Difference between revisions of "2021 AMC 12A Problems/Problem 14"

Problem

What is the value of $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?$$

$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000$

Solution 1 (Condensed Explanation of Logarithmic Identities)

This equals $$\left(\sum_{k=1}^{20}k\log_5{3}\right)\left(\sum_{k=1}^{100}\log_9{25}\right)=\frac{20\cdot21}{2}\cdot\log_5{3}\cdot100\log_3{5}=\boxed{\textbf{(E) }21,000}.$$ ~JHawk0224

Solution 2 (Detailed Explanation of Logarithmic Identities)

We will apply the following logarithmic identity: $$\log_{p^n}{q^n}=\log_{p}{q},$$ which can be proven by the Change of Base Formula: $$\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.$$ Now, we simplify the expressions inside the summations: \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{\left(3^k\right)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} Using these results, we evaluate the original expression: \begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ &= \frac{21\cdot20}{2}\cdot100 \\ &= \boxed{\textbf{(E) }21,000}. \end{align*} ~MRENTHUSIASM

Solution 3 (Properties of Logarithms)

First, we can get rid of the $k$ exponents using properties of logarithms: $$\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.$$ (Leaving the single $k$ in the exponent will come in handy later). Similarly, $$\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.$$ Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\ &= \log_{5} 3^{210}, \\ \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ &= \log_{9} 5^{2(100)} \\ &= \log_{9} 5^{200}. \end{align*} In $(*),$ we use the triangular numbers equation: $$1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.$$ Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: $$\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.$$ Thus, \begin{align*} \left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) &= \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right) \\ &= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\ &= (200)(105) \\ &= \boxed{\textbf{(E) }21,000}. \end{align*} ~Joeya (Solution)

~MRENTHUSIASM (Reformatting)

Solution 4 (Estimations and Answer Choices)

In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$

In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$

We have the inequality $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1,900,$$ which eliminates choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{\textbf{(E) }21,000}$ by either an educated guess or a continued approximation:

Since $3^3=27\approx25,$ it follows that $9^{3/2}\approx25.$ By an extremely rough underestimation, $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)\approx\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2,850.$$ From here, it should be safe to guess that the answer is $\textbf{(E)}.$

As an extra guaranty, note that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19.$ Therefore, we must have $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>>2,850.$$

~MRENTHUSIASM

~IceMatrix