Multiplying and dividing powers with the same base
When two powers of the same number are multiplied or divided, what happens to the exponents, and why?
Where you are 3 in Laws of exponents
Why this exists
An exponent is shorthand for a long product: 2^{10} stands for ten 2’s multiplied together. The shorthand only earns its keep if you can calculate with it without unpacking it every time.
Suppose a culture of bacteria doubles every hour. After 10 hours it has grown by a factor of 2^{10}, and over the next 6 hours it grows by a further factor of 2^6. Its total growth is the product 2^{10} \cdot 2^6. You could write out all sixteen 2’s and count them, but once the exponents run into the hundreds, or a letter stands where a number was, writing the factors out stops being an option.
This lesson gives the two shortcuts. When powers of the same base are multiplied, the exponents add; when they are divided, the exponents subtract. Both come straight from what an exponent counts, and both need the bases to match. They are the first two of the laws of exponents, and the later laws are built from them.
Definition
For a number a and a positive whole number n, the power a^n is the product of n factors, each equal to a (this is exponent notation):
The number a is the base and n is the exponent. A single factor is a power too: a^1 = a. So a base written with no exponent, such as 7 or x, has exponent 1.
Two powers have the same base when their bases are the same number or the same letter. 5^3 and 5^8 have the same base, and so do x^4 and x^9. 2^5 and 3^5 do not, and neither do x^4 and y^4, even though their exponents match.
In plain words, the two rules are these.
- Product of powers. To multiply powers with the same base, keep the base and add the exponents.
- Quotient of powers. To divide a power by a lower power of the same nonzero base, keep the base and subtract the exponents.
Every exponent in this lesson is a positive whole number: 1, 2, 3 and so on. Exponents of zero and below are given a meaning of their own in zero and negative exponents.
Intuition
An exponent counts factors, so each rule comes down to one question: how many factors are there at the end?
Multiplying a^3 by a^4 puts a row of three a’s next to a row of four. Nothing is lost and nothing is added. The two rows simply join into one row of 3 + 4 = 7 factors, which is a^7.
Dividing undoes multiplying. In \displaystyle \frac{a^6}{a^2}, each a on the bottom pairs off with an a on top, and each pair is \displaystyle \frac{a}{a} = 1, which changes nothing in a product. Two pairs go, and 6 - 2 = 4 factors are left: a^4.
Neither picture works when the factors are different numbers. A row of 2’s followed by a row of 3’s is not a row of any one number, so there is no single base whose exponent could count it.
Results
Product of powers. For any number a and positive whole numbers m and n,
The same holds for three or more factors: a^m \cdot a^n \cdot a^p = a^{m+n+p}.
Quotient of powers. For any nonzero number a and positive whole numbers m and n with m > n,
Why it is true
The product. Write each power as its row of factors:
The brackets can come off because of the associative property of multiplication: the factors of a product can be regrouped without changing the result. No reordering is needed, since the m factors of the first power already sit in front of the n factors of the second. What remains is one unbroken row of m + n factors of a.
With three powers, use the rule twice: a^m \cdot a^n \cdot a^p = a^{m+n} \cdot a^p = a^{m+n+p}. Any number of powers works the same way.
The quotient. Take m > n, so that m - n is a positive whole number. By the product rule,
So a^{m-n} is the number that gives a^m when it is multiplied by a^n. That is exactly what a^m \div a^n means, so \displaystyle \frac{a^m}{a^n} = a^{m-n}. The division is allowed because a^n \neq 0, which holds because a \neq 0: a product of nonzero numbers is never zero.
The canceling in Figure 2 is the same argument done one factor at a time. Each pair \displaystyle \frac{a}{a} equals 1 only because a \neq 0, and n pairs cancel, which leaves m - n factors on top.
Worked examples
The lesson is taught in two parts. Examples 1 to 4 belong to the first: multiplying powers with the same base. Examples 5 to 7 belong to the second: dividing them, and deciding when the bases really are the same.
Multiplying powers of a number
Write each product as a single power, then find its value.
(a) 5^3 \cdot 5^4 \qquad (b) 7 \cdot 7^5 \qquad (c) 10^2 \cdot 10^3 \cdot 10
Solution(a) Both factors have base 5, so the product of powers applies and the exponents add: 3 + 4 = 7.
As a check, 5^3 = 125 and 5^4 = 625, and 125 \cdot 625 = 78{,}125. The base stays 5: the product is seven 5’s multiplied together, so it is 5^7 and not 25^7.
