Power and Roots

By Stage 9 the three laws of indices should be second nature. The new skill is using them together — a single expression often needs more than one law before it is fully simplified.

The three laws, all needing the same base:

• Multiplying: $a^m \times a^n = a^{m+n}$ — add the indices.
• Dividing: $a^m \div a^n = a^{m-n}$ — subtract the indices.
• Power of a power: $(a^m)^n = a^{mn}$ — multiply the indices.

Take $\frac{2^5 \times 2^4}{2^3}$. Multiply on the top first: $2^5 \times 2^4 = 2^9$. Then divide: $2^9 \div 2^3 = 2^6$. Naming which law you are using at each step keeps a longer simplification under control.

What does a power of zero mean? You cannot multiply a number "zero times", so the meaning comes from the dividing law instead.

Think about $a^4 \div a^4$. A number divided by itself is $1$. But the dividing law says $a^4 \div a^4 = a^{4-4} = a^0$. Both lines describe the same calculation, so $a^0$ must equal $1$.

So $7^0 = 1$, $100^0 = 1$ and even $(-3)^0 = 1$. The base does not matter, as long as it is not zero — $0^0$ is left undefined and you will not be asked for it.

This is genuinely useful inside the index laws. When the indices subtract to zero, you can write the answer as $1$ straight away, instead of leaving it as $a^0$.

Carry the halving pattern past zero. After $2^1 = 2$ and $2^0 = 1$, dividing by 2 again gives $2^{-1} = \frac{1}{2}$, then $2^{-2} = \frac{1}{4}$. A negative index means a reciprocal.

The rule is $a^{-n} = \frac{1}{a^n}$.

So $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$ and $10^{-3} = \frac{1}{1000}$. A negative index does not make the answer negative — it makes a fraction less than 1.

The index laws all still work. $2^5 \div 2^8 = 2^{5-8} = 2^{-3}$, which is $\frac{1}{8}$. Subtracting a bigger index simply lands you on a negative one.

A root undoes a power. The square root asks "what was squared?" — since $9^2 = 81$, $\sqrt{81} = 9$. The cube root asks "what was cubed?" — since $4^3 = 64$, $\sqrt[3]{64} = 4$.

Most numbers are not perfect squares, so their roots are not whole numbers. The Stage 9 skill is estimating such a root by trapping it between two perfect squares you know.

To estimate $\sqrt{50}$: since $7^2 = 49$ and $8^2 = 64$, the answer lies between 7 and 8. As 50 is very close to 49, $\sqrt{50}$ is a little above 7 — about $7.1$.

Knowing the squares to $15^2 = 225$ and the cubes to $5^3 = 125$ makes both finding and estimating roots far quicker.

The toughest Stage 9 index questions ask you to simplify an expression that needs several laws in turn. The cure is the same as always: one step per line.

Try $\frac{(2^3)^2 \times 2^4}{2^{12}}$.

• Power of a power first: $(2^3)^2 = 2^6$.
• Multiply on the top: $2^6 \times 2^4 = 2^{10}$.
• Divide: $2^{10} \div 2^{12} = 2^{-2}$.
• Write the negative index as a fraction: $2^{-2} = \frac{1}{4}$.

When you finish, sense-check the index: a negative final index means an answer below 1, a positive one means an answer above 1. If the size feels wrong, an index has slipped.

Confident work with powers and roots opens up a great deal of later maths:

• Standard form depends entirely on powers of 10, and small numbers use the negative indices you have just met.
• Algebra uses the index laws constantly — simplifying $x^5 \div x^2$ to $x^3$ is the very same rule you used with numbers.
• Area and volume rest on squares and cubes, and roots reverse them to find a missing length.
• Calculator work asks you to estimate first, so trapping a root between perfect values is a real safety net.

If a later topic with powers feels shaky, a quick refresh of the three laws, the zero index and negative indices usually clears it up. Accuracy beats speed every time.