2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(1)
Problem 1(1)
The largest one among natural numbers that are less than \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020} is \fbox { A } .
Solution
The key idea is the following relationship that holds for positive numbers p, q, and b. \log_p{q} = \frac{\log_b{q}}{\log_b{p}}, where p\neq1 and b\neq1.
For convenience, let L \equiv \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}. Selecting b=e, we obtain \begin{align} L &=\frac{\cancel{\log{3}}}{\log{2}}\cdot \frac{\cancel{\log{4}}}{\cancel{\log{3}}}\cdot \frac{\cancel{\log{5}}}{\cancel{\log{4}}} \cdots\cdots \frac{\log{2020}}{\cancel{\log{2019}}}\\ &=\frac{\log{2020}}{\log{2}}\\ &=\log_2{2020}, \end{align} where \log is the natural logarithm (base e). This is equivalent to 2^L = 2020.
Noticing that \begin{align} 2^{10} &= 1024 \\ 2^{11} &=2048, \end{align} we obtain \begin{align} 1024 \leq 2020 \leq 2048 &\Leftrightarrow 2^{10} \leq 2^L \leq 2^{11}\\ &\Leftrightarrow \log_2{2^{10}} \leq \log_2{2^L} \leq \log_2{2^{11}}\\ &\Leftrightarrow 10 \leq L \leq 11. \end{align} because the exponential and logarithmic functions are both monotonic and increasing. Thus, the largest natural number less than L is 10. ■
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