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Use the rules for exponents, including zero and negative exponents, by counting factors.
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In this lesson you learn the rules for exponents: add them when multiplying powers of the same base, subtract when dividing, and multiply for a power of a power. You see why anything nonzero to the power zero is one and why a negative exponent means a reciprocal. These are the rules that make roots and scientific notation work, so the lesson ties the unit together.
You know that an exponent is shorthand for repeated multiplication: $2^4$ means $2 \times 2 \times 2 \times 2$, which is $16$. The number being multiplied is the base and the small raised number counts how many times it appears. You can cancel a common factor from the top and bottom of a fraction, so $\dfrac{6 \times 7}{6}$ is just $7$. You also know that multiplication can be done in any order and grouped any way you like. This lesson uses only those three facts. Every rule for exponents is a shortcut for writing the factors out and counting them, so if you ever forget a rule, you can rebuild it in a few seconds with a small example.
| Term | What it means |
|---|---|
| Base | The number or letter being multiplied: the $x$ in $x^5$. |
| Exponent | The raised number that counts the factors: the $5$ in $x^5$. Also called the power or the index. |
| Power | A base with an exponent, such as $3^4$, or the value it names, $81$. |
| Reciprocal | One divided by a number. The reciprocal of $8$ is $\frac{1}{8}$. |
| Zero exponent | The exponent $0$. Any nonzero base to the power $0$ equals $1$. |
| Negative exponent | An exponent below zero, meaning a reciprocal: $b^{-n} = \frac{1}{b^n}$. |
An exponent counts how many times a base is used as a factor. Keep that one idea in mind and every rule follows from it.
Multiplying two powers of the same base puts their factor lists side by side. $x^3 \cdot x^4$ is three $x$'s followed by four more, seven in all, so the exponents add: $x^3 \cdot x^4 = x^7$.
Dividing two powers of the same base cancels factors in pairs. In $\dfrac{x^6}{x^2}$, the two $x$'s below cancel two of the six above, leaving four, so the exponents subtract: $x^6 \div x^2 = x^4$.
A power of a power repeats a whole group. $(x^3)^2$ is the group $x^3$ used twice, which is $2$ groups of $3$ factors, so the exponents multiply: $(x^3)^2 = x^6$.
Two more rules come from asking what division means when the numbers run out. A power divided by itself is $1$, and the subtraction rule gives exponent $0$, so $x^0 = 1$. Divide by more factors than there are on top and the subtraction gives a negative exponent. The leftover factors sit in the denominator, so a negative exponent means a reciprocal: $x^{-2} = \dfrac{1}{x^2}$.
All of these rules need the same base. $2^3 \cdot 5^2$ has no single-power shortcut, because the factors are different numbers.
Another way: table
A ladder of powers of $3$, from high to low: $3^3 = 27$, $3^2 = 9$, $3^1 = 3$, $3^0 = 1$, $3^{-1} = \frac{1}{3}$, $3^{-2} = \frac{1}{9}$. Each step down divides by $3$. The step below $3^1$ lands on $1$, and the steps below that give fractions.
Another way: written out
$a^2 \cdot a^3 = (a \cdot a)(a \cdot a \cdot a) = a^5$. Count the letters: two and three make five. Writing the factors out is slow but never wrong, so it is the way to check any rule.
Try it with numbers you can check. $2^3 \cdot 2^4$ means $(2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2)$, which is seven $2$'s multiplied, or $2^7 = 128$. Check: $2^3 = 8$, $2^4 = 16$, and $8 \times 16 = 128$. The rule gave the right answer without the multiplying.
In symbols, $b^m \cdot b^n = b^{m+n}$ for any base $b$. When numbers stand in front of the powers, as in $4x^2 \cdot 5x^3$, rearrange the product so the numbers are together and the powers are together: $(4 \cdot 5)(x^2 \cdot x^3) = 20x^5$. The numbers multiply as usual, and only the exponents of the matching letter add.
A letter with no exponent written has exponent $1$: $x = x^1$. So $x^5 \cdot x = x^6$, not $x^5$. Forgetting that hidden $1$ is one of the most common slips.
In a quotient of powers of one base, each factor below the line cancels one factor above it. $\dfrac{5^6}{5^4}$ leaves $6 - 4 = 2$ factors of $5$, so it is $5^2 = 25$. In symbols, $\dfrac{b^m}{b^n} = b^{m-n}$, with the exponent of the bottom taken away from the exponent of the top.
Now divide a power by itself: $\dfrac{5^4}{5^4}$. As ordinary numbers this is $625 \div 625 = 1$. The rule says $5^{4-4} = 5^0$. Both answers describe the same division, so they must be equal: $5^0 = 1$. The same argument works for every base except $0$, so $b^0 = 1$ whenever $b \neq 0$. ($0^0$ would mean $0 \div 0$, which has no value.)
This is not a special rule someone made up. It is the only value for $b^0$ that lets the division rule keep working, and that is the reason mathematicians agreed on it.
