Logarithmic Value Range

CAT 2020 Slot 2 · Quantitative Ability · Hard · Algebra

The value of log⁡aab+log⁡bba\log_a \frac{a}{b} + \log_b \frac{b}{a}, for 1<a≤b1 < a \le b cannot be equal to

  1. A.

    -0.5

  2. B.

    1

  3. C.

    0

  4. D.

    -1

Answer

B

Explanation

Simplify the given expression: log⁡aab+log⁡bba=(1−log⁡ab)+(1−log⁡ba)=2−(log⁡ab+log⁡ba)\log_a \frac{a}{b} + \log_b \frac{b}{a} = (1 - \log_a b) + (1 - \log_b a) = 2 - \left(\log_a b + \log_b a\right) Let t=log⁡abt = \log_a b. Since 1<a≤b1 < a \le b, t≥1t \ge 1. By AM-GM, for t≥1t \ge 1, t+1t≥2t + \frac{1}{t} \ge 2. Therefore: 2−(t+1t)≤2−2=02 - \left(t + \frac{1}{t}\right) \le 2 - 2 = 0 So the value must be ≤0\le 0. Among the given options, 1>01 > 0, so it cannot be equal to 1.

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