South African NSC Mathematics

## Laws of exponents For a base $a > 0$ and rational exponents $m$ and $n$, the laws are these. $$ \begin{aligned} a^m \times a^n &= a^{m+n} \\ a^m \div a^n &= a^{m-n} \\ (a^m)^n &= a^{mn} \\ (ab)^n &= a^n b^n \\ \left(\frac{a}{b}\right)^n &= \frac{a^n}{b^n} \\ a^0 &= 1 \\ a^{-n} &= \frac{1}{a^n} \end{aligned} $$ A **rational exponent** is a root written as a power: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$. So $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 4$ and $x^{-\frac{1}{2}} = \dfrac{1}{\sqrt{x}}$. Taking the root first keeps the numbers small. The laws apply only when the bases are the same, or can be made the same. The sum $2^x + 2^{x+1}$ cannot be combined with the product law, but it can be factorised: $2^x + 2^{x+1} = 2^x(1 + 2) = 3 \cdot 2^x$. ### Making the bases the same Write every base as a product of primes and then collect the powers. To simplify $\dfrac{2^{n+2} \cdot 3^{n}}{6^{n+1}}$: $$ \begin{aligned} \frac{2^{n+2} \cdot 3^{n}}{6^{n+1}} &= \frac{2^{n+2} \cdot 3^{n}}{2^{n+1} \cdot 3^{n+1}} \\ &= 2^{(n+2)-(n+1)} \cdot 3^{n-(n+1)} \\ &= 2^{1} \cdot 3^{-1} = \frac{2}{3} \end{aligned} $$ To simplify $\dfrac{9^{x} \cdot 27^{x-1}}{3^{5x-2}}$, write $9 = 3^2$ and $27 = 3^3$: $$ \begin{aligned} \frac{3^{2x} \cdot 3^{3x-3}}{3^{5x-2}} &= 3^{2x + 3x - 3 - 5x + 2} \\ &= 3^{-1} = \frac{1}{3} \end{aligned} $$ When a power has a sum in its exponent and the answer must be a single term, split it with the product law first, for example $5^{x+2} - 5^{x} = 5^x \cdot 25 - 5^x = 5^x(25 - 1)…