Why exponents add when powers multiply, the seven properties of exponents, evaluating numerical expressions, scientific notation in a large counting problem, simplifying algebraic expressions to positive exponents, and comparing volumes by scaling.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Use Properties of Exponents
Objectives
Five outcomes. The first explains all seven rules; the rest apply them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-333 — the lesson these objectives are drawn from
Warm-up
Chapter 4 is finished, and Chapter 5 raises the degree. Before polynomials, the arithmetic of powers.
Discussion prompt
Write out 2 to the third and 2 to the fifth as repeated multiplications, then multiply them together. What single power is the result, and where did its exponent come from?
Hint: Count how many factors of 2 there are altogether.
Answer:
\[ 2^3 \cdot 2^5 = (2\cdot2\cdot2)(2\cdot2\cdot2\cdot2\cdot2) = 2^8 \]
Three factors and five factors make eight factors, so the exponents add. Every property in this lesson is that kind of bookkeeping, which means none of them has to be memorised blind — each can be rebuilt by writing the factors out.
Concept
The properties of exponents look like a long list, but they all follow from counting factors. Multiplying powers of the same base adds exponents because the factors are pooled; dividing subtracts them because factors cancel in pairs; and everything else follows from those two together with the definition of a negative exponent.
scientific notation — A number written as c times 10 to the n, where c is at least 1 and less than 10 and n is an integer. It separates a number's significant digits from its size.
\[ a^m \cdot a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n} \]
A negative exponent is defined so that the subtraction rule keeps working past zero, and a zero exponent is defined so that a power divided by itself gives 1. Neither is arbitrary; both are forced.
Figure (svg): Two powers of two written out as repeated factors and combined into a single power
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330
Section
Section 1
Concept
Powers with the same base combine by adding exponents when multiplied and subtracting when divided. A power raised to a power multiplies the exponents, and a power of a product or a quotient distributes across the factors.
\[ a^m a^n = a^{m+n}, \; (a^m)^n = a^{mn}, \; (ab)^m = a^m b^m, \; a^{-m} = \tfrac{1}{a^m}, \; a^0 = 1 \]
Note what is not on the list: nothing distributes a power over a sum. The quantity a plus b, squared, is not a squared plus b squared, as Lesson 4.3's perfect square trinomial showed.
Figure (svg): The seven properties of exponents, each with a definition and a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330 — Properties of Exponents
Picture it
Each property with its definition and an instance.
Figure (svg): The seven properties of exponents, each with a definition and a worked instance
The first two rows are about combining, the middle three about rewriting, and the last two about dividing. Reading them as three groups rather than seven facts makes them far easier to hold.
Worked example
Both by counting factors, as in the warm-up.
\[ \text{Explain why } a^m a^n = a^{m+n} \text{ and } \tfrac{a^m}{a^n} = a^{m-n}. \]
Write the first as repeated factors
Why: A to the m is m copies of a, and a to the n is n copies.
Count them together
Why: Multiplying pools the factors, giving m plus n copies in all.
\[ a ^{m + n} \]
Write the second as a fraction of factors
Why: M copies above and n copies below.
Cancel in pairs
Why: Each factor below cancels one above, leaving m minus n copies.
\[ a ^{m - n} \]
Figure (svg): The solution to Worked example derive two of the rules shown as a ladder of expressions, one row per algebraic move
\[ a^m a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n} \]
Verify: test the subtraction rule past zero
Why: If m is 3 and n is 5, the counting picture gives three factors above and five below, leaving two below — that is 1 over a squared. The rule gives a to the negative 2. So the negative exponent must mean 1 over the positive power, which is exactly how the property is defined. The definition was chosen to keep the rule true.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330
Matching
Name the single property each step uses.
Match the pairs
Why: The first and last both involve negative exponents in the working but are named for the operation being performed. In the last, subtracting negative 6 from negative 3 gives positive 3, which is worth doing slowly: negative 3 minus negative 6 is negative 3 plus 6.
