Multiplying same bases
Joining repeated factors adds their counts:
3² × 3⁴ = (3 × 3)(3 × 3 × 3 × 3) = 3⁶
So aᵐ × aⁿ = aᵐ⁺ⁿ for the same non-zero base.
Think first: Can you evaluate 2⁷ × 2⁵ without writing twelve factors of 2?
Open practice worksheetJoining repeated factors adds their counts:
3² × 3⁴ = (3 × 3)(3 × 3 × 3 × 3) = 3⁶
So aᵐ × aⁿ = aᵐ⁺ⁿ for the same non-zero base.
Cancel matching factors:
5⁶ ÷ 5² = 5⁴
For m ≥ n, aᵐ ÷ aⁿ = aᵐ⁻ⁿ. The bases must match.
(2³)⁴ means four groups of 2³, giving twelve factors of 2:
(2³)⁴ = 2³ˣ⁴ = 2¹²
Expand both sides to verify the numerical relationship:
(3 × 5)² = (3 × 5)(3 × 5) = (3 × 3)(5 × 5) = 3² × 5²
Therefore (ab)² = a²b² for numerical bases.
Same base and multiplication: add indices. 7³⁺⁵ = 7⁸.
Same base and division: subtract indices. 10⁹⁻⁴ = 10⁵.
Power of a power: multiply indices. 4²ˣ³ = 4⁶.
The left side is 14² = 196. The right side is 4 × 49 = 196, so the relationship is verified.
Model response: 6¹¹.
Model response: 9⁴.
Model response: 5¹². Four groups of three factors give 3 × 4 = 12 factors of 5.
Model response: The add-indices law cannot be used because the bases differ. 3² × 4² = 9 × 16 = 144 (also (3 × 4)² = 12²).
Model response: The left side is 24² = 576. The right side is 16 × 36 = 576. Expanding (4 × 6)(4 × 6) and regrouping gives (4 × 4)(6 × 6).
Model response: For 3⁵ ÷ 3², cancel two of the five factors to leave 3³, so division subtracts indices. For (2²)³, three groups of two factors give six factors, so (2²)³ = 2⁶ and a power of a power multiplies indices.