1
The laws are counting, not memory
x² · x³ is two xs multiplied by three xs, which is five xs. If you cannot recall whether the exponents add or multiply, write the factors out for a small case and read the answer off.
Lesson · Foundations of Algebra · Exponents and radicals
Integer Exponents
Three laws, each of which is counting factors rather than remembering a rule. A zero exponent and a negative exponent are then defined — not derived — and the definitions are chosen as the only ones that keep the three laws true for every integer.
xᵐ · xⁿ = xᵐ⁺ⁿ exponents add
xᵐ / xⁿ = xᵐ⁻ⁿ exponents subtract
(xᵐ)ⁿ = xᵐⁿ exponents multiply
x⁰ = 1 and x⁻ⁿ = 1/xⁿ both require x ≠ 0
Three laws and two definitions
Core ideas
What this lesson establishes
The laws are short enough to memorise and short enough to rederive. Rederiving them is better, because it also tells you when they do not apply.
2
x⁰ and x⁻ⁿ are choices, made to keep the laws
"Multiply x by itself zero times" is not a meaningful instruction. The quotient law is what forces the value: extended to equal exponents it makes x³ / x³ into x⁰, and arithmetic makes it 1, so x⁰ must be 1. The definition is designed backwards from the law it has to preserve.
3
Every law is about a single base
x² · y³ does not combine, and 2³ · 3² is 72 — a number, not a power of anything you can read off. The laws move exponents around when the base is the same and say nothing otherwise.
The material
Three laws, then two definitions
Where each law comes from, why zero and negative exponents have the values they do, and where the distribution stops.
Definition · Power with a positive integer exponent
For an integer n ≥ 1, xⁿ means the product of n factors of x. x is the base and n is the exponent.
x¹ = x. Note that this definition says nothing at all about n = 0 or n negative; those cases are handled below, and they have to be handled separately.
The three laws, from counting factors
Theorem · Product, quotient and power laws
For a base x and positive integers m and n: xᵐ · xⁿ = xᵐ⁺ⁿ; xᵐ / xⁿ = xᵐ⁻ⁿ when x ≠ 0 and m > n; and (xᵐ)ⁿ = xᵐⁿ.
The first: xᵐ is m factors and xⁿ is n factors, so their product is m + n factors, which is xᵐ⁺ⁿ.
The second: in xᵐ / xⁿ each of the n factors below cancels one of the m above, leaving m − n factors.
The third: (xᵐ)ⁿ is n copies of a block of m factors, so mn factors in all.
□
x² · x³ = (x·x)(x·x·x) = x⁵
x⁵ / x² = (x·x·x·x·x)/(x·x) = x³
(x²)³ = (x·x)(x·x)(x·x) = x⁶
Two more, and they are the ones that get overextended: (xy)ⁿ = xⁿyⁿ and (x/y)ⁿ = xⁿ/yⁿ for y ≠ 0. An exponent distributes over a product and over a quotient. It does not distribute over a sum: (x + y)² is not x² + y², and at x = 1, y = 3 those are 16 and 10.
Zero and negative exponents
Neither case is covered by the laws as stated, because those were proved for positive exponents by counting factors. So ask instead what the laws would say if they were extended, and let arithmetic say what the value would then have to be:
if the quotient law is to hold when the exponents are equal
x³ / x³ = x³⁻³ = x⁰
and arithmetic gives
x³ / x³ = 1 for x ≠ 0
so x⁰ = 1
if the product law is to hold for a negative exponent
x³ · x⁻³ = x³⁺⁽⁻³⁾ = x⁰ = 1
so x⁻³ = 1 / x³
Definition · Zero and negative exponents
For x ≠ 0 and a positive integer n: x⁰ = 1 and x⁻ⁿ = 1/xⁿ.
These are definitions, not consequences. They are the only values that keep the three laws true for every integer exponent, which is the reason they were chosen. 0⁰ is left undefined on this course, because both arguments above need x ≠ 0.
A negative exponent does not make the value negative. 2⁻³ is 1/8, a positive number. The exponent's sign says which side of the fraction bar the power belongs on; the value's sign comes from the base.
That gives a useful move: a factor may cross the fraction bar if its exponent changes sign. x⁻²/y⁻³ is y³/x². It is not a new rule — it is the definition applied twice.
