Index Laws

When the bases match, multiplying adds indices and dividing subtracts them: $x^7 \times x^4 = x^{11}$ and $x^7 \div x^4 = x^3$.

Raising a power to a power multiplies the indices, because you are repeating the whole block: $(x^3)^4$ means four copies of $x^3$, which is $x^{12}$.

A product inside a bracket distributes to every factor: $(2x^3)^4 = 2^4 x^{12} = 16x^{12}$. Forgetting to raise the coefficient is the single most common slip here.

Same base: multiply means add indices, divide means subtract indices, power of a power means multiply indices. Everything inside a bracket gets the outside index, including the coefficient.

Pause, copy the three core laws ($a^m a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, $(a^m)^n = a^{mn}$) and one worked example showing the coefficient being raised, such as $(2x^3)^4 = 16x^{12}$, into your book.

We just saw that dividing powers of the same base subtracts the indices. That raises a question: what happens when the subtraction gives 0, or a negative number? This card answers it → the division law forces $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$, so neither is an arbitrary rule.

Take $a^3 \div a^3$. By the division law that is $a^0$. But anything non-zero divided by itself is 1. So $a^0 = 1$ is not a convention someone chose, it is what the law forces.

Now take $a^2 \div a^5$. The law gives $a^{-3}$. Writing it out gives $\frac{a \times a}{a \times a \times a \times a \times a} = \frac{1}{a^3}$. So $a^{-n} = \frac{1}{a^n}$.

A negative index therefore moves a factor across the fraction bar. $\frac{2x^{-3}}{y^{-1}} = \frac{2y}{x^3}$. The sign of the answer is untouched.

$a^0 = 1$ because $a^n \div a^n = a^0$ and equals 1. $a^{-n} = \frac{1}{a^n}$, so a negative index means reciprocal, never a negative answer. Negative indices move factors across the fraction bar.

Pause, copy the derivation of $a^0 = 1$ from $a^3 \div a^3$, the meaning of $a^{-n} = \frac{1}{a^n}$, and the warning that $2^{-3}$ is $\frac{1}{8}$ and not $-8$, into your book.

We just saw that negative indices are reciprocals, forced by the division law. That raises a question: what could a fractional index such as $a^{1/2}$ possibly mean? This card answers it → the power-of-a-power law forces $a^{1/n}$ to be the $n$th root of $a$.

If $a^{1/2}$ is to obey the laws, then $\left(a^{1/2}\right)^2 = a^{1} = a$. The number that gives $a$ when squared is $\sqrt{a}$, so $a^{1/2} = \sqrt{a}$. The same argument gives $a^{1/n} = \sqrt[n]{a}$.

For a general fraction, the denominator is the root and the numerator is the power: $a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$.

Take the root first when working by hand. $8^{2/3}$ is easier as $\left(\sqrt[3]{8}\right)^2 = 2^2 = 4$ than as $\sqrt[3]{64}$.

$a^{1/n} = \sqrt[n]{a}$ and $a^{m/n} = \left(\sqrt[n]{a}\right)^m$. The denominator is the root, the numerator is the power. Take the root first to keep the numbers small. A negative fractional index means reciprocal as well.

Pause, copy $a^{m/n} = \left(\sqrt[n]{a}\right)^m$, the root-first strategy, and the fully worked $27^{-2/3} = \frac{1}{9}$ showing both the reciprocal and the root, into your book.

1. Simplify $a^9 \div a^4 \times a^2$.
2. Simplify $(5m^3n^2)^2$.
3. Evaluate $81^{1/2} + 2^{-2}$.
4. Write $\dfrac{3}{x^{-5}}$ with a positive index.