Worked example
The same argument run at m equal to n.
\[ \text{Explain why } a^0 = 1 \text{ for every non-zero } a. \]
Divide a power by itself
Why: Any non-zero number divided by itself is 1.
\[ a ^{5} / a ^{5} = 1 \]
Apply the quotient rule to the same expression
Why: Subtracting the exponents gives 5 minus 5.
\[ a ^{5} / a ^{5} = a ^{0} \]
Compare the two results
Why: Both describe the same quantity, so they must be equal.
\[ a ^{0} = 1 \]
Note the exclusion
Why: The argument divides by a to the fifth, which requires a to be non-zero.
\[ a\text{ is not } 0 \]
Figure (svg): The solution to Worked example why a to the zero is 1 shown as a ladder of expressions, one row per algebraic move
\[ a^0 = 1, \quad a \neq 0 \]
Verify: check it against the product rule too
Why: A to the zero times a to the third should be a to the zero plus 3, which is a to the third — so a to the zero must act as a multiplicative identity, and the only such number is 1. Two independent rules force the same definition, which is a good sign that the choice is not arbitrary.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330
Trap
\[ 3^2 \cdot 3^5 \]
Multiply the bases and add the exponents
Why: Both parts of the expression are combined.
\[ 9^7 \quad \text{(wrong)} \]
Three squared times 3 to the fifth is 9 times 243, or 2187, which is 3 to the seventh. Nine to the seventh is more than four million.
\[ 3^2 \cdot 3^5 = 3^{2+5} = 3^7 = 2187 \]
Keep the base and add only the exponents
Why: The factors being pooled are all copies of 3, so the base does not change.
\[ 3^2 \cdot 3^5 \neq 9^7 \]
The book flags this in an Avoid Errors note. Writing out the factors settles it instantly: there are seven 3s, not seven 9s.
Sorting
What happens to the exponents in each case?
Sort into buckets
Sort each expression by what the properties do to its exponents.
The sum is the important one to notice: a to the third plus a to the fifth cannot be combined into a single power at all, and there is no rule that would let it.
Fill the middle
Guided Practice 5.
Fill in the blanks
x^2x^___x^___ = x^___ = x^___}
Why: Negative 6 plus 5 plus 3 is 2, so the answer is x squared. The product rule extends to any number of factors, because pooling three groups of factors is no different from pooling two — the rule is stated for two only because that is enough to build the rest by repetition.
Prediction
Commit before reasoning.
Predict first
Is the quantity a plus b, squared, equal to a squared plus b squared?
Correct: No — squaring a sum produces a middle term.
\[ (a+b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2 \]
Why: The quantity a plus b, squared, expands to a squared plus 2ab plus b squared, which is exactly the perfect square trinomial of Lesson 4.3. A quick numerical test settles it: 3 plus 4, squared, is 49, while 9 plus 16 is 25. Powers distribute over products and quotients only, and this is the single most common error involving exponents.
Section
Section 2
Concept
For a numerical expression involving powers, applying the properties first usually leaves a much smaller calculation than evaluating each power separately. Naming the property at each step keeps a long chain checkable.
\[ (2^4 \cdot 2^5)^2 = (2^4)^2 (2^5)^2 = 2^8 \cdot 2^{10} = 2^{18} \]
Which property to use first is rarely forced. Different orders reach the same answer, and choosing the one that shrinks the numbers fastest is a matter of practice rather than rule.
Figure (svg): The seven properties of exponents, each with a definition and a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330 — Evaluate numerical expressions
Picture it
Each row is a move available at any step.
Figure (svg): The seven properties of exponents, each with a definition and a worked instance
In a numerical problem the same seven rules apply, but the answer is a number rather than an expression, so the last step is always an evaluation.
Worked example
Example 1. Each line names the property it uses.
\[ \text{Evaluate } (2^4 \cdot 2^5)^2 \text{ and } \left(\tfrac{11^5}{11^8}\right)^{-1}. \]
First: power of a product
Why: The exponent 2 distributes across the two factors inside.
\[ (2 ^{4}) ^{2}(2 ^{5}) ^{2} \]
First: power of a power
Why: Four times 2 is 8 and 5 times 2 is 10, and the product of powers then gives 18.
\[ 2 ^{18} \]
Second: negative exponent
Why: An exponent of negative 1 flips the fraction.
\[ 11 ^{8} / 11 ^{5} \]
Second: quotient of powers
Why: Eight minus 5 is 3, and 11 cubed is 1331.
\[ 11 ^{3} = 1331 \]
Figure (svg): The solution to Worked example evaluate two expressions shown as a ladder of expressions, one row per algebraic move
\[ 2^{18} \qquad \text{and} \qquad 11^3 = 1331 \]
Verify: check the second a different way
Why: Applying the quotient rule inside first gives 11 to the 5 minus 8, or 11 to the negative 3; raising that to the negative 1 multiplies the exponents to give 11 to the third. Same answer by a different route, which is what the properties being consistent means in practice.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-330
Fill the middle
Guided Practice 1.