Look at the powers of $10$ as you step down: $10^3 = 1000$, $10^2 = 100$, $10^1 = 10$, $10^0 = 1$. Each step divides by $10$. Keep going and the pattern gives $10^{-1} = \dfrac{1}{10}$, $10^{-2} = \dfrac{1}{100}$ and $10^{-3} = \dfrac{1}{1000}$.
So $b^{-n} = \dfrac{1}{b^n}$: a negative exponent tells you how many factors of the base sit in the denominator. The division rule agrees: $\dfrac{7^2}{7^5}$ leaves three factors of $7$ below the line, which is $\dfrac{1}{7^3}$, and the rule gives $7^{2-5} = 7^{-3}$.
A negative exponent never makes a number negative. $2^{-3} = \frac{1}{8}$, which is positive and small. The minus sign moves the factors to the other side of the fraction bar. It works backward too: $\dfrac{1}{4^{-2}} = 4^2 = 16$, because a factor that was in the denominator moves up.
Every rule still works with negative exponents. $6^5 \cdot 6^{-2} = 6^{5 + (-2)} = 6^3$: five factors of $6$ on top and two below, so three survive.
$(b^m)^n$ means the group $b^m$ used $n$ times, so there are $n$ groups of $m$ factors: $(b^m)^n = b^{mn}$. For example, $(10^2)^3 = 10^2 \cdot 10^2 \cdot 10^2 = 10^6$, a million.
A power of a product gives the exponent to every factor inside. $(2x)^3 = 2x \cdot 2x \cdot 2x = 2^3 x^3 = 8x^3$. The $2$ is cubed as well as the $x$. Likewise $(5a^2)^2 = 25a^4$.
Be careful with sums: the exponent does not share out over addition. $(3 + 4)^2 = 7^2 = 49$, but $3^2 + 4^2 = 9 + 16 = 25$. The rule is about factors, and $3 + 4$ is not a product.
The rules need one base. $x^2 \cdot y^3$ cannot be written as a single power, and neither can $2^4 \cdot 3^2$. You can still work each out: $16 \times 9 = 144$.
The rules are about multiplying and dividing, not adding. $x^2 + x^3$ does not simplify: two $x$'s multiplied and three $x$'s multiplied are different kinds of term, like squares and cubes. And $x^3 + x^3 = 2x^3$, not $x^6$: that is two copies of $x^3$ added, which is collecting like terms.
One useful exception runs the other way. When the exponents match, the bases can be multiplied: $2^3 \cdot 5^3 = (2 \cdot 5)^3 = 10^3 = 1000$. That is the power of a product read backward, and it can save a lot of arithmetic.
Why the moves are allowed. Each rule is a count of factors, and multiplication can be reordered and regrouped freely, so counting factors cannot change the value.
How to check. Put a small number, such as $2$, in for the letter, and work out both the question and your answer. If $\dfrac{x^8}{x^2}$ became $x^6$, check with $x = 2$: $256 \div 4 = 64$ and $2^6 = 64$. They match. If you had written $x^4$, the check gives $16$, and the mistake shows at once. Avoid checking with $x = 1$, since every power of $1$ is $1$ and the check cannot catch anything.
The NCAA men's basketball tournament has $68$ teams. Four early games cut the field to a bracket of $64$, which is $2^6$. Every game knocks out one team, so each round halves the field. After $r$ rounds, $\dfrac{2^6}{2^r} = 2^{6-r}$ teams are left. After $4$ rounds that is $2^2 = 4$ teams, the Final Four. After all $6$ rounds it is $2^0 = 1$ team, the champion: the zero exponent gives exactly the right answer. Counting games is easy the same way: every team but one loses once, so the bracket of $64$ plays $63$ games.
Metric units are named by powers of ten. Kilo means $10^3$, milli means $10^{-3}$ and micro means $10^{-6}$. How many millimeters are in a kilometer? Divide: $\dfrac{10^3}{10^{-3}} = 10^{3-(-3)} = 10^6$, so a million. A human hair is roughly $100$ micrometers thick, which is $10^2 \times 10^{-6} = 10^{-4}$ meters, or one ten-thousandth of a meter. The exponent rules turn a unit conversion into adding or subtracting small whole numbers.
In ideal laboratory conditions, the bacterium E. coli can divide about every $20$ minutes, so a colony doubles three times an hour. In $3$ hours it doubles $9$ times, growing by a factor of $2^9 = 512$. How much more does it grow in the next $3$ hours? Another $2^9$, so over $6$ hours the growth is $2^9 \cdot 2^9 = 2^{18}$, more than $260{,}000$ times. That is why food left out on a warm counter becomes unsafe so quickly.
Multiplying the exponents when you should add them. $x^3 \cdot x^5 = x^8$, not $x^{15}$. Three factors and five more make eight.
Multiplying the bases. $3^2 \cdot 3^4 = 3^6$, not $9^6$. The base stays the same; only the count of factors changes.
Reading a negative exponent as a negative number. $4^{-2} = \frac{1}{16}$, not $-16$.
Thinking $b^0 = 0$. Any nonzero base to the power $0$ is $1$.
Forgetting the number in a power of a product. $(3x)^2 = 9x^2$, not $3x^2$.
