ICSE Class 8 Maths Algebra Chapter 03 Exponents

Chapter 3: Exponents

In this chapter, we will revise what you have learnt about exponents in your previous class and discuss fractional indices.

You know that \(x \times x = x^2\), which is read as x squared or x raised to the power 2 or x to the power 2. Here, x is the base and 2 is the exponent or index.

An exponent (or index) is a number written to the right and a little above the base. It indicates the number of times the base occurs in a product.

Examples

(i) In \(x^4\) (= \(x \times x \times x \times x\)), read as x to the power 4, the exponent is 4.

(ii) In \(x^m\) (= \(x \times x \times x \times ... \, m\) times), read as x to the power m, the exponent is m.

Reciprocal Of A Power

\(x \div x = 1\), that is, \(x \times \frac{1}{x} = 1\); \(\frac{1}{x}\) is called the reciprocal of x, and is written as \(x^{-1}\).

Similarly, \(\frac{1}{x^2}\) is the reciprocal of \(x^2\) and is written as \(x^{-2}\).

\(\frac{1}{x^3}\) is the reciprocal of \(x^3\) and is written as \(x^{-3}\).

In general, \(\frac{1}{x^m} = x^{-m}\), \(\frac{1}{x^{-m}} = x^m\)

because \(x^m \times x^{-m} = x^{m-m} = x^0 = 1\).

Some Laws Of Indices

(i) \(x^m \times x^n = x^{m+n}\) and \(x^m \times x^n \times x^p = x^{m+n+p}\)

(ii) \(x^m \div x^n = x^{m-n}\)

(iii) \((x^m)^n = x^{mn}\)

(iv) \((x \times y)^m = x^m \times y^m\)

(v) \(\left(\frac{x}{y}\right)^m = \frac{x^m}{y^m}\)

(vi) If \(x^m = x^n\) then m = n, so long as x is a positive number other than 1.

Some Important Results

(i) \(1 = a^m \div a^m = a^{m-m} = a^0\), so \(a^0 = 1\). (Remember: \(a \neq 0\))

(ii) \(\{(a^m)^n\}^p = (a^{mn})^p = a^{mnp}\), where a is a nonzero number and m, n and p are integers.

(iii) If n is an even integer, \((-1)^n = (-1)^{2m} = \{(-1)^2\}^m = \{(-1) \times (-1)\}^m = 1^m = 1\). If n is an odd integer, \((-1)^n = (-1)^{2m+1} = (-1) \times (-1)^{2m} = -1 \times 1 = -1\).

Fractional Indices

\(x^m \times x^n = x^{m+n}\), so \(x^{1/2} \times x^{1/2} = x^{1/2+1/2} = x\).

But \(x^{1/2} \times x^{1/2} = (x^{1/2})^2\), so \((x^{1/2})^2 = x\).

Taking the square root of both sides, \(x^{1/2} = \sqrt{x}\).

Again, \(x^{1/3} \times x^{1/3} \times x^{1/3} = x^{1/3+1/3+1/3} = x\).

In other words, \((x^{1/3})^3 = x\).

Taking the cube root of both sides, \(x^{1/3} = \sqrt[3]{x}\).

Similarly, \((x^{1/4})^4 = x\), so \(x^{1/4} = \sqrt[4]{x}\).

In general, \((x^{1/n})^n = x\) or \(x^{1/n} = \sqrt[n]{x}\)

Now, \(x^{2/3} \times x^{2/3} \times x^{2/3} = x^{2/3+2/3+2/3} = x^2\).

In other words, \((x^{2/3})^3 = x^2\).

Taking the cube root of both sides, \(x^{2/3} = (x^2)^{1/3}\).

Also, \((x^{1/3})^2 = x^{1/3} \times x^{1/3} = x^{1/3+1/3} = x^{2/3}\).

So, \(x^{2/3} = (x^2)^{1/3} = (x^{1/3})^2\).

In general, \(x^{m/n} = (x^m)^{1/n} = (x^{1/n})^m\)

Solved Examples

Example 1

Evaluate each of the following expressions.

(i) \((-8)^0\)

(ii) \((6 + 4 + 1)^0\)

(iii) \(6^0 + 4^0 + 1^0\)

(iv) \((6x)^0, x \neq 0\)

(v) \(6x^0, x \neq 0\)

(vi) \((7^0)^{-2}\)

(vii) \([(x^4)^0]^{-3}, x \neq 0\)

(viii) \([(x^{-5})^0]^2, x \neq 0\)

Solution

(i) \((-8)^0 = 1\). [Since \(x^0 = 1, x \neq 0\)]

(ii) \((6 + 4 + 1)^0 = 11^0 = 1\).

(iii) \(6^0 + 4^0 + 1^0 = 1 + 1 + 1 = 3\).