Fill in the blanks
(4^2)^3 = 4^6 = 4^___} = 4096
Why: Two times 3 is 6, because three copies of a group of two factors is six factors. This is the one property where the exponents multiply rather than add, and confusing it with the product rule would give 4 to the fifth, or 1024, instead.
Worked example
Guided Practice 1 to 4.
\[ \text{Evaluate } (4^2)^3, \; (-8)(-8)^3, \; \left(\tfrac{2}{9}\right)^3, \; \frac{6 \times 10^{-4}}{9 \times 10^7}. \]
First: power of a power
Why: Two times 3 is 6, and 4 to the sixth is 4096.
\[ 4 ^{6} = 4096 \]
Second: product of powers
Why: The first factor is negative 8 to the first, so the exponents 1 and 3 add to 4.
\[ (-8) ^{4} = 4096 \]
Third: power of a quotient
Why: The exponent distributes to numerator and denominator: 8 over 729.
\[ \frac{8}{729} \]
Fourth: separate the parts
Why: Six over 9 is two thirds, and 10 to the negative 4 over 10 to the seventh is 10 to the negative 11.
\[ (\frac{2}{3}) x 10 ^{-11} \]
Figure (svg): The solution to Worked example four more evaluations shown as a ladder of expressions, one row per algebraic move
\[ 4096, \; 4096, \; \tfrac{8}{729}, \; \tfrac{2}{3}\times 10^{-11} \]
Verify: notice the coincidence in the first two
Why: Four to the sixth and negative 8 to the fourth are both 4096, because 4 is 2 squared and 8 is 2 cubed, so both expressions are 2 to the twelfth in disguise — and the fourth power of a negative number is positive. Rewriting everything with base 2 is a useful trick when two powers look unrelated.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-331
Error analysis
A student evaluates a power of a negative number.
Annotate
On: \( (-8)(-8)^3 = -8^4 = -4096 \)
Brackets around a negative base are part of the expression, not decoration. An even exponent on a negative base always gives a positive result, and an odd one a negative.
Sorting
Look at the base's sign and whether the exponent is even.
Sort into buckets
Sort each power.
The third item is the trap: without brackets the exponent binds tighter than the minus sign, so the power is computed first and then negated.
Ranking
Evaluating a numerical expression with powers.
Put in order
Why: The evaluation goes last on purpose: combining exponents first keeps the numbers small, whereas evaluating 2 to the fourth and 2 to the fifth separately and then squaring the product means multiplying 512 by itself. The properties are labour-saving devices as much as they are rules.
Prediction
Commit before reasoning.
Predict first
For the quantity 2 to the fourth times 2 to the fifth, all squared, which route involves the smallest numbers?
Correct: Combine the exponents first, then square, then evaluate once.
\[ (2^4 \cdot 2^5)^2 = (2^9)^2 = 2^{18} \]
Why: Adding 4 and 5 gives 2 to the ninth; squaring gives 2 to the eighteenth; one evaluation follows. The first route computes 16 times 32, or 512, and then squares it — three multiplications of larger numbers. All routes give 262,144, but the properties exist to postpone arithmetic until it is unavoidable, which is exactly the habit this lesson is building.
Section
Section 3
Concept
A number in scientific notation is written as a number between 1 and 10 times a power of ten. Multiplying or dividing two such numbers means handling the two parts separately, using the product or quotient of powers on the tens.
\[ c \times 10^n, \quad 1 \le c < 10 \]
If the leading part comes out at 10 or more, it must be rewritten: 10.2 times 10 to the tenth becomes 1.02 times 10 to the eleventh, which is one more application of the product rule.
Figure (svg): A scientific-notation multiplication carried out in stages, from two measured quantities to one answer
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-331 — Use scientific notation in real life
Picture it
Example 2: 85 million per square kilometre over 1200 square kilometres.
Figure (svg): A scientific-notation multiplication carried out in stages, from two measured quantities to one answer
The arithmetic reduces to 8.5 times 1.2 and 7 plus 3, both of which are easy. Without scientific notation the same calculation is an eleven-digit multiplication.
