Zero Exponent

Learn why any nonzero number to the power of zero equals 1, and how to handle expressions like $5x^0$ and $(-3)^0$.

Intermediate20 min18 problems
Practice Problems

Definition

The zero exponent rule says that any nonzero number raised to the power of 00 equals 11:
a0=1(a≠0)a^0 = 1 \quad (a \neq 0)
So 70=17^0 = 1, 1000=1100^0 = 1, (−5)0=1(-5)^0 = 1 and (23)0=1\left(\frac{2}{3}\right)^0 = 1.
Why? Look at the pattern. Every time the exponent goes down by 11, you divide by the base:
PowerValue
242^41616
232^388
222^244
212^122
202^02÷2=12 \div 2 = 1
The quotient rule agrees. Dividing a power by itself gives 11, and the quotient rule gives exponent 00:
anan=an−n=a0andanan=1\frac{a^n}{a^n} = a^{n-n} = a^0 \quad \text{and} \quad \frac{a^n}{a^n} = 1
Watch what the exponent belongs to. An exponent only applies to the base directly in front of it:
  • 5x0=5×x0=5×1=55x^0 = 5 \times x^0 = 5 \times 1 = 5, but (5x)0=1(5x)^0 = 1
  • −30=−(30)=−1-3^0 = -(3^0) = -1, but (−3)0=1(-3)^0 = 1
Special case: 000^0 is left undefined at this level. The rule a0=1a^0 = 1 is only for a≠0a \neq 0.

Try it now

What is 909^0?

Worked Examples

Use the pattern of powers of 33 to find 303^0.
1

Start with a power you know

33=3×3×33^3 = 3 \times 3 \times 3 → 33=273^3 = 27
2

Lower the exponent by 1: divide by 3

27÷3=927 \div 3 = 9 → 32=93^2 = 9
3

Lower it again: divide by 3

9÷3=39 \div 3 = 3 → 31=33^1 = 3
4

One more step down to exponent 0

3÷3=13 \div 3 = 1 → 30=13^0 = 1

Common Mistakes

Writing a0=0a^0 = 0, for example 90=09^0 = 0

Why it's wrong: The exponent 00 does not mean multiplying by zero. Going down the pattern means dividing by the base, and 9÷9=19 \div 9 = 1.

Correct: 90=19^0 = 1, because 9191=90\frac{9^1}{9^1} = 9^0 and 99=1\frac{9}{9} = 1

Writing a0=aa^0 = a, for example 90=99^0 = 9

Why it's wrong: That is the rule for exponent 11: 91=99^1 = 9. Exponent 00 is one more step down, so you divide by 99 once more.

Correct: 91=99^1 = 9 but 90=19^0 = 1

Writing 5x0=15x^0 = 1

Why it's wrong: The exponent belongs only to xx, not to the coefficient 55. Only x0x^0 becomes 11.

Correct: 5x0=5×1=55x^0 = 5 \times 1 = 5, while (5x)0=1(5x)^0 = 1

Writing −30=1-3^0 = 1

Why it's wrong: Without brackets, the base is 33, not −3-3. The minus sign stays in front of the result.

Correct: −30=−(30)=−1-3^0 = -(3^0) = -1, while (−3)0=1(-3)^0 = 1

Interactive Visual

Exponents as repeated factors

Quotient rule

Same base, divide: subtract the exponents.

5353\frac{5^{3}}{5^{3}}

Write each power as repeated factors

55·55·55
55·55·55

3 common factors cancel, top and bottom

All factors cancel: the result is 1, so the exponent is 0

1=50\displaystyle 1 = 5^{0}

Result

5353\displaystyle \frac{5^{3}}{5^{3}}==53−3\displaystyle 5^{3 - 3}==50\displaystyle 5^{0}==1\displaystyle 1

Value: 50=15^{0} = 1

This works for every base except 0

2³ = 8
3

See how powers of a number grow on the number line. Change the base and exponent.

Interactive Sandbox

Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
What is 909^0?

Why It Matters

The zero exponent turns up wherever powers are used:
  • Place value: the ones place is 100=110^0 = 1, so 345=3×102+4×101+5×100345 = 3 \times 10^2 + 4 \times 10^1 + 5 \times 10^0
  • Computers: the rightmost digit of a binary number is worth 20=12^0 = 1
  • Growth models: in P=500×2tP = 500 \times 2^t, the starting value at t=0t = 0 is 500×20=500500 \times 2^0 = 500
  • Algebra: simplifying x4x4\frac{x^4}{x^4} or 3x03x^0 needs the rule every time
If a0a^0 were 00, the ones place would be worth nothing and every growth model would start at zero. The value 11 is the only one that keeps all the exponent rules working.

Real World Applications

Place Value and Powers of Ten

Every place in our number system is a power of ten. The ones place is $10^0 = 1$.

