An algebra problem by أحمد الحلاق

Algebra Level 3

log 10 2 0 2 log 20 3 0 2 log 30 4 0 2 log 990 100 0 2 log 10 1 1 2 log 11 1 2 2 log 12 1 3 2 log 99 10 0 2 = ? \large \dfrac{ \log_{10} 20^2 \cdot \log_{20} 30^2 \cdot \log_{30} 40^2 \cdots \log_{990} 1000^2 }{ \log_{10} 11^2 \cdot \log_{11} 12^2 \cdot \log_{12} 13^2 \cdots \log_{99} 100^2} =\, ?


The answer is 768.

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1 solution

Chew-Seong Cheong
Sep 10, 2016

Q = log 10 2 0 2 log 20 3 0 2 log 30 4 0 2 log 990 100 0 2 log 10 1 1 2 log 11 1 2 2 log 12 1 3 2 log 99 10 0 2 = 2 log 20 log 10 2 log 30 log 20 2 log 40 log 30 2 log 1000 log 990 2 log 11 log 10 2 log 12 log 11 2 log 13 log 12 2 log 100 log 99 = 2 99 log 1000 log 10 2 90 log 100 log 10 = 2 99 3 2 90 2 = 768 \begin{aligned} Q & = \frac {\log_{10}20^2\cdot \log_{20}30^2\cdot \log_{30}40^2\cdots \log_{990}1000^2}{\log_{10}11^2\cdot \log_{11}12^2\cdot \log_{12}13^2\cdots \log_{99}100^2} \\ & = \frac {\frac {2 \cancel{\log 20}}{\log 10}\cdot \frac {2 \cancel{\log 30}}{\cancel{\log 20}}\cdot \frac {2\cancel{\log 40}}{\cancel{\log 30}}\cdots \frac {2\log 1000}{\cancel{\log 990}}}{\frac {2 \cancel{\log 11}}{\log 10}\cdot \frac {2 \cancel{\log 12}}{\cancel{\log 11}}\cdot \frac {2\cancel{\log 13}}{\cancel{\log 12}}\cdots \frac {2\log 100}{\cancel{\log 99}}} \\ & = \frac {2^{99}\frac {\log 1000}{\log 10}}{2^{90}\frac {\log 100}{\log 10}} = \frac {2^{99}\cdot 3}{2^{90}\cdot 2} = \boxed{768} \end{aligned}

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