Example · An expression with every law in it
(2x⁻³y²)⁻². The outer exponent distributes over the product: 2⁻² = 1/4, (x⁻³)⁻² = x⁶ and (y²)⁻² = y⁻⁴. Collecting, that is x⁶ / (4y⁴), valid for x ≠ 0 and y ≠ 0.
Interactive
The exponent laws, twice each
By the law, and by writing the powers out
Method
Simplifying a power expression
Outermost first, then one base at a time, then the negatives. Changing that order is what produces stray factors.
1
Deal with the outermost exponent first
An exponent on a bracket distributes over every factor inside it, including the numerical one: in (2x³)⁴ the 2 is raised too, giving 16x¹². It distributes over a quotient as well, and over a sum never.
2
Collect one base at a time
Gather all the powers of x, then all the powers of y, then the plain numbers. Each base is tracked separately and they never mix, so this turns one hard problem into three easy ones.
3
Convert negative exponents last
Do the arithmetic on the exponents while they are still exponents — adding −6 and −1 is easier than manipulating two nested fractions. Move factors across the bar only once the exponents are final.
4
State the restrictions
Every quotient law and every negative exponent needs its base to be nonzero. Writing x ≠ 0 beside the answer is not pedantry; in Rational and Radical Expressions the excluded values become the first thing you are asked for.
Powers
Score 0 / 0Choose the answer you can justify.
One expression, every law used once
(2x³y⁻²)⁻² · (4x⁻¹y³), with x ≠ 0 and y ≠ 0.
(2x³y⁻²)⁻² · (4x⁻¹y³)
outer exponent, over each factor of the bracket
2⁻² = 1/4 (x³)⁻² = x⁻⁶ (y⁻²)⁻² = y⁴
→ (1/4)x⁻⁶y⁴ · 4x⁻¹y³
the numbers
(1/4) · 4 = 1
the x factors
x⁻⁶ · x⁻¹ = x⁻⁶⁺⁽⁻¹⁾ = x⁻⁷
the y factors
y⁴ · y³ = y⁴⁺³ = y⁷
→ x⁻⁷y⁷
negative exponent last
x⁻⁷ = 1/x⁷
answer y⁷ / x⁷
Four exponent arithmetics decided the whole thing: 3 · (−2) = −6 and (−2)(−2) = 4 inside the first bracket, then −6 + (−1) = −7 and 4 + 3 = 7. Every one of them is signed integer addition or multiplication, which is why the bracket-and-sign care of “Order of Operations” matters here.
Check at x = 2 and y = 3. The first bracket is 2 · 8 · (1/9) = 16/9, and (16/9)⁻² = 81/256. The second factor is 4 · (1/2) · 27 = 54. Their product is 4374/256, which reduces to 2187/128. And y⁷/x⁷ is 2187/128. Agreement at one pair of values is not a proof, but a disagreement would have been decisive.
For the faded pass, simplify (3x⁻²)²x⁵ with x ≠ 0. Keep the exponent arithmetic visible: the bracket gives 9x⁻⁴, then −4 + 5 = 1, so the answer is 9x. An answer of 9x⁹ added before applying the outer power; 3x forgot that the outer square also applies to the coefficient.
Common mistakes
1
Multiplying the exponents when the bases are multipliedx² · x³ = x⁵, not x⁶. Exponents multiply for a power of a power, (x²)³ = x⁶, and the two rules get swapped under pressure. Writing out (x·x)(x·x·x) settles it in five seconds.
2
Applying a law across two different bases2³ · 3² is 8 · 9 = 72. There is no law that combines it into a single power, because every law here begins "for a base x". Different bases stay apart.
3
Reading a negative exponent as a negative value2⁻³ is 1/8, and (−2)³ is −8. The exponent's sign controls which side of the fraction bar the power sits on; the base's sign controls whether the value is negative.
Check your understanding
✓
Finish when you can produce x⁰ = 1 from the quotient law instead of recalling it.
Derive x⁰ = 1 from a quotient, explain where x ≠ 0 enters, and simplify (2a⁻³b²)² · 3ab⁻¹ with restrictions stated. The test is whether each addition or multiplication of exponents can be tied to one law, not whether the final expression looks compact.
Note: “Scientific Notation” uses these laws on powers of ten and nothing else, which makes them concrete. “Rational Exponents” pushes the exponent out of the integers altogether: if x^(1/2) · x^(1/2) is to equal x¹, then x^(1/2) has no choice but to be a square root. Every extension of the exponent on this path is made by that same argument.