Lesson 5, Topic 3
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# Law 3: Negative Indices (Powers)

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This law applies when the power is negative.

$$\scriptsize a^{-n} = \normalsize \frac{1}{a^n}$$

In other words, an index number with a negative power is equal to the inverse of its index number with positive power.

$$\scriptsize 3^{-3}= \normalsize \frac{1}{3^{3}}$$

$$\scriptsize 2^{-4}= \normalsize \frac{1}{2^{4}}$$

### Example 1

Simplify the following

a. $$\scriptsize 2^4 \: \div \: 2^8$$

b. y2  ÷  y6

c. x5 ÷ x7

Solution

a. $$\scriptsize 2^4 \: \div \: 2^8 \\= \frac{\not{2} \: \times \: \not{2} \: \times \: \not{2}\: \times \: \not{2}}{\not{2} \: \times \: \not{2} \: \times \: \not{2}\: \times \: \not{2}\: \times \: 2\: \times \: 2\: \times \: 2 \: \times \: 2} \\= \frac{1}{2^4}$$

Alternatively, using the law:

$$\scriptsize 2^4 \: \div \: 2^8 \\ \scriptsize = 2^{4 \: – \: 8} \\ \scriptsize = 2^{-4}$$

= $$\frac{1}{2^4}$$

b. $$\scriptsize y^2 \: \div \: y^6 \\ = \frac{\not{y} \: \times \: \not{y}}{\not{y}\: \times \: \not{y} \: \times \: y \: \times \: y \: \times \: y\: \times \: y} \\= \frac{1}{y^4}$$

Alternatively, using the law:

$$\scriptsize y^2 \: \div \: y^6 \\ \scriptsize = y^{2 \: – \: 6} \\ \scriptsize = y^{-4}$$

= $$\frac{1}{y^4}$$

c. $$\scriptsize x^5 \: \div \: x^7 \\= \frac{ \not{x} \: \times \: \not{x}\: \times \: \not{x} \: \times \: \not{x} \: \times \: \not{x}}{\not{x} \: \times \: \not{x} \: \times \: \not{x}\: \times \: \not{x}\: \times \: \not{x}\: \times \: x\: \times \: x } \\= \frac{1}{x^2}$$

Alternatively, using the law:

$$\scriptsize x^5 \: \div \: x^7 \\ \scriptsize = x^{5 \: – \: 7} \\ \scriptsize = x^{-2}$$

= $$\frac{1}{x^2}$$

### Example 2

Simplify the following;

a. $$\scriptsize 5^{-2}$$

b. $$\scriptsize 3^{-5}$$

c. $$\left(\frac{2}{3} \right)^{-1}$$

d. $$\left(\frac{4}{5} \right)^{-2}$$

e. $$\scriptsize 2^5 \: \times \: 2^{-3} \: \times \: 2^{-8}$$

f. $$\scriptsize \left( \scriptsize 5y \right)^{-2}$$

g. $$\scriptsize 5y^{-2}$$

h. $$\scriptsize 5^{-2}y$$

Solution

a. $$\scriptsize 5^{-2}$$ (negative index)

= $$\frac{1}{5^2}$$ (positive index)

b. $$\scriptsize 3^{-5}$$ (negative index)

= $$\frac{1}{3^5}$$ (positive index)

c. $$\left(\frac{2}{3} \right)^{-1}$$ (negative index)

= $$\frac{1}{\frac{2}{3}}$$

= $$\scriptsize 1 \: \div \: \frac{2}{3}$$

= $$\scriptsize 1 \: \times\: \frac{3}{2}$$

= $$\frac{3}{2}$$ (positive index)

d. $$\left(\frac{4}{5} \right)^{-2}$$ (negative index)

= $$\frac{1}{\left(\frac{4}{5}\right)^2}$$

= $$\scriptsize 1 \: \div \: \left(\frac{4}{5}\right)^2$$

= $$\scriptsize 1 \: \times\: \left(\frac{5}{4}\right)^2$$

= $$\left(\frac{5}{4}\right)^2$$ (positive index)

= $$\frac{25}{16}$$

e. $$\scriptsize 2^5 \: \times \: 2^{-3} \: \times \: 2^{-8}$$ (same base)

= $$\scriptsize 2^{5 +(-3)+(-8)}$$

= $$\scriptsize 2^{5 -3-8}$$

= $$\scriptsize 2^{5 -11}$$

= $$\scriptsize 2^{-6}$$

=$$\frac{1}{2^6}$$

f. $$\scriptsize \left( \scriptsize 5y \right)^{-2}$$

Note: 5y is enclosed in the bracket. The index affects both 5 and y.

=$$\frac{1}{(5y)^2}$$

= $$\frac{1}{5y \: \times \: 5y}$$

= $$\frac{1}{25y^2}$$

g. $$\scriptsize \scriptsize 5y^{-2}$$

Note: The negative index affects only y in this case

= $$\scriptsize 5 \: \times \: y^{-2}$$

= $$\scriptsize 5 \: \times \: \normalsize \frac{1}{y^2}$$

= $$\frac{5}{y^2}$$

h. $$\scriptsize \scriptsize 5^{-2}y$$

Note: The negative index affects only 5 in this case

= $$\scriptsize 5^{-2} \: \times \: y$$

= $$\frac{1}{5^2} \scriptsize \: \times \: y$$

= $$\frac{1}{25} \scriptsize \: \times \: y$$

= $$\frac{y}{25}$$

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