Laws of Indices (Exponents): Every Rule With Examples

Quick answer: The laws of indices are shortcuts for powers. When you multiply, add the powers (a^m × a^n = a^(m+n)); when you divide, subtract them (a^m ÷ a^n = a^(m-n)); a power of a power multiplies them ((a^m)^n = a^(mn)). From those three you get the rest: anything to the power 0 is 1, a negative power means "one over" (a^-n = 1/a^n), and a fractional power is a root (a^(1/n) is the nth root of a). They only work when the bases are the same.

Indices were the first bit of IGCSE maths that made me feel like I was getting away with something. Instead of writing 2 × 2 × 2 × 2 × 2, you write 2^5, and suddenly a whole page of multiplication collapses into two symbols. The rules that make that work are short, they never change, and once they click they hand you free marks on almost every non-calculator paper. Here is the whole set, the way I actually use them, plus the mistakes that used to cost me.

The three core laws (learn these first)

Everything else is built from these. The base has to be the same for the first two to work.

• Multiplying: a^m × a^n = a^(m+n). So p^5 × p^3 = p^8. You add the powers.
• Dividing: a^m ÷ a^n = a^(m-n). So p^7 ÷ p^2 = p^5. You subtract.
• Power of a power: (a^m)^n = a^(mn). So (p^4)^3 = p^12. You multiply.

One more that examiners love to hide inside a bracket: a power outside a bracket hits everything inside it. (ab)^n = a^n × b^n. That is why (2x)^3 = 2^3 × x^3 = 8x^3, not 2x^3. The 2 gets cubed too.

A quick algebra one, so you see them combine: simplify (2x^3)^4 ÷ x^5. First (2x^3)^4 = 16x^12, then divide by x^5 to subtract powers: 16x^(12-5) = 16x^7. Two rules, one line.

Zero, negative and fractional indices (where the marks hide)

This is the part that separates a grade 5 from a grade 8, and it is worth knowing why each one is true, because "why" is how you stop forgetting them under pressure.

Anything to the power 0 is 1. Take a^3 ÷ a^3. By the division law that is a^(3-3) = a^0. But any number divided by itself is 1. So a^0 = 1. It has to be. (2026^0 = 1. So does (-7)^0.)

A negative power means the reciprocal — flip it. Take a^2 ÷ a^5 = a^(2-5) = a^-3. But writing the same division as a fraction gives a^2/a^5 = 1/a^3. So a^-3 = 1/a^3. A negative index never makes the answer negative; it moves the term to the bottom of a fraction.

A fractional power is a root. a^(1/2) means the number that, multiplied by itself, gives a — because a^(1/2) × a^(1/2) = a^1 = a. So a^(1/2) is the square root. In general a^(1/n) is the nth root of a. And a^(m/n) combines the two: the bottom is the root, the top is the power. a^(m/n) = (nth root of a)^m.

The one order trick that makes fractions easy

For something like 27^(2/3), do the root first, then the power. Cube root of 27 is 3, then 3^2 = 9. If you square first you get 729 and then have to cube-root that back to 9 — same answer, uglier journey. Small numbers, fewer mistakes.

Which tier this shows up on

Positive whole-number indices and the three core laws turn up everywhere, including Foundation GCSE and IGCSE Core. The negative and fractional stuff is Higher tier on GCSE (AQA, Edexcel and OCR all treat it the same way) and Extended on Cambridge IGCSE. It is also a near-guaranteed WAEC/WASSCE topic, usually paired with surds. If you are aiming above a grade 5, treat negative and fractional indices as non-optional.

Common mistakes I see (and used to make)

• Adding the bases. 2^3 × 2^4 is 2^7, not 4^7. The base stays put; only the powers move.
• Using the laws on addition. 3^2 + 3^2 is not 3^4. The laws are for multiply and divide only. This is just 9 + 9 = 18.
• Forgetting the coefficient in a bracket. (3x)^2 = 9x^2, not 3x^2.
• Thinking a negative power flips the sign. x^-2 = 1/x^2, a positive quantity. Nothing turns negative.
• Leaving the answer unfinished. If a question says "evaluate", it wants a number. Writing 16^(3/4) and stopping scores nothing where the mark scheme demands 8. "Simplify" lets you leave it in index form; "evaluate" does not. Read the command word.

Worked example: evaluate (8/27)^(-2/3)

This one packs a negative and a fractional index into one fraction, which is exactly what an Extended or Higher paper likes. Label every step so an examiner can follow you.

1. Kill the negative power by flipping the fraction. A negative index means reciprocal, so (8/27)^(-2/3) = (27/8)^(2/3).
2. Split the fractional power. The bottom number, 3, is the root; the top, 2, is the power: (27/8)^(2/3) = ((27/8)^(1/3))^2.
3. Take the cube root, top and bottom. Cube root of 27 is 3, cube root of 8 is 2, so (27/8)^(1/3) = 3/2.
4. Apply the remaining power. (3/2)^2 = 9/4.

Answer: 9/4 (which is 2.25). Notice we never touched a calculator — that is the point of doing the root before the power.

If you want to see this method run on your own numbers, drop the expression into Math Solver and it shows each line of working, not just the final number.

Test yourself

Try these on paper first, then check.

1. Simplify: (2p^3)^4
2. Evaluate: 16^(-3/4)
3. Simplify (x and y positive): (x^6 y^-2)^(1/2)

Quick answers: (1) 2^4 × p^12 = 16p^12. (2) 16^(-3/4) = 1 ÷ (16^(1/4))^3 = 1 ÷ 2^3 = 1/8. (3) Halve each power: x^3 y^-1 = x^3/y.

Got one wrong? That is the useful one — turn it into a quick set with the quiz generator to drill the exact type, and check any messy working line by line in Math Solver. Two short sessions and the pattern sticks.

FAQ

Why does anything to the power of zero equal 1?

Because a^m ÷ a^m must be both a^(m-m) = a^0 and, since it is a number over itself, exactly 1. The two have to agree, so a^0 = 1. It is a consequence of the division law, not a random rule.

What does a negative index actually mean?

"One over." a^-n = 1/a^n. So 5^-2 = 1/25 and (2/3)^-1 = 3/2. The index going negative just sends the term to the denominator; it never makes the value negative.

How do I work out a fractional power like 27^(2/3)?

The denominator is the root, the numerator is the power. Do the root first: cube root of 27 is 3, then square it to get 9. Root-first keeps the numbers small.

Do the index laws work if the bases are different?

No. a^m × a^n = a^(m+n) needs the same base. You cannot combine 2^3 × 5^2 into one power — just work them out separately (8 × 25 = 200). The only cross-base rule is the bracket one, (ab)^n = a^n b^n, and that shares the same power, not the same base.

In short: add powers to multiply, subtract to divide, multiply for a power of a power; then zero gives 1, a negative index flips it into a fraction, and a fractional index is a root over a power. Same base every time, root before power, and always finish the sum when it says "evaluate". Make flashcards of the five laws, drill them in Math Solver, and they become the easiest marks on the paper.