(b) The lone factor 7 is 7^1, so the exponents are 1 and 5:
(c) There are three factors with base 10, and the last is 10^1. The exponents add to 2 + 3 + 1 = 6:
In ordinary numbers this is 100 \cdot 1000 \cdot 10, and the zeros add up the same way: 2 + 3 + 1 = 6 of them. Counting zeros when multiplying powers of ten is the product rule in disguise.
Multiplying powers of a letter
Simplify.
(a) y^6 \cdot y^9 \qquad (b) m \cdot m^8 \qquad (c) p^2 \cdot p^5 \cdot p^3
Solution(a) Both factors have base y, so the exponents add: 6 + 9 = 15.
(b) The factor m is m^1, so the exponents are 1 and 8:
(c) Three factors with base p; the exponents add to 2 + 5 + 3 = 10:
These hold for every value of the letter, because the counting argument never used what the letter stands for. A spot check at one value is still reassuring. At p = 2, part (c) gives 4 \cdot 32 \cdot 8 = 1024 on the left, and 2^{10} = 1024 on the right.
Writing factors as powers of one base
(a) Write 8 \cdot 2^5 as a single power of 2.
(b) Write 9 \cdot 3^4 \cdot 27 as a single power of 3.
(c) Simplify b^{3n} \cdot b^{n+2}, where n is a positive whole number.
Solution(a) The rule needs both factors written as powers of the same base, and 8 is not written that way yet. But 8 = 2 \cdot 2 \cdot 2 = 2^3, so
As a check, 8 \cdot 32 = 256 and 2^8 = 256.
(b) Here 9 = 3^2 and 27 = 3^3, so all three factors have base 3 and the exponents add to 2 + 4 + 3 = 9:
As a check, 9 \cdot 81 \cdot 27 = 729 \cdot 27 = 19{,}683, and 3^9 = 19{,}683.
(c) The exponents are expressions this time, but each still counts factors of b: 3n of them in the first power and n + 2 in the second. The count of the whole row is their sum, 3n + (n + 2) = 4n + 2:
At n = 1 this says b^3 \cdot b^3 = b^6, which agrees with the rule for numbers.
Doubling bacteria
Brigid’s culture starts with 2^6 = 64 bacteria, and if nothing slows it, the number doubles every 20 minutes. Write the number of bacteria after 3 hours as a power of 2 and as an ordinary number.
SolutionThree hours is 180 minutes, which is 180 \div 20 = 9 periods of 20 minutes. Each period multiplies the number by 2, so the nine periods together multiply it by 2^9. The number after 3 hours is
The exponent keeps a running count of doublings. The starting 64 is what one cell would become after 6 doublings; 9 more make 15 in all.
Dividing powers
Simplify. The letters stand for nonzero numbers.
(a) 6^9 \div 6^7 \qquad (b) \dfrac{z^{12}}{z^5} \qquad (c) \dfrac{w^3 \cdot w^{10}}{w^8}
Solution(a) The base is 6 in both, and 9 > 7, so the quotient of powers applies and the exponents subtract: 9 - 7 = 2.
Written out, seven of the nine 6’s on top cancel with the seven below, and two are left.
(b) The base is z in both and 12 > 5:
(c) Simplify the top first with the product rule, then divide:
Counted in one go: there are 3 + 10 = 13 factors of w on top and 8 below, so 13 - 8 = 5 are left.
Same base or not
Decide which of these can be written as a single power using the two rules, and simplify the ones that can.
(a) 2^5 \cdot 3^5 \qquad (b) \dfrac{4^7}{4^3} \qquad (c) \dfrac{x^8}{y^2} \qquad (d) \dfrac{64}{2^4}
Solution(a) The bases are 2 and 3, so neither rule applies. Written out, the product is five 2’s followed by five 3’s, and no single number repeats along the whole row. Adding the exponents would give an answer with exponent 10, and every choice of base for it is wrong: the true value is 32 \cdot 243 = 7776, while 2^{10} = 1024 and 3^{10} = 59{,}049. Pairing each 2 with a 3 does give 6^5 = 7776, but that is a different rule, the power of a product, which belongs to powers of powers, products and quotients.
(b) Same base, 4, with 7 > 3:
(c) The bases are different letters, so no factor on the bottom matches one on top and nothing cancels. The quotient \displaystyle \frac{x^8}{y^2} is already as simple as it gets.
(d) The top is not written as a power, but 64 = 2^6, so the bases can be made to match:
As a check, 64 \div 16 = 4.
So (b) and (d) can be written as single powers, 4^4 and 2^2, and (a) and (c) cannot.
Grains on a chessboard
Put one grain of rice on the first square of a chessboard, two on the second, four on the third, doubling on each square. Then square number k holds 2^{k-1} grains.
(a) How many times as many grains does the last square, square 64, hold as square 40?
