CAT 2020 Quant Question [Slot 2] with Solution 07

Question

The value of \({\log _a}\left( {\frac{a}{b}} \right) + {\log _b}\left( {\frac{b}{a}} \right),\) for \(1 < a \le b\) cannot be equal to

  1. -0.5
  2. 1
  3. 0
  4. -1
Option: 2
Solution:

\({\log _a}\left( {\frac{a}{b}} \right) + {\log _b}\left( {\frac{b}{a}} \right)\)

\( = {\log _a}a - {\log _a}b + {\log _b}b - {\log _b}a\)

\( = 1 - {\log _a}b + 1 - {\log _b}a\quad \left[ {{{\log }_n}n = 1} \right]\)

since \(\left( {{{\log }_a}b + {{\log }_a}b} \right) \ge 2\)

\(\therefore \) The above value is \( \le 0\).

\(\therefore 1\) can't be the answer.

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CAT 2020 Quant questions with Solutions