Question

Difficulty: MediumOperations with Variable Exponents

An algebraic expression is given as:

(3a4b−2)2(2a−3b5)6a2b \frac{(3a^4b^{-2})^2 (2a^{-3}b^5)}{6a^2b}

What is the fully simplified form of this expression for all non-zero values of aa and bb?

  1. 3a33a^3Answer
  2. B
    a3a^3
  3. C
    3ab43ab^4
  4. D
    3a3b83a^3b^8

Answer

The simplified form of the expression is 3a33a^3.
The expression is correctly simplified by first applying the power to all terms inside the parentheses (resulting in 9a8b−49a^8b^{-4}), then multiplying by the adjacent term in the numerator (resulting in 18a5b18a^5b), and finally dividing by the denominator (giving 3a33a^3).

Step-by-Step Solution

1
Simplify the squared term in the numerator using the power of a product and power of a power rules.
9a8b−49a^8b^{-4}
When a product is raised to a power, raise each factor to that power (32=93^2 = 9). For variables, multiply the inner and outer exponents: (a4)2=a8(a^4)^2 = a^8 and (b−2)2=b−4(b^{-2})^2 = b^{-4}.
2
Multiply the simplified term by the second factor in the numerator, (2a−3b5)(2a^{-3}b^5).
18a5b18a^5b
Multiply the coefficients (9×2=189 \times 2 = 18). For variables with the same base, add their exponents: a8+(−3)=a5a^{8+(-3)} = a^5 and b−4+5=b1=bb^{-4+5} = b^1 = b.
3
Divide the resulting numerator by the denominator, 6a2b6a^2b.
3a33a^3
Divide the coefficients (18÷6=318 \div 6 = 3). For variables with the same base, subtract the denominator's exponent from the numerator's exponent: a5−2=a3a^{5-2} = a^3. The terms b1b^1 and b1b^1 cancel out since b1−1=b0=1b^{1-1} = b^0 = 1.

Key Concept

Operations with Variable Exponents
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