Worked example
Example 2. A verbal model, then the properties.
\[ \text{A swarm holds } 85 \text{ million locusts per km}^2 \text{ over } 1200 \text{ km}^2. \text{ How many locusts?} \]
Write the verbal model
Why: The number of locusts is the density times the area.
\[ 85, 000, 000 x 1200 \]
Put both numbers in scientific notation
Why: Eighty-five million is 8.5 times 10 to the seventh; 1200 is 1.2 times 10 cubed.
\[ (8.5 x 10 ^{7}) (1.2 x 10 ^{3}) \]
Regroup and apply the product rule
Why: Eight point five times 1.2 is 10.2, and 7 plus 3 is 10.
\[ 10.2 x 10 ^{10} \]
Rewrite the leading part in range
Why: Ten point two is not between 1 and 10, so it becomes 1.02 times 10, absorbed into the power.
\[ 1.02 x 10 ^{11} \]
Figure (svg): The solution to Worked example count the locusts shown as a ladder of expressions, one row per algebraic move
\[ 1.02 \times 10^{11} \approx 102{,}000{,}000{,}000 \]
Verify: estimate the answer roughly
Why: Eighty-five million is close to 10 to the eighth, and 1200 is a bit more than 10 cubed, so the product should be a bit above 10 to the eleventh. The answer, 1.02 times 10 to the eleventh, is exactly that. A rough estimate of the exponent is the fastest way to catch a slipped power of ten.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-331
Matching
Move the decimal point and count.
Match the pairs
Why: Numbers greater than 10 have positive exponents and numbers less than 1 have negative ones, with the exponent counting how many places the decimal point moved. The third is the only one below 1, and its exponent is the only negative one.
Worked example
Guided Practice 4. Division uses the quotient rule instead.
\[ \text{Evaluate } \frac{6 \times 10^{-4}}{9 \times 10^{7}}. \]
Separate the two parts
Why: The digits divide and the powers of ten divide.
\[ (\frac{6}{9}) (10 ^{-4} / 10 ^{7}) \]
Simplify the digits
Why: Six over 9 reduces to two thirds, about 0.667.
\[ \text{about } 0.667 \]
Apply the quotient rule to the tens
Why: Negative 4 minus 7 is negative 11.
\[ 10 ^{-11} \]
Put the leading part back in range
Why: Point six six seven is less than 1, so it becomes 6.67 times 10 to the negative 1.
\[ 6.67 x 10 ^{-12} \]
Figure (svg): The solution to Worked example dividing in scientific notation shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{2}{3}\times 10^{-11} \approx 6.67 \times 10^{-12} \]
Verify: check the direction of the exponent
Why: Dividing a very small number by a very large one should give something far smaller still, and negative 12 is indeed smaller than negative 4. Note that pushing the leading digit up from 0.667 to 6.67 lowered the exponent by one — moving the decimal point right always costs one from the exponent.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-331
Error analysis
A student finishes a scientific-notation multiplication.
Annotate
On: \( (8.5 \times 10^7)(1.2 \times 10^3) = 10.2 \times 10^{10} \)
The last step of any scientific-notation calculation is checking that the leading part is in range. It is easy to skip because the value is already right.
Fill the middle
Example 2, at the product rule.
Fill in the blanks
10^7 \times 10^3 = 10^10 = 10^___}
Why: Seven plus 3 is 10, so the powers of ten give 10 to the tenth. The whole convenience of scientific notation is that the size of a number is carried in a single exponent, so multiplying sizes becomes adding small whole numbers.
Prediction
Commit before reasoning.
Predict first
A calculator handles 85,000,000 times 1200 without complaint. What does scientific notation add?
Correct: It keeps significant digits and magnitude separate, and makes the arithmetic small.
\[ 1.02 \times 10^{11}: \; \text{three significant digits, magnitude } 10^{11} \]
Why: Writing 1.02 times 10 to the eleventh says both how precise the figure is — three significant digits — and how big it is, in one glance. It also reduces every multiplication to a small product plus an addition of exponents, which matters when the numbers run to twenty digits, as they routinely do in astronomy and chemistry. Precision is not gained; it is made visible.
Comparison
Fill the blanks. Same split, two rules.
Comparison matrix
| Step | Multiplying | Dividing |
|---|---|---|
| The leading numbers | multiply them | divide them |
| The powers of ten | add the exponents | subtract the exponents |
| Property used | product of powers | quotient of powers |
| Final check | leading part between 1 and 10 | leading part between 1 and 10 |
The last row is identical for both, and it is the step most often forgotten in each.