Example:

6072=6×103+0×102+7×101+2×1006072 = 6 \times 10^3 + 0 \times 10^2 + 7 \times 10^1 + 2 \times 10^0, and the last term is 2×1=22 \times 1 = 2.

1Try It Yourself

A teacher writes 60726072 in expanded form: 6×103+0×102+7×101+2×1006 \times 10^3 + 0 \times 10^2 + 7 \times 10^1 + 2 \times 10^0.

What is the value of the last term, 2×1002 \times 10^0?

Step 1: Write the mathematical expression

Write the last term as a calculation:

Binary Numbers in Computers

Computers store numbers in binary. Each binary place is a power of two, and the rightmost place is $2^0 = 1$.

Example:

Binary 10111011 means 1×23+0×22+1×21+1×20=8+0+2+1=111 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 11.

2Try It Yourself

A sensor sends the binary number 101101, which means 1×22+0×21+1×201 \times 2^2 + 0 \times 2^1 + 1 \times 2^0.

What is this number in our usual decimal system?

Step 1: Write the mathematical expression

Write the sum of the place values:

Starting Value of a Growth Model

Biologists model a population that doubles every hour with $P = P_0 \times 2^t$, where $t$ is the time in hours.

Example:

For P=200×2tP = 200 \times 2^t, after 3 hours P=200×23=1600P = 200 \times 2^3 = 1600, and at the start P=200×20=200P = 200 \times 2^0 = 200.

3Try It Yourself

A culture of bacteria grows according to P=500×2tP = 500 \times 2^t, where tt is the number of hours since the start.

How many bacteria are there at the start, when t=0t = 0?

Step 1: Write the mathematical expression

Put t=0t = 0 into the formula:

Key Takeaways

  • 1For any nonzero number aa: a0=1a^0 = 1
  • 2Pattern: each step down in the exponent divides by the base, so 21=22^1 = 2 becomes 20=2÷2=12^0 = 2 \div 2 = 1
  • 3Quotient rule: anan=an−n=a0\frac{a^n}{a^n} = a^{n-n} = a^0, and a number divided by itself is 11
  • 4The exponent belongs only to the base right in front of it: 5x0=55x^0 = 5 but (5x)0=1(5x)^0 = 1
  • 5Brackets matter with negatives: −30=−1-3^0 = -1 but (−3)0=1(-3)^0 = 1
  • 6a0a^0 is never 00 and never aa; 000^0 is left undefined at this level

Frequently Asked Questions

Because each step down in the exponent divides by the base. From 51=55^1 = 5, one more step down gives 5÷5=15 \div 5 = 1. The quotient rule says the same thing: 5353=50\frac{5^3}{5^3} = 5^0, and any nonzero number divided by itself is 11.
Because each step down in the exponent divides by the base. From 51=55^1 = 5, one more step down gives 5÷5=15 \div 5 = 1. The quotient rule says the same thing: 5353=50\frac{5^3}{5^3} = 5^0, and any nonzero number divided by itself is 11.
At this level 000^0 is left undefined. The pattern argument needs us to divide by the base, and you cannot divide by 00. That is why the rule is written a0=1a^0 = 1 for a≠0a \neq 0.
Yes, as long as the base is not zero and the whole base is in brackets: (−8)0=1(-8)^0 = 1 and (34)0=1\left(\frac{3}{4}\right)^0 = 1.
The calculator follows the order of operations: the power comes before the minus sign, so it computes −(30)=−1-(3^0) = -1. Type (−3)0(-3)^0 with brackets to get 11.

Glossary

Zero exponent
An exponent equal to 00. For any nonzero base, a0=1a^0 = 1.
Base
The number that is raised to a power. In (−3)0(-3)^0 the base is −3-3, in −30-3^0 it is 33.
Quotient rule
When dividing powers with the same base, subtract the exponents: aman=am−n\frac{a^m}{a^n} = a^{m-n}.
Coefficient
The number in front of a variable. In 5x05x^0 the coefficient is 55, and it is not affected by the exponent.
Undefined
An expression that is given no value. At this level 000^0 is left undefined.

Formula Card

Zero exponent rule

a0=1(a≠0)a^0 = 1 \quad (a \neq 0)

Any nonzero number to the power of zero equals 1

Why it works

anan=an−n=a0=1\frac{a^n}{a^n} = a^{n-n} = a^0 = 1

The quotient rule and dividing a number by itself give the same result

Coefficient in front

c⋅x0=c(x≠0)c \cdot x^0 = c \quad (x \neq 0)

Only the variable becomes 1, the coefficient stays

Brackets and negatives

(−a)0=1,−a0=−1(-a)^0 = 1, \quad -a^0 = -1

With brackets the base is negative, without brackets the minus sign stays outside

Today's challengeOne problem, three tries

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