NSW HSC Mathematics Advanced

## Index laws For a base $a \neq 0$ and whole-number or fractional indices, the laws follow from what an index counts: $a^m$ is $m$ copies of $a$ multiplied together, so combining them just counts copies. $$ a^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}, \qquad (ab)^n = a^n b^n, \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} $$ Dividing $a^n$ by itself gives $1$ by the second law and $a^{n-n} = a^0$, so $a^0 = 1$ for $a \neq 0$. Dividing $a^0$ by $a^n$ in the same way gives a negative index: $$ a^{-n} = \frac{1}{a^n}, \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n} $$ A fractional index is a root. Since $\left(a^{1/n}\right)^n = a$, the number $a^{1/n}$ must be the $n$th root of $a$, and combining this with the power law gives $$ a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m} $$ Taking the root first keeps the numbers small, so $8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$, and $16^{-3/4} = \dfrac{1}{\left(\sqrt[4]{16}\right)^3} = \dfrac{1}{8}$. Worked example: simplify $\dfrac{(2x^3)^2 \times x^{-5}}{\sqrt{x}}$. $$ \begin{aligned} \frac{(2x^3)^2 \times x^{-5}}{x^{1/2}} &= \frac{4x^6 \times x^{-5}}{x^{1/2}} \\ &= \frac{4x^{1}}{x^{1/2}} \\ &= 4x^{1/2} \end{aligned} $$ An equation with the unknown in the index is solved by writing both sides with the same base. To solve $4^{x} = 32$, write $4 = 2^2$ and $32 = 2^5$, so $2^{2x} = 2^5$, which gives $2x = 5$ and $x =…