Section
Section 4
Concept
The same properties simplify expressions containing variables. An expression counts as simplified when it contains only positive exponents, so every negative exponent must be moved across the fraction bar and every zero exponent replaced by 1.
\[ \left(\frac{r^{-2}}{s^3}\right)^{-3} = \frac{r^6}{s^{-9}} = r^6 s^9 \]
A negative exponent in a numerator moves to the denominator and vice versa. That is not a separate rule; it is the negative exponent property read in whichever direction is useful.
Figure (svg): A single algebraic expression simplified one property at a time down to positive exponents
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-332 — Simplify expressions
Picture it
Example 4: the quantity x to the negative 2, y cubed, all cubed, over x to the fourth y to the ninth.
Figure (svg): A single algebraic expression simplified one property at a time down to positive exponents
Six lines, six named properties, and each line is checkable on its own. Doing two properties in one line is where errors hide.
Worked example
Example 3, all three parts.
\[ \text{Simplify } b^{-4}b^6b^7, \; \left(\frac{r^{-2}}{s^3}\right)^{-3}, \; \frac{16m^4n^{-5}}{2n^{-5}}. \]
First: product of powers
Why: Negative 4 plus 6 plus 7 is 9.
\[ b ^{9} \]
Second: power of a quotient, then power of a power
Why: The outer negative 3 distributes, giving r to the 6 over s to the negative 9.
\[ r ^{6} / s ^{-9} \]
Second: negative exponent
Why: An s to the negative 9 in the denominator becomes s to the ninth in the numerator.
\[ r ^{6} s ^{9} \]
Third: quotient of powers
Why: Sixteen over 2 is 8, and negative 5 minus negative 5 is 0, so the n term is n to the zero.
\[ 8 m ^{4} n ^{0} = 8 m ^{4} \]
Figure (svg): The solution to Worked example three simplifications shown as a ladder of expressions, one row per algebraic move
\[ b^9, \quad r^6 s^9, \quad 8m^4 \]
Verify: substitute numbers into the third
Why: With m equal to 1 and n equal to 2, the original is 16 times 1 times 2 to the negative 5, over 2 times 2 to the negative 5, which is 16 over 2, or 8. The simplified form gives 8 times 1, also 8. The n terms cancelled entirely, which is what a zero exponent records.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 331-331
Sorting
Look for zero or negative exponents anywhere.
Sort into buckets
Sort each expression.
Both failures are equal in value to their neighbours. Simplified is a statement about form, not about value, exactly as it was for radicals in Lesson 4.5.
Worked example
Guided Practice 5 to 8.
\[ \text{Simplify } x^{-6}x^5x^3, \; (7y^2z^5)(y^{-4}z^{-1}), \; \left(\frac{s^3}{t^{-4}}\right)^2, \; \left(\frac{x^4y^{-2}}{x^3y^6}\right)^3. \]
First: add the exponents
Why: Negative 6 plus 5 plus 3 is 2.
\[ x ^{2} \]
Second: combine each base separately
Why: For y, 2 plus negative 4 is negative 2; for z, 5 plus negative 1 is 4.
\[ 7 y ^{-2} z ^{4} = 7 z ^{4} / y ^{2} \]
Third: distribute the square
Why: S cubed becomes s to the sixth; t to the negative 4 becomes t to the negative 8, which moves up as t to the eighth.
\[ s ^{6} t ^{8} \]
Fourth: simplify inside first, then cube
Why: Inside, x is 4 minus 3, or 1, and y is negative 2 minus 6, or negative 8; cubing gives x cubed and y to the negative 24.
\[ x ^{3} / y ^{24} \]
Figure (svg): The solution to Worked example four more shown as a ladder of expressions, one row per algebraic move
\[ x^2, \; \tfrac{7z^4}{y^2}, \; s^6t^8, \; \tfrac{x^3}{y^{24}} \]
Verify: check the fourth by the other order
Why: Distributing the cube first gives x to the twelfth y to the negative 6, over x to the ninth y to the eighteenth; subtracting exponents gives x cubed y to the negative 24 — the same answer. Simplifying inside first was quicker, but both orders are legal, which is a general feature of these problems.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 333-333
Trap
\[ \frac{7y^{-2}z^4}{1} \]
Move the whole term down because it has a negative exponent
Why: The 7 and the z to the fourth travel with the y.
