ICSE Class 8 Maths Chapter 12 Exponents

Chapter 12

Exponents

You already know that 3 × 3 × 3 × 3 × 3 can be written as 3⁵. Here 3 is the base and 5 is the exponent or index. 3⁵ is read as '3 raised to the power 5' or '3 to the power 5' or simply '3 power 5'. In general, we have:

If a is any real number and n is a natural number, then aⁿ = a × a × a.... n times where a is called the base, n is called the exponent or index and aⁿ is the exponential form. aⁿ is read as 'a raised to the power n' or 'a to the power n' or simply 'a power n'.

For zero power, we have: a⁰ = 1 (where a ≠ 0).

For example:

(i) 7⁰ = 1

(ii) \(\left(-\frac{2}{3}\right)^0 = 1\)

(iii) \((\sqrt{7})^0 = 1\)

For negative powers, we have:

\(a^{-n} = \frac{1}{a^n}\) and \(\frac{1}{a^{-n}} = a^n\) (where a ≠ 0).

For example:

(i) \(5^{-3} = \frac{1}{5^3} = \frac{1}{125}\)

(ii) \((-3)^{-4} = \frac{1}{(-3)^4} = \frac{1}{81}\)

(iii) \(\frac{1}{2^{-5}} = 2^5 = 32\)

(iv) \(\frac{1}{(-7)^{-3}} = (-7)^3 = -243\)

For fractional indices, remember that:

\(\sqrt[n]{a} = a^{\frac{1}{n}}\) and \(\sqrt[n]{a^m} = (a^m)^n = a^{\frac{m}{n}}\)

For example:

(i) \(\sqrt{3} = 3^{\frac{1}{2}}\)

(ii) \(\sqrt[3]{8} = 8^{\frac{1}{3}}\)

(iii) \(\sqrt[4]{5^3} = 5^{\frac{3}{4}}\)

Laws Of Exponents

If a and b are any two real numbers and m and n are any two integers, then the following results hold:

1. \(a^m \times a^n = a^{m+n}\)

2. \(a^m \div a^n = a^{m-n}\) (a ≠ 0)

3. \((a^m)^n = a^{mn}\)

4. \((ab)^m = a^m b^m\)

5. \(\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\) (b ≠ 0)

6. \(a^n = a^m \Rightarrow n = m\), provided a > 0 and a ≠ 1

Remark

If a is any real number and n is a natural number, then

\((-a)^n = (-1 \times a)^n = (-1)^n a^n = \begin{cases} a^n \text{ if } n \text{ is even} \\ -a^n \text{ if } n \text{ is odd} \end{cases}\)

Teacher's Note

Exponents are used everywhere in real life - from calculating compound interest in bank accounts to determining how viruses spread exponentially during a pandemic. Understanding powers helps us work with very large and very small numbers efficiently.

Example 1

Use the laws of exponents to simplify the following:

(i) \([(2^3)^4]^5\)

(ii) \([3^6 \div 3^4]^3\)

(iii) \((2^4)^3 \times 2 \times 3^0\)

(iv) \((81)^{-1} \times 3^5\)

(v) \(\left(\frac{2}{3}\right)^0 + \left(\frac{2}{3}\right)^{-2}\)

(vi) \((3^{-2} \times 5^4)^4\)

Solution.

(i) \([(2^3)^4]^5 = [2^{3 \times 4}]^5 = [2^{12}]^5 = 2^{12 \times 5} = 2^{60}\)

(ii) \([3^6 \div 3^4]^3 = [3^{6-4}]^3 = [3^2]^3 = 3^{2 \times 3} = 3^6\)

(iii) \((2^4)^3 \times 2 \times 3^0 = 2^{12} \times 2^1 \times 1 = 2^{12+1} = 2^{13}\)

(iv) \((81)^{-1} \times 3^5 = (3^4)^{-1} \times 3^5 = 3^{-4} \times 3^5 = 3^{-4+5} = 3^1 = 3\)

(v) \(\left(\frac{2}{3}\right)^0 + \left(\frac{2}{3}\right)^{-2} = 1 + \frac{1}{\left(\frac{2}{3}\right)^2} = 1 + \frac{1}{\frac{4}{9}} = 1 + \frac{9}{4} = \frac{4+9}{4} = \frac{13}{4}\)

(vi) \((3^{-2} \times 5^4)^4 = (3^{-2})^4 \times (5^4)^4 = 3^{-2 \times 4} \times 5^{4 \times 4} = 3^{-8} \times 5^{12}\)

Teacher's Note

When simplifying expressions with exponents, remembering the order of operations and which law applies to which situation is like learning traffic rules - each rule has its specific place to apply.

Example 2

Simplify the following:

(i) \(\frac{(-2)^5 \times (3^5)^2}{12 \times 9^7}\)

(ii) \(\frac{(2^3 \times 3^4)^3 \times (-5)^3}{60 \times (-2)^5}\)

(iii) \(\frac{(2^{-4})^2 \times 2^{-5}}{2^{-6}}\)

Solution.

(i) \(\frac{(-2)^5 \times (3^5)^2}{12 \times 9^7} = \frac{(-2)^5 \times 3^{5 \times 2}}{2 \times 2 \times 3 \times 3^7} = -\frac{2^5 \times 3^{10}}{2^2 \times 3^{7+1}}\)

\(= -2^{5-2} \times 3^{10-8} = -2^3 \times 3^2 = -8 \times 9 = -72\)

(ii) \(\frac{(2^3 \times 3^4)^3 \times (-5)^3}{60 \times (-2)^5} = \frac{(2^3)^3 \times (3^4)^3 \times (-5)^3}{2 \times 2 \times 3 \times 5 \times (-2)^5} = -\frac{2^8 \times 3^{12} \times 5^3}{2^2 \times 3 \times 5 \times 2^5}\)

\(= 2^{8-2-5} \times 3^{12-1} \times 5^{3-1} = 2^1 \times 3^{11} \times 5^2\)

(iii) \(\frac{(2^{-4})^2 \times 2^{-5}}{2^{-6}} = \frac{2^{-8} \times 2^{-5}}{2^{-6}} = 2^{-8+(-5)-(-6)} = 2^{-7} = \frac{1}{2^7} = \frac{1}{128}\)

Teacher's Note

Breaking down complex expressions into simpler parts using exponent laws is like breaking down a complex project into manageable tasks - each step builds on the previous one.

Example 3

Simplify and write in the exponential form:

\(2^3 \times 3^2 + (-11)^2 + 2^{-5} \div 2^{-8} - \left(\frac{2}{5}\right)^0\)

Solution.

\(2^3 \times 3^2 + (-11)^2 + 2^{-5} \div 2^{-8} - \left(\frac{2}{5}\right)^0\)

\(= 8 \times 9 + 11^2 + 2^{-5-(-8)} - 1 = 72 + 121 + 2^3 - 1\)

\(= 72 + 121 + 8 - 1 = 200\)

\(= 2 \times 2 \times 2 \times 5 \times 5 = 2^3 \times 5^2\)

Teacher's Note

Converting final answers into exponential form makes them compact and easier to compare with other values, just as using scientific notation helps scientists communicate very large or very small measurements.

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