CAT 2018 Question Paper | Quants Slot 2
CAT Previous Year Paper | CAT Logarithm Questions | Question 34
CAT Questions This is a question from the topic of Logarithms, which got rolled out in CAT 2018 Question Paper. This topic is heavily tested in the CAT Exam. For the past few years, there have been at least 2 questions from this topic in the CAT Question Paper. Every CAT Exam aspirant's CAT Online Preparation is incomplete without mastering this topic. We suggest you to solve CAT Previous Year Paper to ace this topic. Try one right here!
Question 34 : \\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}) = ?
- 0
- \\frac{1}{2})
- -4
- 10
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Explanatory Answer
Method of solving this CAT Question from Logarithm
We know that \\frac{1}{log_a b}) = logb a
⟹ log100 2 - log100 4 + log100 5 - log100 10 + log100 20 - log100 25 + log100 50
We also know that (- loga b) = loga (\\frac{1}{b}))
⟹ log100 (2 × \\frac{1}{4}) × 5 × \\frac{1}{10}) × 20 × \\frac{1}{25}) × 50)
⟹ log100 10 = \\frac{1}{2})
The question is "\\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}) = ?"
Hence, the answer is \\frac{1}{2})
Choice B is the correct answer