\[ \frac{1}{7y^2z^{-4}} \quad \text{(wrong)} \]
Only the factor with the negative exponent moves. The 7 and the z to the fourth had positive exponents and belong where they were.
\[ 7y^{-2}z^4 = \frac{7z^4}{y^2} \]
Move each factor independently, according to its own exponent
Why: The negative exponent property applies to one base at a time.
\[ a^{-m} = \frac{1}{a^m} \quad \text{for that base only} \]
Testing with numbers settles any doubt: at y equal to 2 and z equal to 1 the expression is 7 over 4, and so is the simplified form. The wrong version gives 1 over 28.
Fill the middle
Example 3b, at the last step.
Fill in the blanks
\frac9___} = r^6 s^___}
Why: An s to the negative 9 in the denominator is the reciprocal of s to the negative 9, which is s to the ninth. Crossing the fraction bar flips the sign of the exponent, and that is the negative exponent property read backwards rather than a new rule.
Ranking
Simplifying an algebraic expression with powers.
Put in order
Why: Clearing negative exponents is left until last because doing it early usually creates fractions that then have to be recombined. The zero exponents are cleared just before, since they only appear once exponents have been subtracted.
Prediction
Commit before reasoning.
Predict first
Example 4 could distribute the outer cube first, or simplify inside the bracket first. Do the two routes agree?
Correct: Yes, always — the properties are consistent.
\[ \left(\tfrac{x^4y^{-2}}{x^3y^6}\right)^3 = (xy^{-8})^3 = x^3y^{-24} = \tfrac{x^3}{y^{24}} \]
Why: The properties are true statements about numbers, so any sequence of them applied to the same expression must produce equal results. What differs is how much arithmetic each route involves, and simplifying inside first is usually shorter because it keeps the exponents small. Choosing the shorter route is a matter of taste; being confident that both are legal is what lets you choose.
Section
Section 5
Concept
When two quantities differ by a scale factor, forming their ratio lets the unknown quantities cancel by the quotient of powers, leaving a pure number. This is why a comparison can often be made without knowing either quantity.
\[ \frac{\tfrac{4}{3}\pi(1500r)^3}{\tfrac{4}{3}\pi r^3} = 1500^3 \]
The result generalises: scaling every length by k multiplies areas by k squared and volumes by k cubed, which is why a modest change in size produces an enormous change in volume.
Figure (svg): Two spheres of very different radii with the ratio of their volumes computed algebraically
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 332-332 — Compare real-life volumes
Picture it
Example 5: a radius 1500 times as large.
Figure (svg): Two spheres of very different radii with the ratio of their volumes computed algebraically
Neither radius was ever needed. The ratio is 1500 cubed, about 3.4 billion, whatever r happens to be.
Worked example
Example 5. Let the unknown stand and watch it cancel.
\[ \text{Betelgeuse's radius is } 1500 \text{ times the sun's. How many times its volume?} \]
Name the unknown
Why: Let r be the sun's radius, so Betelgeuse's is 1500r.
\[ r\text{ and } 1500 r \]
Write the ratio of volumes
Why: Each volume is four thirds pi times the cube of the radius.
\[ (\frac{4}{3}) \pi(1500 r) ^{3}\text{ over } (\frac{4}{3}) \pi r ^{3} \]
Apply the power of a product
Why: The quantity 1500r, cubed, is 1500 cubed times r cubed.
\[ 1500 ^{3} r ^{3}\text{ over } r ^{3} \]
Cancel with the quotient of powers
Why: The four thirds pi cancels, and r cubed over r cubed is r to the zero, which is 1.
\[ 1500 ^{3} \]
Evaluate
Why: Fifteen hundred cubed is 3,375,000,000.
\[ \text{about } 3.4\text{ billion} \]
Figure (svg): The solution to Worked example compare two volumes shown as a ladder of expressions, one row per algebraic move
\[ 1500^3 = 3{,}375{,}000{,}000 \]
Verify: sanity-check the scaling
Why: A factor of 1500 in length cubes to a factor of about 3.4 billion in volume. Checking the magnitude: 1500 is 1.5 times 10 cubed, so its cube is 3.375 times 10 to the ninth — which is 3.375 billion, matching. Estimating with scientific notation is faster than the long multiplication and catches a slipped zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 332-332
Fill the middle
Example 5, at the quotient rule.
