a \times b)^m = a^m \times b^m\) and \(\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\)

e.g. (i) \((2 \times 3)^5 = 2^5 \times 3^5\) (ii) \(\left(\frac{2}{3}\right)^5 = \frac{2^5}{3^5}\) and so on.

2. If \(a \neq 0\) and \(n\) is a positive integer, then \(\sqrt[n]{a} = a^{1/n}\)

e.g. \(\sqrt[3]{a} = a^{1/3}\); \(\sqrt[4]{a} = a^{1/4}\); \(\sqrt[8]{a} = a^{1/8}\) and so on.

Also, \(\sqrt{a} = a^{1/2}\) i.e. \(\sqrt{2} = 2^{1/2}\), \(\sqrt{10} = 10^{1/2}\) and so on.

3. \(a^{m/n} = \sqrt[n]{a^m}\); where \(a \neq 0\).

e.g. \(a^{4/5} = \sqrt[5]{a^4}\); \(5^{2/3} = \sqrt[3]{5^2}\) and so on.

Conversely: \(\sqrt[n]{a^m} = a^{m/n}\) i.e. \(\sqrt[5]{a^5} = a^{5/5}\), \(\sqrt[3]{3^8} = 3^{8/5}\) and so on.

4. For any non-zero number \(a\),

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

e.g. \(a^7 = \frac{1}{a^{-7}}\); \(a^{-3} = \frac{1}{a^3}\); \(a^4 = \frac{1}{a^{-4}}\) and so on.

5. Any non-zero number raised to the power zero is always equal to unity (i.e. 1).

e.g. \(a^0 = 1\); \(5^0 = 1\); \(2^0 = 1\) and so on.

\((-a)^m = a^m\); if \(m\) is an even number.

\((-a)^m = -a^m\); if \(m\) is an odd number.

e.g. \((-2)^4 = 2^4\); \((-2)^5 = -2^5\) and so on.

Simplification Of Expressions

Evaluate: (i) \(27^{-1/3}\) (ii) \(9^{3/2} - 3(5)^0 - \left(\frac{1}{81}\right)^{-2}\) (iii) \(\left(\frac{64}{125}\right)^{-2/3} \div \frac{1}{(256)^{1/4}} \times \sqrt[3]{64}\)

Solution:

(i) \(27^{-1/3} = (3^3)^{-1/3} = 3^{3 \times -1/3} = 3^{-1} = \frac{1}{3}\) Ans.

(ii) \(= (3^2)^{3/2} - 3 \times 1 - (81)^{-1/2}\) [- \(5^0 = 1\) and \(\left(\frac{1}{81}\right)^{-1/2} = (81)^{1/2}\)]

\(= 3^3 - 3 - 9^{2 \times 1/2} = 27 - 3 - 9 = 15\) Ans.

(iii) \(= \left(\frac{125}{64}\right)^{2/3} + \left(\frac{625}{256}\right)^{1/4} \times \frac{5}{\sqrt[3]{4^3}}\)

\(= \left[\left(\frac{5}{4}\right)^{3}\right]^{2/3} \times \left(\frac{256}{625}\right)^{1/4} \times \frac{5}{4} = \left(\frac{5}{4}\right)^2 \times \frac{4}{5} \times \frac{5}{4}\)

\(= \frac{25}{16} = 1\frac{9}{16}\) Ans.

Simplify:

(i) \((27)^{4/3} + (32)^{0.8} + (0.8)^{-1}\) (ii) \(27^{-1/3}\left(27^{1/3} - 27^{2/3}\right)\) (iii) \(\left[5\left(8^{1/3} + 27^{1/3}\right)^3\right]^{1/4}\)

Solution:

(i) \(= (3^3)^{4/3} + (2^5)^{8/10} + \left(\frac{8}{10}\right)^{-1}\)

\(= 3^4 + 2^4 + \frac{10}{8} = 81 + 16 + 1.25 = 98.25\) Ans.

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

\(= 3^{-1}(3^1 - 3^2) = \frac{1}{3}(3 - 9) = \frac{1}{3} \times -6 = -2\) Ans.

(iii) \(= \left[5\left(2^{3 \times 1/3} + 3^{3 \times 1/3}\right)^3\right]^{1/4}\)

\(= [5(2+3)^3]^{1/4} = (5 \times 5^3)^{1/4} = (5^4)^{1/4} = 5\) Ans.

Given: \(1176 = 2^p \cdot 3^q \cdot 7^r\), find:

(i) the numerical values of \(p\), \(q\) and \(r\). (ii) the value of \(2^p \cdot 3^q \cdot 7^{-r}\) as a fraction.

Solution:

(i) \(1176 = 2^p \cdot 3^q \cdot 7^r\)

\(\Rightarrow 2^3 \times 3^1 \times 7^2 = 2^p \cdot 3^q \cdot 7^r\) [1176 = 2 x 2 x 2 x 3 x 7 x 7]

\(\Rightarrow p = 3, q = 1\) and \(r = 2\) Ans.

(ii) \(2^p \cdot 3^q \cdot 7^{-r} = 2^3 \cdot 3^1 \cdot 7^{-2}\)

\(= \frac{8 \times 3}{7^2} = \frac{24}{49}\) Ans.

Simplify: (i) \(\frac{3^{a+2} - 3^{a+1}}{4 \times 3^a - 3^a}\) (ii) \(\left(\frac{a^m}{a^n}\right)^{m+n} \cdot \left(\frac{a^n}{a^l}\right)^{n+l} \cdot \left(\frac{a^l}{a^m}\right)^{l+m}\)

Solution:

(i) The given expression \(= \frac{3^a \cdot 3^2 - 3^a \cdot 3^1}{4 \times 3^a - 3^a}\) [\(3^a \cdot 3^2 = 3^{a+2}\)]

\(= \frac{3^a(3^2 - 3^1)}{3^a(4 - 1)} = \frac{9 - 3}{3} = 2\) Ans.

(ii) The given expression \(= (a^{m-n})^{m+n} \cdot (a^{n-l})^{n+l} \cdot (a^{l-m})^{l+m}\)

\(= a^{m^2-n^2} \cdot a^{n^2-l^2} \cdot a^{l^2-m^2}\)

\(= a^{m^2-n^2+n^2-l^2+l^2-m^2} = a^0 = 1\) Ans.

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