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Powers of real numbers: base, exponent and the product and power laws
Mathematics Prep 2 Term 1, Unit 1, Lesson 5
GoalRead and evaluate powers with real bases, and use the product law and the power laws.
New words
base · the number that is multiplied by itselfexponent · how many times the base is used as a factorpower · the whole expression aⁿsame base · the product law works only when the bases are equal
Explanation
In aⁿ the base a is multiplied by itself n times: 2⁴ = 2 × 2 × 2 × 2 = 16. The base can be a root or a fraction: (√2)⁵ = √2 × √2 × √2 × √2 × √2 = 4√2 and (2/3)³ = 8/27. Brackets matter: (−5)⁴ = 625, but −5⁴ = −625 because only 5 is raised to the power. The product law: aᵐ × aⁿ = aᵐ⁺ⁿ, because 2³ × 2⁴ = (2 × 2 × 2)(2 × 2 × 2 × 2) = 2⁷. The power laws: (aᵐ)ⁿ = aᵐⁿ and (ab)ⁿ = aⁿbⁿ. The product law needs the same base: 2³ × 3³ is not 6⁶, but it is (2 × 3)³ = 6³. And aᵐ + aⁿ is not aᵐ⁺ⁿ.
| Law | Example |
|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | 3⁴ × 3² = 3⁶ = 729 |
| (aᵐ)ⁿ = aᵐⁿ | (2³)² = 2⁶ = 64 |
| (ab)ⁿ = aⁿbⁿ | (2 × 5)³ = 8 × 125 = 1000 |
Worked examples
Compare (−5)⁴ and −5⁴. (−5)⁴ = (−5)(−5)(−5)(−5) = 625, since an even number of negative factors gives a positive answer. −5⁴ = −(5 × 5 × 5 × 5) = −625. Now (−2)³ = −8.
ƒFormula
(√2)⁵ = 4√2
(√2)² = 2 makes the work short.
Use the laws on 2³ × 3³: both powers have exponent 3, so 2³ × 3³ = (2 × 3)³ = 6³ = 216. Check: 8 × 27 = 216. And (2⁴)³ = 2¹² = 4096.
Class activity
Powers on a chessboard. A chessboard has 8 × 8 = 64 squares. Write 64 as 8², as 4³ and as 2⁶. Show with the power law that (2³)² = 2⁶ = 64.
Practice
1
Find (−3)⁴ and −3⁴.
(−3)⁴ = 81; −3⁴ = −81.
2
Simplify 5³ × 5².
5⁵ = 3125.
3
Simplify (2⁴)³.
2¹² = 4096.
4
Find (√5)⁶.
((√5)²)³ = 5³ = 125.
5
Karim says 2³ + 2⁴ = 2⁷. Is he right?
No. 2³ + 2⁴ = 8 + 16 = 24 but 2⁷ = 128. The product law is for multiplying.