Fill in the blanks
\frac0___ = 1500^3 r^___ = 1500^3 r^___} = 1500^3
Why: Three minus 3 is 0, and r to the zero is 1, so the unknown radius disappears entirely. That is the point of forming a ratio: a quantity nobody measured cancels, and the comparison survives without it.
Worked example
The same argument one dimension down.
\[ \text{If one circle's radius is } 20 \text{ times another's, compare their areas.} \]
Name the unknown and write the ratio
Why: Let r be the smaller radius; the areas are pi r squared and pi times the quantity 20r, squared.
\[ \pi(20 r) ^{2}\text{ over } \pi r ^{2} \]
Distribute the square
Why: The quantity 20r, squared, is 400 r squared.
\[ 400 r ^{2}\text{ over } r ^{2} \]
Cancel
Why: The pi cancels and r squared over r squared is 1.
\[ 400 \]
Compare with the volume case
Why: A length factor of 20 gives an area factor of 400 and would give a volume factor of 8000.
\[ k, k ^{2}, k ^{3} \]
Figure (svg): The solution to Worked example scaling areas instead shown as a ladder of expressions, one row per algebraic move
\[ 20^2 = 400 \]
Verify: check the pattern across dimensions
Why: Length scales by k, area by k squared, volume by k cubed, because area is a product of two lengths and volume of three. The exponent is the number of dimensions, which is why doubling a shape's size multiplies its volume by 8 rather than by 2 — a fact worth carrying well beyond this lesson.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 332-332
Error analysis
A student compares the two stars.
Annotate
On: \( \text{the radius is } 1500 \text{ times as big, so the volume is } 1500 \text{ times as big} \)
Whenever a quantity depends on a power of a length, a scale factor must be raised to that power. Writing the ratio out algebraically makes the exponent impossible to miss.
Matching
The exponent is the number of dimensions.
Match the pairs
Why: In every case the scale factor is raised to the number of dimensions: squared for areas and cubed for volumes. The third is the one worth remembering as a rule of thumb — doubling a solid's dimensions multiplies its volume by eight, not by two.
Prediction
Commit before reasoning.
Predict first
Example 5 never uses the sun's actual radius. Why is the answer still exact?
Correct: The ratio depends only on the scale factor, so the radius cancels.
\[ \frac{V_B}{V_S} = 1500^3 \text{ for every } r > 0 \]
Why: Both volumes contain the same r cubed and the same four thirds pi, and dividing removes them. Whatever value r has, the ratio is 1500 cubed. This is a general and very useful pattern: comparative questions often need far less information than absolute ones, and forming the ratio symbolically is what reveals how little is needed.
Comparison
Fill the blanks. One needs more information than the other.
Comparison matrix
| Question | What you need | Why |
|---|---|---|
| How big is Betelgeuse? | the sun's radius, in kilometres | an absolute answer needs an absolute input |
| How many times bigger? | only the scale factor | the unknown cancels in the ratio |
| The property used | power of a product | quotient of powers |
| The answer | a measurement | a pure number |
Asking the comparative question instead of the absolute one is often the difference between a problem you can solve and one you cannot.
Comparison
Fill the blanks. Seven rules grouped by what they change.
Comparison matrix
| Group | Properties | Effect on exponents |
|---|---|---|
| Combining like bases | product and quotient of powers | add or subtract them |
| Nested powers | power of a power | multiply them |
| Distributing | power of a product and of a quotient | copy the exponent to each factor |
| Rewriting | negative and zero exponent | move across the bar, or replace by 1 |
No property applies to a sum. That absence is as important as the seven that are there.
Pattern
One routine for any expression built from powers.
Do one property per line and name it. A six-line simplification with named steps is checkable; a two-line one is not.
OpenStax Algebra and Trigonometry 2e, §1.2 Exponents and Scientific Notation §1.2
Check
Evaluating. Watch which rule applies.
Check your understanding
Evaluate (2^4 x 2^5)^2.
Answer: A
Why: Inside, the exponents add to 9; squaring multiplies by 2, giving 2^18.
Check
Simplifying. Only positive exponents.
Check your understanding
Simplify (x^-2 y^3)^3 / (x^4 y^9).
Answer: A
Why: The numerator is x^-6 y^9; dividing gives x^-10 y^0, which is 1/x^10.
