2021 AMC 12A Spring Problem 14

Attempt Problem 14 of the 2021 AMC 12A Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

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14.

What is the value of (k=120log5k3k2)(k=1100log9k25k)? \begin{aligned} &\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right) \\ &\quad {}\cdot \left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)? \end{aligned}

2121

100log53100\log_5 3

200log35200\log_3 5

2,2002{,}200

21,00021{,}000

Answer: E
Concepts:logarithmsummation
Difficulty rating: 1780
Solution:

For the first sum, log5k3k2=k2klog53=klog53,\log_{5^k} 3^{k^2} = \dfrac{k^2}{k}\log_5 3 = k\log_5 3, so k=120klog53=20212log53=210log53. \begin{aligned} &\sum_{k=1}^{20} k\log_5 3 = \frac{20\cdot 21}{2}\log_5 3 \\ &= 210\log_5 3. \end{aligned}

For the second sum, log9k25k=log925=log35,\log_{9^k} 25^k = \log_9 25 = \log_3 5, independent of k,k, so the sum is 100log35.100\log_3 5.

Since log53log35=1,\log_5 3 \cdot \log_3 5 = 1, the product is 210100=21,000.210 \cdot 100 = 21{,}000.

Thus, the correct answer is E.

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