(b) Which square holds 2^{10} times as many grains as square 25?
Solution(a) Square 64 holds 2^{63} grains and square 40 holds 2^{39}. “How many times as many” is a division, and the bases match with 63 > 39:
The last square holds nearly seventeen million times as many grains as square 40, and nobody had to work out either number.
(b) Square 25 holds 2^{24} grains. Multiplying by 2^{10} gives
A square holding 2^{34} grains has k - 1 = 34, so it is square 35. That fits the picture of the board: each square doubles the one before, so ten squares later means ten doublings.
Boundaries
The bases must match. Both rules rest on one row of a single repeated factor. With different bases there is no such row: 2^4 \cdot 5^2 = 16 \cdot 25 = 400, and no power of 2, of 5 or of 10 with exponent 6 equals 400. Different letters behave the same way, so x^3 y^5 is already fully simplified.
Products and quotients only, never sums. A sum of powers is not a row of factors, so nothing adds in the exponent: 3^4 + 3^2 = 81 + 9 = 90, while 3^6 = 729. Differences are the same: 2^5 - 2^3 = 32 - 8 = 24, which is not 2^2.
Dividing needs a nonzero base. The product rule holds for every base, 0 included, since 0^m \cdot 0^n = 0 = 0^{m+n}. The quotient rule does not: \displaystyle \frac{0^5}{0^2} asks for a division by 0, which has no answer.
The quotient rule needs m > n. When the bottom has more factors, canceling leaves the extra ones on the bottom:
and when m = n every factor cancels and the quotient is 1. In both cases m - n is zero or negative, and a^0 or a^{-2} means nothing yet under the definition here, which counts factors. Zero and negative exponents gives those symbols the meanings that keep \displaystyle \frac{a^m}{a^n} = a^{m-n} true in every case.
Whole-number exponents. The arguments here count factors, so they need the exponents to be counts. Exponents such as \displaystyle \frac{1}{2} are defined later, in rational exponents, chosen so that these same rules stay true.
Where this goes next
A power of a power is a product of equal powers: (a^3)^4 is a^3 four times over, so by the product rule its exponent is 3 + 3 + 3 + 3 = 12. Repeated adding is multiplying, and that is the next law, in powers of powers, products and quotients. Asking the quotient rule to keep working when m = n forces a^0 = 1, which is how zero and negative exponents begins.
In algebra, single terms carry a number in front and often several letters. Multiplying and dividing those combines the numbers as numbers and uses the rules of this lesson on each letter separately; that is monomials and the exponent rules.
The rules with base 10 are how scientific notation works: in calculating in scientific notation, multiplying two numbers adds their powers of 10 and dividing subtracts them.
Much later, the product rule becomes a fact about exponential functions. For f(x) = 2^x it says f(x + y) = f(x) \cdot f(y): adding inputs multiplies outputs, which is exactly what makes doubling, halving and compound growth work.
Exercises
Exercises 1 to 12 go with the first part of the lesson (Examples 1 to 4), Exercises 13 to 22 with the second (Examples 5 to 7), and the reasoning group draws on both. Every letter stands for a nonzero number.
Part 1: powers of numbers
Write each product as a single power, then find its value.
- 4^3 \cdot 4^6
- 9 \cdot 9^3
- 10^5 \cdot 10^2 \cdot 10^4
Part 1: powers of letters
Simplify.
- t^8 \cdot t^3
- n^7 \cdot n
- k^2 \cdot k^9 \cdot k
- c^{2n} \cdot c^{3n}, where n is a positive whole number
Part 1: one base, and growth
- Write 25 \cdot 5^6 as a single power of 5.
- Write 16 \cdot 2^3 \cdot 4 as a single power of 2, then find its value.
- Find the whole number k for which 7^k \cdot 7^4 = 7^{11}.
- Each fold in half doubles the number of layers in a sheet of paper. Dilnoza folds a sheet in half 4 times, so it has 2^4 layers, and then folds it in half 3 more times. How many layers are there now? Give the answer as a power of 2 and as a number.
- Emeka’s sample holds 3^4 yeast cells, and the number triples every hour. Write the number of cells after 6 more hours as a power of 3 and as a number.
Part 2: dividing powers
Simplify. Where the base is a number, find the value too.
