Powers and Their Laws
An exponent counts repeated factors, and every exponent law — including a⁰ = 1 and a⁻ⁿ = 1/aⁿ — follows from counting the factors or continuing the dividing pattern. · 12 min
Write a⁵ and you have written five a's multiplied together: a · a · a · a · a. That is the whole definition — the exponent counts factors. Every law in this folio follows from that counting, which means there is nothing here to memorize that you cannot rebuild in ten seconds by writing the factors out. Readers who learn the counting never mix up the laws; readers who memorize the laws mix them up for years.
Guess before you learn
Before any rule is stated: 2³ · 2⁴ = ?
Write the factors out: (2 · 2 · 2) · (2 · 2 · 2 · 2) is seven 2's multiplied together — 2⁷. If you chose 2¹², you multiplied the exponents; that move is real, but it belongs to a different pattern, and this folio will show you exactly which.
9–12
3–5
An exponent is a tally. 2⁴ means multiply four 2's: 2 × 2 × 2 × 2 = 16. Careful — it does not mean 2 × 4. The tally says how many copies, and the copies are multiplied, not added. That is why powers grow so fast: 2¹⁰ is already 1,024.
When you multiply powers of the same number, the tallies pool. Three 2's times two more 2's is five 2's: 2³ × 2² = 2⁵. Count the copies and the rule writes itself.
6–8
aᵐ means m factors of a. From that: aᵐ · aⁿ pools the factors — m of them, then n more — giving aᵐ⁺ⁿ. And aᵐ ÷ aⁿ cancels n factors out of m, leaving aᵐ⁻ⁿ. Write out (a · a · a · a · a) ÷ (a · a) once, and the quotient law stops being a rule and becomes a fact you saw.
(aᵐ)ⁿ counts groups of groups: n groups, each holding m factors, is mn factors in all — so (a³)² = a⁶, not a⁵. The commonest error in this unit is pooling exponents when the pattern calls for counting groups. When unsure, write four factors out; the answer is sitting there.
9–12
The three laws — product aᵐ · aⁿ = aᵐ⁺ⁿ, quotient aᵐ/aⁿ = aᵐ⁻ⁿ, power (aᵐ)ⁿ = aᵐⁿ — are theorems about counting factors, each provable in one line by writing the factors out. But counting only defines aⁿ for positive whole n. What should a⁰ or a⁻² mean? Nothing forces a meaning; we choose one.
The choice is made for us if we ask the quotient law to keep working: a³/a³ should be a⁰, and it plainly equals 1; a²/a⁵ should be a⁻³, and writing the factors gives 1/a³. So a⁰ = 1 and a⁻ⁿ = 1/aⁿ are not new facts — they are the only definitions that keep the old pattern honest, for a ≠ 0.
K–2
Fold a paper once: 2 layers. Fold again: 4. Again: 8. Each fold doubles the pile. Three folds make 2 × 2 × 2 = 8 layers.
We write 2³ for 2 × 2 × 2. The little 3 counts how many 2's we multiply. It counts; it does not multiply.
Undergrad
For fixed a ≠ 0, the map n ↦ aⁿ from the integers under addition to numbers under multiplication is pinned down by one demand: f(m + n) = f(m) · f(n). That homomorphism property forces f(0) = 1 — it must be the multiplicative identity — and f(−n) = 1/f(n), since inverses must map to inverses. The school 'dividing pattern' is this structure seen from below.
The extensions continue: demanding the same law on the rationals makes a^(1/q) the positive qth root of a, and continuity then extends the map uniquely to all real exponents — which is where the exponential function comes from. Algebra I's counting argument is the first rung of that ladder.
Postgrad
aⁿ is the image of n under the unique monoid homomorphism ℕ → (R, ·) sending 1 ↦ a; the exponent laws restate that free commutative monoids carry well-defined multiplicities. Extending along ℕ ↪ ℤ is group completion, available exactly when a is a unit — hence the a ≠ 0 proviso attached to negative exponents.
Analytically, a^x := exp(x log a) is the unique continuous homomorphism from (ℝ, +) to the positive reals under multiplication sending 1 ↦ a; measurability alone suffices for uniqueness, by Cauchy's functional equation. The pattern-continuation argument is a shadow of a rigidity theorem — the homomorphism condition leaves no freedom at 0 or at −n.
base and exponent
In aⁿ, the base a is the factor being repeated; the exponent n is the count of factors. Read it 'a to the n'.
Now the laws, by counting. Product: a³ · a² is (a · a · a)(a · a) — five factors — so a³ · a² = a⁵, and in general aᵐ · aⁿ = aᵐ⁺ⁿ. Quotient: a⁵/a² cancels two factors, leaving a³; in general aᵐ/aⁿ = aᵐ⁻ⁿ. Power of a power: (a³)² means two groups of three factors — six in all — so (aᵐ)ⁿ = aᵐⁿ. Add when factors pool; multiply when groups nest. One caution: every law needs the same base. 2³ · 5² pools nothing.
Simplify (x⁵ · x²) / x³ — the steps fade as you master them
x⁷ / x³
x⁴
4 factors of x remain
Counting factors defines a⁵, a², a¹ — but what could a⁰ mean? Zero copies of a? The pattern answers. Read down the powers of 2: 2³ = 8, 2² = 4, 2¹ = 2. Each step down divides by 2. Keep stepping: 2⁰ must be 1. Step again: 2⁻¹ = 1/2, then 2⁻² = 1/4, and in general a⁻ⁿ = 1/aⁿ. These are definitions, chosen so the quotient law never breaks — not mysteries. And note what 2⁰ is not: it is not 0. An empty product starts at 1, the way an empty sum starts at 0.
Why is this true?
Why is a⁰ defined as 1 rather than 0?
Because a³/a³ must equal both a⁰ (by the quotient law) and 1 (anything nonzero divided by itself). Defining a⁰ as 0 would break the law; defining it as 1 keeps every pattern intact.
Powers now behave: pool, cancel, nest, and continue the pattern past zero. Keep the counting argument in your pocket — any time a law goes blurry, write three factors out and it sharpens again. Next folio, the last one: the unknown multiplies itself, and a single new principle cracks the equation that balancing alone cannot.
Practice — new ink and old, interleaved
1.Put these values in order from least to greatest.
- 2⁰
- 2⁻²
- 2³
- 2²
2.Evaluate (−3)² − 3².
3.2 + 3 × 4² = ?
4.Which expression equals x¹⁰?
5.From memory: what does a⁻ⁿ mean, and why?
a⁻ⁿ = 1/aⁿ, because continuing the divide-by-a pattern below a⁰ = 1 makes n more divisions by a.
How close were you? Grade yourself honestly — it sets your review date.
6.Match each expression to its simplified form.
7.Simplify, then evaluate: 3⁵ / 3³.
8.Put the moves of the substitution method in working order.
- Replace y in the other equation with (x + 3)
- Solve one equation for y
- Solve the one-variable equation for x
- Back-substitute to find y
- Check the pair in both original equations
9.From memory: state the slope formula, and name what positive, negative, zero, and undefined slopes look like.
m = (y₂ − y₁)/(x₂ − x₁); positive climbs left to right, negative falls, zero is horizontal, undefined is vertical.
How close were you? Grade yourself honestly — it sets your review date.