Question:
\(\frac{1}{{{{\log }_2}100}} - \frac{1}{{{{\log }_4}100}} + \frac{1}{{{{\log }_5}100}} - \frac{1}{{{{\log }_{10}}100}} + \frac{1}{{{{\log }_{20}}100}} - \frac{1}{{{{\log }_{25}}100}} + \frac{1}{{{{\log }_{50}}100}} = ?\)
- 1/2
- 0
- 10
- −4
Correct Answer: 1
We know that $\frac{1}{{{\log }_{b}}a}={{\log }_{a}}b$, therefore,
$\frac{1}{\log _{2} 100}-\frac{1}{\log _{4} 100}+\frac{1}{\log _{5} 100}-\frac{1}{\log _{10} 100}+\frac{1}{\log _{20} 100}-\frac{1}{\log _{25} 100}+\frac{1}{\log _{50} 100}$
= $\log _{100} 2-\log _{100} 4+\log _{100} 5-\log _{100} 10+\log _{100} 20-\log _{100} 25+\log _{100} 50$
=${{\log }_{100}}\left( \frac{2}{4}\times \frac{5}{10}\times \frac{20}{25}\times 50 \right)$
=${{\log }_{100}}10$
Using the relation ${{\log }_{{{a}^{m}}}}b=\frac{1}{m}{{\log }_{a}}b$
${{\log }_{100}}10={{\log }_{{{10}^{2}}}}10=\frac{1}{2}{{\log }_{10}}10=\frac{1}{2}$
We know that $\frac{1}{{{\log }_{b}}a}={{\log }_{a}}b$, therefore,
$\frac{1}{\log _{2} 100}-\frac{1}{\log _{4} 100}+\frac{1}{\log _{5} 100}-\frac{1}{\log _{10} 100}+\frac{1}{\log _{20} 100}-\frac{1}{\log _{25} 100}+\frac{1}{\log _{50} 100}$
= $\log _{100} 2-\log _{100} 4+\log _{100} 5-\log _{100} 10+\log _{100} 20-\log _{100} 25+\log _{100} 50$
=${{\log }_{100}}\left( \frac{2}{4}\times \frac{5}{10}\times \frac{20}{25}\times 50 \right)$
=${{\log }_{100}}10$
Using the relation ${{\log }_{{{a}^{m}}}}b=\frac{1}{m}{{\log }_{a}}b$
${{\log }_{100}}10={{\log }_{{{10}^{2}}}}10=\frac{1}{2}{{\log }_{10}}10=\frac{1}{2}$
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