- 8^7 \div 8^5
- \dfrac{10^9}{10^3}
- \dfrac{r^{15}}{r^6}
- h^{10} \div h
- \dfrac{2^6 \cdot 2^5}{2^8}
- \dfrac{q^7 \cdot q^2}{q^3 \cdot q^4}
Part 2: same base, and memory
- Which of these can be written as a single power using the rules of this lesson? Simplify the ones that can. (a) 7^4 \cdot 7^5 \quad (b) 3^4 \cdot 5^4 \quad (c) \dfrac{m^6}{n^2} \quad (d) 11^8 \div 11^3
- Cormac’s memory card holds 2^{35} bytes, and each song file takes up 2^{22} bytes. How many song files fit on the card? Give the answer as a power of 2 and as a number.
- On the chessboard of Example 7, square k holds 2^{k-1} grains. How many times as many grains does square 33 hold as square 25?
- Find the whole number k for which 5^{12} \div 5^k = 5^7.
Reasoning
- Adebayo writes 2^3 \cdot 2^4 = 4^7. Find the value of each side, and explain what the correct single power is and why.
- Explain why 5^3 + 5^2 cannot be written as 5^5, and find the value of each.
- Freyja wants to use the quotient of powers on \dfrac{2^3}{2^5}. Write out the factors and cancel to find the value. Then explain why the rule as stated in this lesson does not apply.
Answers
- 4^9 = 262{,}144. The base is 4 in both factors, so the exponents add: 3 + 6 = 9.
- 9^4 = 6561. The lone 9 is 9^1, and 1 + 3 = 4.
- 10^{11} = 100{,}000{,}000{,}000. The exponents add: 5 + 2 + 4 = 11.
- t^{11}, since 8 + 3 = 11.
- n^8. The lone n is n^1, and 7 + 1 = 8.
- k^{12}, since 2 + 9 + 1 = 12.
- c^{5n}. The exponents still count factors of c, and 2n + 3n = 5n.
- 5^8. Since 25 = 5^2, the product is 5^2 \cdot 5^6 = 5^{2+6} = 5^8.
- 2^9 = 512. Since 16 = 2^4 and 4 = 2^2, the product is 2^4 \cdot 2^3 \cdot 2^2, and 4 + 3 + 2 = 9. Check: 16 \cdot 8 \cdot 4 = 512.
- k = 7. The exponents add, so k + 4 = 11.
- 2^7 = 128 layers. The 3 further folds multiply the layers by 2^3, and 2^4 \cdot 2^3 = 2^{4+3} = 2^7.
- 3^{10} = 59{,}049 cells. Six triplings multiply the number by 3^6, and 3^4 \cdot 3^6 = 3^{10}.
- 8^2 = 64. The base is 8 in both and 7 > 5, so 7 - 5 = 2.
- 10^6 = 1{,}000{,}000, since 9 - 3 = 6.
- r^9, since 15 - 6 = 9.
- h^9. The bottom is h^1, and 10 - 1 = 9.
- 2^3 = 8. The top is 2^{6+5} = 2^{11}, and 11 - 8 = 3.
- q^2. The top is q^{7+2} = q^9, the bottom is q^{3+4} = q^7, and 9 - 7 = 2.
- (a) and (d) can: (a) 7^4 \cdot 7^5 = 7^9 and (d) 11^8 \div 11^3 = 11^5. (b) cannot, since the bases 3 and 5 differ; its value is 81 \cdot 625 = 50{,}625. (c) cannot either, since m and n are different letters and nothing cancels.
- 2^{13} = 8192 song files. “How many fit” is a division: \displaystyle \frac{2^{35}}{2^{22}} = 2^{35-22} = 2^{13}.
- 2^8 = 256 times as many. Square 33 holds 2^{32} grains and square 25 holds 2^{24}, and \displaystyle \frac{2^{32}}{2^{24}} = 2^{32-24} = 2^8.
- k = 5. The exponents subtract, so 12 - k = 7.
- The left side is 8 \cdot 16 = 128 and 4^7 = 16{,}384, so the two sides are not equal. The product is three 2’s followed by four 2’s, a row of seven 2’s, so it is 2^3 \cdot 2^4 = 2^7 = 128. The exponents add, but the base stays 2; nothing in the row is a 4.
- 5^3 + 5^2 = 125 + 25 = 150, while 5^5 = 3125. The product rule adds exponents because a product of powers is one long row of factors. A sum is not a product, so there is no row of factors to count, and the rule says nothing about it.
- \displaystyle \frac{2^3}{2^5} = \frac{2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{1}{2 \cdot 2} = \frac{1}{4}. Three pairs cancel, leaving two 2’s on the bottom, so the quotient is \displaystyle \frac{1}{2^2}. The rule here needs m > n, but 3 < 5: subtracting would give exponent 3 - 5 = -2, which does not count any factors, so 2^{-2} has no meaning yet. Zero and negative exponents are defined so that \displaystyle 2^{-2} = \frac{1}{2^2}, which matches what canceling found.