Laws of Exponents

This section covers the laws governing exponents, including their properties and how they apply to negative exponents.

The section explains the basic laws of exponents, how they apply to negative numbers, and provides examples of simplifying exponents through various mathematical operations. It highlights the significance of exponent rules in expressing mathematical expressions clearly and concisely.

Detailed Summary of Laws of Exponents

In this section, we explore the key laws of exponents, which are rules used in the manipulation and simplification of expressions involving exponents.

1. Product of Powers Law: For any non-zero integer a, the product of powers states that:
   a^m × a^n = a^(m+n). This rule applies even when m and n are negative integers, confirming that this law is universal across positive, negative, and zero exponents.

2. Quotient of Powers Law: This law states that:
   a^m ÷ a^n = a^(m-n). Thus, when dividing powers with the same base, we subtract the exponents regardless of their signs.

3. Power of a Power Law: When raising a power to another power, the rule is:
   (a^m)^n = a^(m*n). This rule also holds for negative and positive exponents.

4. Zero Exponent Rule: Any non-zero number raised to the power of zero is equal to one:
   a^0 = 1.

5. Negative Exponent Rule: A negative exponent indicates a reciprocal:
   a^(-m) = 1/(a^m). This means we can rewrite the negative exponent as a positive exponent in the denominator.

The section provides several examples for each law, ensuring clarity in understanding how to simplify expressions that involve these rules. Additionally, exercises are included to reinforce understanding and application of these laws.

Example : Simplify

(i) $\left( \frac{4}{5} \right)^{-3} \left( \frac{2}{3} \right)^{2}$
(ii) $\left( \frac{6}{7} \right)^{-4} \left( \frac{9}{10} \right)^{-2}$

Solution:

(i) $\left( \frac{4}{5} \right)^{-3} = \frac{5^{3}}{4^{3}} = \frac{125}{64}$
$\left( \frac{2}{3} \right)^{2} = \frac{4}{9}$
$\frac{125}{64} \times \frac{4}{9} = \frac{500}{576} = \frac{125}{144}$

(ii) $\left( \frac{6}{7} \right)^{-4} = \frac{7^{4}}{6^{4}} = \frac{2401}{1296}$
$\left( \frac{9}{10} \right)^{-2} = \frac{10^{2}}{9^{2}} = \frac{100}{81}$
$\frac{2401}{1296} \times \frac{100}{81} = \frac{240100}{104976}$

In general, $\left( \frac{a}{b} \right)^{m} = \frac{b^{m}}{a^{m}}$

New Question: Simplify

(i) $\left( \frac{3}{4} \right)^{-2} \left( \frac{5}{6} \right)^{3}$
(ii) $\left( \frac{8}{9} \right)^{-5} \left( \frac{11}{12} \right)^{1}$

Solution:

(i) $\left( \frac{3}{4} \right)^{-2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$
$\left( \frac{5}{6} \right)^{3} = \frac{125}{216}$
$\frac{16}{9} \times \frac{125}{216} = \frac{2000}{1944} = \frac{500}{486}$

(ii) $\left( \frac{8}{9} \right)^{-5} = \frac{9^{5}}{8^{5}} = \frac{59049}{32768}$
$\left( \frac{11}{12} \right)^{1} = \frac{11}{12}$
$\frac{59049}{32768} \times \frac{11}{12} = \frac{649539}{393216}$

In general, $\left( \frac{a}{b} \right)^{m} = \frac{b^{m}}{a^{m}}$