Question

Difficulty: HardExponents, Powers, and Square Roots
If kk is a positive integer, which of the following expressions is equivalent to 2k+2⋅32k−1+6k⋅3k−112k⋅2−1\frac{2^{k+2} \cdot 3^{2k-1} + 6^k \cdot 3^{k-1}}{12^k \cdot 2^{-1}} for all values of kk?
  1. A
    53(32)k\frac{5}{3} \left(\frac{3}{2}\right)^k
  2. 103(32)k\frac{10}{3} \left(\frac{3}{2}\right)^kAnswer
  3. C
    53⋅6k\frac{5}{3} \cdot 6^k
  4. D
    5(32)k5 \left(\frac{3}{2}\right)^k
  5. E
    103⋅3k\frac{10}{3} \cdot 3^k

Answer

The expression simplifies to 103(32)k\frac{10}{3} \left(\frac{3}{2}\right)^k.
Factoring the terms in the numerator into base 18k18^k gives 43⋅18k+13⋅18k=53⋅18k\frac{4}{3} \cdot 18^k + \frac{1}{3} \cdot 18^k = \frac{5}{3} \cdot 18^k. The denominator equals 12⋅12k\frac{1}{2} \cdot 12^k. Dividing numerator by denominator yields 5/31/2⋅(1812)k=103(32)k\frac{5/3}{1/2} \cdot \left(\frac{18}{12}\right)^k = \frac{10}{3} \left(\frac{3}{2}\right)^k.

Step-by-Step Solution

1
Rewrite each term in the numerator using prime base factorization
2k+2⋅32k−1=2k⋅22⋅(32)k⋅3−1=4⋅2k⋅9k⋅13=43⋅18k2^{k+2} \cdot 3^{2k-1} = 2^k \cdot 2^2 \cdot (3^2)^k \cdot 3^{-1} = 4 \cdot 2^k \cdot 9^k \cdot \frac{1}{3} = \frac{4}{3} \cdot 18^k, and 6k⋅3k−1=(2⋅3)k⋅3k⋅3−1=2k⋅9k⋅13=13⋅18k6^k \cdot 3^{k-1} = (2 \cdot 3)^k \cdot 3^k \cdot 3^{-1} = 2^k \cdot 9^k \cdot \frac{1}{3} = \frac{1}{3} \cdot 18^k.
Converting all powers to base 18 allows terms with identical exponential factors to be combined.
2
Combine the terms in the numerator
43⋅18k+13⋅18k=(43+13)⋅18k=53⋅18k\frac{4}{3} \cdot 18^k + \frac{1}{3} \cdot 18^k = \left(\frac{4}{3} + \frac{1}{3}\right) \cdot 18^k = \frac{5}{3} \cdot 18^k.
Adding coefficients of like exponential terms.
3
Simplify the denominator expression
12k⋅2−1=12⋅12k12^k \cdot 2^{-1} = \frac{1}{2} \cdot 12^k.
Applying the negative exponent rule a−n=1ana^{-n} = \frac{1}{a^n}.
4
Divide the numerator by the denominator
53⋅18k12⋅12k=5/31/2⋅(1812)k=(53⋅2)⋅(32)k=103(32)k\frac{\frac{5}{3} \cdot 18^k}{\frac{1}{2} \cdot 12^k} = \frac{5/3}{1/2} \cdot \left(\frac{18}{12}\right)^k = \left(\frac{5}{3} \cdot 2\right) \cdot \left(\frac{3}{2}\right)^k = \frac{10}{3} \left(\frac{3}{2}\right)^k
Dividing fractions by multiplying by the reciprocal and applying quotient rule for powers with the same exponent.

Key Concept

Prime base factorization and laws of exponents
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