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CAT Previous Year Paper | CAT Exponents and Logarithms Questions | Question 23

This is a question from Logarithms. Logarithms is a favorite in CAT Exam and appears more often than expected in the CAT Quantitative Aptitude section in the CAT Exam. With every extra hour you log in for this topic, it becomes exponentially simpler.

Question 23 : The value of log0.008√5 + log√381 – 7 is equal to :

  1. \\frac{1}{3})
  2. \\frac{2}{3})
  3. \\frac{5}{6})
  4. \\frac{7}{6})

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Explanatory Answer

Method of solving this CAT Question from Exponents and Logarithms

Here we have to find the value of log0.008 √5 + log√3 81 – 7.
⟹ log0.008 √5 + log√3 81 – 7
log0.008 √5 can be written in the terms of five and log√3 81 can be written in the terms of 3.
where √5 = 5\^\frac{1}{2}) ⟹ 0.008 = \\frac{2^3}{10^3}) = 5-3
⟹ log0.008 √5 + log√3 81 – 7
⟹ \\frac{log_5 √5}{log_5 0.008}) + \\frac{log_3 81}{log_3 √3}) - 7
⟹ \\frac{\frac{-1}{2}}{-3}) + \\frac{4}{\frac{1}{2}}) - 7
⟹ \\frac{-1}{6}) + 1 = \\frac{5}{6})
Hence the value of log0.008√5 + log√381 – 7 = \\frac{5}{6})

The question is "The value of log0.008√5 + log√381 – 7 is equal to :"

Hence, the answer is \\frac{5}{6})

Choice C is the correct answer

 

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