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CAT Previous Year Paper | CAT Quants Questions | Question 26

Exponents and Logarithms form an important part of the CAT syllabus. Logarithms require a  lot of practice as there is no intuitive way around them. The benefit of preparation is that there could be easy questions that can be attempted because of the preparation.

Question 26 : Let x and y be positive real numbers such that log5(x + y) + log5(x - y) = 3, and log2y - log2x = 1 - log23. Then xy equals

  1. 25
  2. 150
  3. 250
  4. 100

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Explanatory Answer

Given that, log5(x + y) + log5(x - y) = 3
We know log A + log B = log (A x B)
log5(x + y) + log5(x - y) = log5(x2 - y2)
log5(x2 - y2) = 3
x2 - y2 =
x2 - y2 = 125
Similarly, log2y - log2x = 1 - log23
log2= log22 - log23
log2= log2
3y = 2x
(2 - y2 = 125
2 - y2 = 125
2 = 125
2 = 100
y = 10
x = 15
So, xy = 15 x 10 = 150


The question is "Let x and y be positive real numbers such that log5(x + y) + log5(x - y) = 3, and log2y - log2x = 1 - log23. Then xy equals"

Hence, the answer is 150

Choice B is the correct answer.

 

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