Evaluate $5^0$, $5^{-1}$ and $5^{-2}$ from the pattern of powers of $5$. Start with a power you know.
$5^2 = 25$
Two factors of $5$ multiplied.
Step down one exponent by dividing by the base.
$5^1 = 25 \div 5 = 5$
Each lower exponent has one factor of $5$ fewer.
Step down again to exponent $0$.
$5^0 = 5 \div 5 = 1$
The last factor divides out, leaving $1$.
Step down to exponent $-1$.
$5^{-1} = 1 \div 5 = \dfrac{1}{5}$
The pattern keeps dividing by $5$, so the result becomes a fraction.
Step down to exponent $-2$.
$5^{-2} = \dfrac{1}{5} \div 5 = \dfrac{1}{25}$
Two factors of $5$ now sit below the line: $5^{-2}$ is the reciprocal of $5^2$.
Simplify $\dfrac{a^4 \cdot a^6}{a^3}$. Check that every power has the same base.
$a^4, \; a^6, \; a^3$
All three are powers of $a$, so the rules apply.
Multiply the powers on top by adding their exponents.
$a^4 \cdot a^6 = a^{4+6}$
Four factors of $a$ followed by six more.
Finish the addition.
$a^{10}$
The top now holds ten factors of $a$.
Divide by subtracting the exponent below the line.
$\dfrac{a^{10}}{a^3} = a^{10-3}$
Three factors on the bottom cancel three on the top.
Finish the subtraction.
$a^7$
Seven factors of $a$ are left.
Check with $a = 2$.
$\dfrac{16 \times 64}{8} = 128 = 2^7$
The original and the answer give the same number, so the simplifying is right.
Simplify $\dfrac{(3x^2)^3 \cdot x^{-4}}{9x}$. Give the outside exponent to each factor in the bracket.
$(3x^2)^3 = 3^3 \cdot (x^2)^3$
The bracket is a product, and the power reaches every factor of it.
Work out the number part.
$3^3 = 27$
Three factors of $3$.
Multiply the exponents of the power of a power.
$(x^2)^3 = x^6$
Three groups of two factors.
Multiply by $x^{-4}$ by adding exponents.
$27x^6 \cdot x^{-4} = 27x^{6 + (-4)} = 27x^2$
Four of the six factors cancel against the four that the negative exponent puts below the line.
Divide the number parts.
$27 \div 9 = 3$
The numbers divide as ordinary numbers.
Divide the powers, remembering $x = x^1$.
$x^2 \div x^1 = x^1 = x$
A letter with no written exponent has exponent $1$.
Write the result.
$3x$
One number and one power of $x$: nothing more combines.
Check with $x = 2$.
$\dfrac{12^3 \cdot \frac{1}{16}}{18} = \dfrac{108}{18} = 6 = 3 \times 2$
$3x^2 = 12$ when $x = 2$, and the original gives the same value as $3x$.
Subtract the bottom exponent from the top one.
$y^{2-7} = y^{-5}$
Seven factors below cancel two above, and five stay below the line.
Write the negative exponent as a reciprocal.
$y^{-5} = \dfrac{1}{y^5}$
A negative exponent counts factors in the denominator.
Check with $y = 2$.
Compare with the answer.
What is $8^0$? Think of $8^{2} \div 8^{2}$.
Complete the worked solution: simplify $\dfrac{(x^{4})^{3} \cdot x^{7}}{x^{9}}$.
Simplify the power of a power by multiplying its exponents.
$(x^{4})^{3}$ has exponent p
The bracket holds a group of factors, and the group is repeated.
Multiply by the next power of $x$ by adding exponents.
the top is $x$ to the power q
Multiplying puts the two lists of factors side by side.
Divide by subtracting the exponent below the line.
the answer is $x$ to the power r
Each factor below the line cancels one factor above it.
Simplify $3x^{5} \cdot 4x^{2}$.
Answer:
Simplify $\dfrac{8a^{11}}{2a^{3}}$.
Answer:
Simplify $(3y^{5})^{3}$.
Answer:
Write $2^{-3}$ as one divided by a whole number, then as a fraction.
$2^{-3} = 1 \div$ p $=$ v
A sheet of paper doubles in thickness with every fold. Sheet A is folded $10$ times and sheet B, cut from the same pack, is folded $4$ times. How many times thicker is A than B?
A is $2$ to the power e, which is t times thicker.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Simplify $5^{3} \cdot 5^{-5}$ to a single power of $5$, then give its value as a fraction.
The exponent is e and the value is v.
You can use the exponent rules with zero and negative exponents. Explain why $7^0 = 1$ using the division rule, and simplify $\dfrac{x^3}{x^8}$.
18. Your turn: simplify $\dfrac{y^2}{y^7}$, step 3
$\dfrac{4}{128} = \dfrac{1}{32}$
$2^2 = 4$ and $2^7 = 128$.
18. Your turn: simplify $\dfrac{y^2}{y^7}$, step 4
$\dfrac{1}{2^5} = \dfrac{1}{32}$
Both give the same value, so $\frac{y^2}{y^7} = \frac{1}{y^5}$.