Table of Contents

Last modified on October 24th, 2024

A negative exponent is defined as the reciprocal (or multiplicative inverse) of the base, raised to the corresponding positive exponent.

Thus, while positive exponents involve repeated multiplication of a number, negative exponents indicate how many times to divide by that number.

Thus, 5^{-3} = ${\dfrac{1}{125}}$

Here are a few more examples of negative exponents:

- 8
^{-3}= ${\dfrac{1}{8^{3}}}$ = ${\dfrac{1}{512}}$ - 7
^{-2}= ${\dfrac{1}{7^{2}}}$ = ${\dfrac{1}{49}}$ - 3
^{-5}= ${\dfrac{1}{3^{5}}}$ = ${\dfrac{1}{243}}$

Thus, larger negative exponents represent smaller numbers.

We follow a set of rules while simplifying problems involving negative exponents. These rules are applied along with the basic exponent rules.

It states that for base x with the negative exponent -n, we take the reciprocal of the base ${\dfrac{1}{x}}$ and multiply it by itself n times.

**${x^{-n}=\dfrac{1}{x^{n}}}$**

Rule 1 remains the same even when there is a negative exponent in the denominator:

**${\dfrac{1}{x^{-n}}=x\times x\times \ldots \times x}$ (n times) = x ^{n}**

When multiplying exponents, we follow the following rules:

- If the bases are the same, the exponents are added
- If the bases are different but the exponents are the same, the bases are multiplied by keeping the exponents unchanged.

For example,

2^{-3} × 2^{-2} = 2^{-3 + (-2)} = 2^{-5}

When dividing exponents, we follow the following rules:

- If the bases are the same, the exponents are subtracted
- If the bases are different but the exponents are the same, the bases are divided by keeping the exponents unchanged.

For example,

${\dfrac{2^{-3}}{2^{-2}}}$ = 2^{-3 – (-2)} = 2^{-1}

**Evaluate:****a) 3 ^{-3}**

Solution:

a) Here, 3^{-3} = ${\dfrac{1}{3^{3}}}$ = ${\dfrac{1}{27}}$

b) Here, 5^{-1} = ${\dfrac{1}{5^{1}}}$ = ${\dfrac{1}{5}}$

c) Here, 8^{-2} = ${\dfrac{1}{8^{2}}}$ = ${\dfrac{1}{64}}$

**Apply the negative exponent rule to simplify:****a) ${\dfrac{1}{5^{-2}}}$****b) ${\dfrac{1}{2^{-4}}}$****c) ${\dfrac{1}{10^{-3}}}$**

Solution:

a) Here, ${\dfrac{1}{5^{-2}}}$ = 5^{2} = 25

b) Here, ${\dfrac{1}{2^{-4}}}$ = 2^{4} = 16

c) Here, ${\dfrac{1}{10^{-3}}}$ = 10^{3} = 1000

**Convert the following fractions into expressions with positive exponents:****a) ${\dfrac{1}{9^{5}}}$****b) ${\dfrac{1}{12^{2}}}$****c) ${\dfrac{1}{4^{7}}}$**

Solution:

a) Here, ${\dfrac{1}{9^{5}}}$ = 9^{-5}

b) Here, ${\dfrac{1}{12^{2}}}$ = 12^{-2}

c) Here, ${\dfrac{1}{4^{7}}}$ = 4^{-7}

**Apply the exponent rules to multiply the expression with negative exponents 2 ^{-3} × 2^{4} × 2^{-2}**

Solution:

Given, 2^{-3} × 2^{4} × 2^{-2}

Using the product rule,

= 2^{-3 + 4 – 2}

= 2^{-1}

Using the negative exponent rule,

= ${\dfrac{1}{2}}$

Thus, 2^{-3} × 2^{4} × 2^{-2} = ${\dfrac{1}{2}}$

**Divide the expression with negative exponents ${\dfrac{8^{-2}}{8^{-6}}}$**

Solution:

Given, ${\dfrac{8^{-2}}{8^{-6}}}$

Using the quotient rule,

= 8^{-2 – (-6)}

= 8^{4}

**Simplify:****a) ${\dfrac{10^{-2}\times 5^{-3}}{10^{-5}}}$****b) ${\dfrac{4^{-3}\times 2^{-2}}{4^{-1}\times 2^{-4}}}$**

Solution:

a) Given, ${\dfrac{10^{-2}\times 5^{-3}}{10^{-5}}}$

Using the quotient rule,

= ${10^{-2-\left( -5\right) }\times 5^{-3}}$

= ${10^{3}\times 5^{-3}}$

Using the power of product rule,

= ${\left( 2\times 5\right) ^{3}\times 5^{-3}}$

= ${2^{3}\times 5^{3}\times 5^{-3}}$

Using the product rule,

= 2^{3} × 5^{3 + (-3)}

= 2^{3} × 5^{0}

Using the zero exponent rule,

= 2^{3} × 1

= 8

Thus, ${\dfrac{10^{-2}\times 5^{-3}}{10^{-5}}}$ = 8

b) Given, ${\dfrac{4^{-3}\times 2^{-2}}{4^{-1}\times 2^{-4}}}$

Using the power of the quotient rule,

= ${\dfrac{4^{-3}}{4^{-1}}\times \dfrac{2^{-2}}{2^{-4}}}$

Using the quotient rule,

= 4^{-3 – (-1)} × 2^{-2 – (-4)}

= 4^{-2} × 2^{2}

= (2^{2})^{-2} × 2^{2}

Using the power of power rule,

= 2^{-4} × 2^{2}

Using the product rule,

= 2^{-4 + 2}

= 2^{-2}

Using the negative exponent rule,

= ${\dfrac{1}{2^{2}}}$

= ${\dfrac{1}{4}}$

Thus, ${\dfrac{4^{-3}\times 2^{-2}}{4^{-1}\times 2^{-4}}}$ = ${\dfrac{1}{4}}$

**Problem: **Solving **NEGATIVE EXPONENTS **with **FRACTIONS**

**Simplify ${\left( \dfrac{2}{3}\right) ^{-2}}$**

Solution:

Here, ${\left( \dfrac{2}{3}\right) ^{-2}}$

= $[\dfrac{2^{-2}}{3^{-2}}}$

= ${\dfrac{1}{2^{2}}\times 3^{2}}$

= ${\dfrac{9}{4}}$

** **

**Problem: **Simplifying **NEGATIVE FRACTIONAL EXPONENTS**

**Simplify ${\left( \dfrac{8}{27}\right) ^{-\dfrac{2}{3}}}$**

Solution:

Here, ${\left( \dfrac{8}{27}\right) ^{-\dfrac{2}{3}}}$

= ${\left( \dfrac{27}{8}\right) ^{\dfrac{2}{3}}}$

= ${\left( \dfrac{\sqrt[3] {27}}{\sqrt[3] {8}}\right) ^{2}}$

= ${\left( \dfrac{3}{2}\right) ^{2}}$

= ${\dfrac{9}{4}}$

Last modified on October 24th, 2024