(iv) \((6x)^0 = 6^0 \times x^0 = 1 \times 1 = 1\). [Since \((xy)^m = x^m y^m\)]

(v) \(6x^0 = 6 \times 1 = 6\).

(vi) \((7^0)^{-2} = 7^{0 \times (-2)} = 7^0 = 1\). [Since \((x^m)^n = x^{mn}\)]

(vii) \([(x^4)^0]^{-3} = (x^4)^{0 \times (-3)} = (x^4)^0 = x^{4 \times 0} = x^0 = 1\).

(viii) \([(x^{-5})^0]^2 = (x^{-5})^{0 \times 2} = (x^{-5})^0 = x^{(-5) \times 0} = x^0 = 1\).

Example 2

Simplify and express the result with positive index.

(i) \((7^3)^2\)

(ii) \([(5)^{-3}]^6\)

(iii) \((2^8)^{-2}\)

(iv) \([(3)^{-7}]^{-3}\)

(v) \([(-x)^{-4}]^{-3}\)

(vi) \([2a^{-3}]^{-3}\)

Solution

(i) \((7^3)^2 = 7^{3 \times 2} = 7^6\).

(ii) \([(5)^{-3}]^6 = (5)^{-3 \times 6} = (5)^{-18} = \frac{1}{5^{18}}\). [Since \(x^{-m} = \frac{1}{x^m}\)]

(iii) \((2^8)^{-2} = 2^{8 \times (-2)} = 2^{-16} = \frac{1}{2^{16}}\).

(iv) \([(3)^{-7}]^{-3} = (3)^{(-7) \times (-3)} = 3^{21}\).

(v) \([(-x)^{-4}]^{-3} = (-x)^{(-4) \times (-3)} = (-x)^{12} = [(-1) \times x]^{12} = (-1)^{12} \times x^{12} = 1 \times x^{12} = x^{12}\). [Since \((-1)^n = 1\) if n is even]

(vi) \([2a^{-3}]^{-3} = (2)^{-3} \times (a^{-3})^{-3} = \frac{1}{2^3} \times a^{(-3) \times (-3)} = \frac{1}{2^3} \times a^9\).

Example 3

Simplify (i) \([(3^{-2})^3]^{-4}\), (ii) \(\left[\frac{8^3}{4^2}\right]^4\).

Solution

(i) \([(3^{-2})^3]^{-4} = [3^{-2 \times 3}]^{-4} = (3^{-6})^{-4} = 3^{(-6) \times (-4)} = 3^{24}\).

(ii) \(\left[\frac{8^3}{4^2}\right]^4 = \left[\frac{(2^3)^3}{(2^2)^2}\right]^4 = \left[\frac{2^{3 \times 3}}{2^{2 \times 2}}\right]^4 = \left[\frac{2^9}{2^4}\right]^4 = (2^{9-4})^4\). [Since \(\frac{x^m}{x^n} = x^{m-n}\)]

\(= (2^5)^4 = 2^{5 \times 4} = 2^{20}\).

Example 4

Simplify the following.

(i) \(\frac{8x^5 y^7}{12x^8 y^4}\)

(ii) \(\left(\frac{2a^{-3}}{3b^2}\right)^2\)

(iii) \(\left(\frac{-5x^3}{2y^{-4}}\right)^3\)

Solution

(i) \(\frac{8x^5 y^7}{12x^8 y^4} = \frac{8}{12} \times \frac{x^5}{x^8} \times \frac{y^7}{y^4} = \frac{2}{3} \times x^{5-8} \times y^{7-4} = \frac{2}{3} \times x^{-3} \times y^3 = \frac{2y^3}{3x^4}\).

(ii) \(\left(\frac{2a^{-3}}{3b^2}\right)^2 = \frac{(2a^{-3})^2}{(3b^2)^2}\). [Since \(\left(\frac{x}{y}\right)^m = \frac{x^m}{y^m}\)]

\(= \frac{(2)^2 \cdot (a^{-3})^2}{3^2 \cdot (b^2)^2} = \frac{4 \cdot a^{-3 \times 2}}{9 \cdot b^{2 \times 2}} = \frac{4 \cdot a^{-6}}{9 \cdot b^4} = \frac{4}{9a^6 b^4}\).

(iii) \(\left(\frac{-5x^3}{2y^{-4}}\right)^3 = \frac{(-5)^3 \cdot (x^3)^3}{(2)^3 \cdot (y^{-4})^3} = \frac{-125 \cdot x^9}{8 \cdot y^{-12}} = \frac{-8 \times x^{-9}}{125y^{12}} = \frac{-8}{125x^9 y^{12}}\).

Teacher's Note

Exponents are used everywhere in science and engineering - from calculating the area of a square to understanding how populations grow or decay over time.

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