U24 Laws of Indices (II) Essential Formulas

This section extends the fundamental laws of indices to cover rational exponents, negative exponents, and the zero exponent. Mastery of these rules is crucial for simplifying algebraic expressions and solving exponential equations in the DSE exam.

1 Rational and Negative Exponents

Definition of Rational Exponents

For any positive integer $n$ and real number $a$ (where $a \ge 0$ if $n$ is even), the $n$th root of $a$ is defined as $a^{\frac{1}{n}} = \sqrt[n]{a}$. More generally, for integers $m$ and $n$ ($n > 0$), we have $a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$.

$$ a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m $$
Rational Exponent Visualization $x^{\frac{3}{2}} = \sqrt{x^3}$ $8^{\frac{2}{3}} = \sqrt[3]{8^2} = 4$ The exponent m/n indicates a power (m) and a root (n).

Negative Exponents and the Zero Exponent

For any non-zero real number $a$ and integer $n$, a negative exponent represents the reciprocal of the base raised to the positive exponent. The zero exponent rule states that any non-zero base raised to the power of zero equals one.

$$ a^{-n} = \frac{1}{a^n} \quad \text{and} \quad a^{0} = 1 \quad (a \neq 0) $$

2 Combined Application of Index Laws

Simplifying Complex Expressions

The laws of indices can be applied in combination to simplify expressions involving products, quotients, and powers of powers with rational or negative exponents. The key laws are: $a^m \times a^n = a^{m+n}$, $\frac{a^m}{a^n} = a^{m-n}$, and $(a^m)^n = a^{mn}$, where $a \neq 0$ and $m, n$ are rational numbers.

$$ \frac{(x^{\frac{1}{2}} y^{-3})^2}{x^2 y^{\frac{1}{2}}} = x^{1-2} y^{-6-\frac{1}{2}} = x^{-1} y^{-\frac{13}{2}} = \frac{1}{x y^{\frac{13}{2}}} $$

3 Solving Exponential Equations

Equating Exponents with the Same Base

If $a^m = a^n$ for $a > 0$ and $a \neq 1$, then $m = n$. This principle is fundamental for solving equations where the unknown variable is in the exponent. First, express both sides of the equation as powers of the same base.

$$ 8^{x-1} = 16^{2x} \implies (2^3)^{x-1} = (2^4)^{2x} \implies 2^{3x-3} = 2^{8x} $$

Therefore, $3x - 3 = 8x$, which solves to $x = -\frac{3}{5}$.

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