Power Play · Lesson 2 of 7
Exponential Notation and Operations
“Learn exponential notation, exponent laws, repeated growth and combination counting.”
• Read and write exponential notation. • Identify base and exponent. • Derive exponent laws from repeated multiplication. • Apply powers to branching stories, growth patterns and combinations. • Distinguish exponentiation from ordinary multiplication and addition.
Exponential notation shortens repeated multiplication. Instead of 5 × 5 × 5 × 5, we write 5⁴. The base is the repeated factor. The exponent tells how many equal factors are multiplied.
Problem
Find 4³.
- 1.4³ = 4 × 4 × 4.
- 2.4 × 4 = 16.
- 3.16 × 4 = 64.
- 4.Therefore, 4³ = 64.
Problem
Find (−4)³.
- 1.The base is −4 because the negative sign is inside the brackets.
- 2.(−4)³ = (−4)(−4)(−4).
- 3.The result is −64.
4 + 4 + 4 = 12, while 4 × 4 × 4 = 4³ = 64. Exponentiation is repeated multiplication, not repeated addition.
Problem
Write 32400 using powers of prime factors.
- 1.32400 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5.
- 2.Group equal prime factors.
- 3.32400 = 2⁴ × 3⁴ × 5².
The Stones that Shine ...
The story builds repeated groups of three. Three daughters lead to three baskets each, then three keys, rooms and further groups. Every new level contributes another factor of 3.
If seven factors of 3 are split into four factors and three factors, then 3⁷ = 3⁴ × 3³. The exponents add because together there are seven factors of the same base.
Problem
Simplify p⁴ × p⁶.
- 1.p⁴ contains 4 factors of p.
- 2.p⁶ contains 6 factors of p.
- 3.Together there are 10 factors.
- 4.So p⁴ × p⁶ = p¹⁰.
A power can itself be raised to another power. For example, (4³)² contains two copies of 4³, so there are 3 × 2 = 6 factors of 4.
Problem
Simplify (7²)⁴.
- 1.There are 4 copies of 7².
- 2.That gives 2 × 4 = 8 factors of 7.
- 3.Therefore, (7²)⁴ = 7⁸.
Magical Pond
The lotus pond doubles every day. If it is completely covered on day 30, it must be half covered on day 29 because the next day's doubling changes half to full.
The chapter also combines four doublings with four triplings. That creates 2⁴ × 3⁴. Because the exponents match, pair each 2 with a 3 to get (2 × 3)⁴ = 6⁴.
Problem
Simplify 2⁵ × 5⁵.
- 1.The exponent is 5 in both powers.
- 2.2⁵ × 5⁵ = (2×5)⁵.
- 3.Therefore, the expression equals 10⁵.
How Many Combinations
Repeated independent choices are counted by multiplication. Four dresses and three caps give 4 × 3 = 12 combinations because every dress can be paired with every cap.
A numerical lock with five slots has 10 choices for each position, so it has 10⁵ possible codes. A six-slot letter lock has 26 choices at each position, so it has 26⁶ possible codes.
Quiz
What does the 5 in 7⁵ represent?
Simplify x³ × x⁸.
Simplify (a⁴)³.
Simplify 3⁴ × 5⁴.
How many 4-digit numerical codes are possible if repetition is allowed?
Practice Problems
- Write 6 × 6 × 6 × 6 × 6 in exponential form and evaluate it.
- Simplify a⁷ × a⁵ using repeated factors.
- Write 8¹² as a power of a power in two different ways.
- Simplify 4⁶ × 5⁶.
- A code has 5 positions and each can contain any of 8 symbols. Write the number of possible codes in exponential form.
Key Takeaways
• The exponent counts repeated factors of the base. • nᵃ × nᵇ = nᵃ⁺ᵇ for the same base. • (nᵃ)ᵇ = nᵃᵇ. • mᵃnᵃ = (mn)ᵃ and mᵃ/nᵃ = (m/n)ᵃ. • Repeated choices multiply, so code counts naturally become powers. • Exponentiation is repeated multiplication, not repeated addition.