Check
Scaling. Count the dimensions.
Check your understanding
A star's radius is 1500 times the sun's. How many times as great is its volume?
Answer: A
Why: Volume depends on the cube of the radius, and the r^3 terms cancel in the ratio, leaving 1500^3.
Real world
A data centre stores about 4.5 times 10 to the 15th bytes. A single archival tape holds about 1.8 times 10 to the 13th bytes.
Discussion prompt
How many tapes are needed? If each tape is 2.5 centimetres thick, how tall is the stack, in metres? And if the centre's storage grows by a factor of 10 every four years, how many tapes in twelve years?
Hint: Divide in scientific notation, then convert units with another power of ten.
Answer:
\[ \frac{4.5 \times 10^{15}}{1.8 \times 10^{13}} = 2.5 \times 10^{2} = 250 \text{ tapes} \]
\[ 250 \times 2.5 \text{ cm} = 625 \text{ cm} = 6.25 \text{ m} \]
\[ 12 \text{ years} = 3 \text{ doublings of scale} \;\Longrightarrow\; 10^3 \times 250 = 2.5 \times 10^{5} \text{ tapes} \]
Two hundred and fifty tapes today, a stack 6.25 metres tall, growing to 250,000 tapes in twelve years — a stack over six kilometres high.
The last figure is the one worth pausing on. Growing by a factor of ten three times is a factor of 10 cubed, not 30, and that is the product-of-powers rule doing the work. Repeated multiplication grows far faster than intuition suggests, which is exactly the theme Lesson 7.1 will take up as exponential growth.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is 3 squared times 3 to the fifth equal to 9 to the seventh?
Correct: No — the base stays 3, giving 3 to the seventh.
\[ 3^2 \cdot 3^5 = 3^7 = 2187 \neq 9^7 \]
Why: Writing the factors out settles it: two 3s times five 3s is seven 3s, which is 3 to the seventh, or 2187. Nine to the seventh is 4,782,969, more than two thousand times larger. The bases are not multiplied because they are not being combined as numbers — they are being counted as factors, and the count is what the exponent records. The textbook flags this exact error in an Avoid Errors note.
Explain it
They have memorised the rules and keep mixing up when exponents add and when they multiply.
Discussion prompt
In four sentences or fewer, explain the difference by talking about counting factors rather than about rules.
Hint: Write both expressions out.
Answer:
A to the third times a to the fifth pools three factors with five factors, giving eight, so the exponents add. A to the third, all raised to the fifth, is five copies of a group of three factors, giving fifteen, so the exponents multiply.
The difference is whether you are joining two piles or repeating one pile several times. If you forget which is which, write out a small case with a equal to 2 and count — it takes ten seconds and it is never wrong.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For adding against multiplying, write out a small case with base 2 and count the factors. For negative exponents, deal with them last and move one factor at a time. For scientific notation, make checking the leading part your final written line. For distributing, write each factor on its own before applying the exponent to it. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference for the seven properties. Down the left, write each property's name, its definition in letters, and one numerical example you make up and verify by direct calculation. In the middle, take the single expression given by x to the fourth y to the negative 2, over x cubed y to the sixth, all cubed, and simplify it twice — once by distributing the outer cube first and once by simplifying inside first — showing that both routes reach the same answer. On the right, write the two most common errors with exponents, each with a wrong line, a corrected line and a small numerical counterexample. At the bottom, write out why a to the zero must equal 1, using the quotient rule. In a margin, note what happens to area and to volume when every length is multiplied by k.
If your two simplification routes disagree, check the distributing route first: applying an outer exponent to a fraction means applying it to numerator and denominator separately, and both of those are products.
Recap
Five things, and the first one makes the other four rederivable.
| If you see | Then |
|---|---|
| Powers of the same base multiplied | Add the exponents |
| Powers of the same base divided | Subtract the exponents |
| A power raised to a power | Multiply the exponents |
| An exponent outside a product | Copy it onto each factor |
| A negative exponent | Move it across the fraction bar |
| A zero exponent | Replace it by 1 |
| A sum inside a power | No property applies; expand instead |
These rules are the arithmetic of the whole chapter. Lesson 5.2 uses them to define polynomial functions of any degree and to look at what their graphs do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.1 Use Properties of Exponents §5.1, pp. 330-333 — everything on these slides traces back here
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