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CAT 2020 Slot 2 — QA Question 16

LogarithmsEasy

    The value of⁡ loga⁡(ab)+logb⁡(ba), for 1 < a ≤ b cannot be equal to

Answer & solution

  • A

    0

  • B

    -1

  • 1

  • D

    -0.5

Solution

Easy

Split each logarithm of a quotient into a difference. The expression reduces to 22 minus a quantity of the form t+1tt+\tfrac1t which is always ≥2\ge 2, so the whole expression is ≤0\le 0. Any positive option is impossible.

1

Expand each term. Use log⁡aab=log⁡aa−log⁡ab=1−log⁡ab\log_a\tfrac{a}{b}=\log_a a-\log_a b = 1-\log_a b, and likewise for the second term.

log⁡a ⁣(ab)+log⁡b ⁣(ba)=(1−log⁡ab)+(1−log⁡ba)⇒ =2−(log⁡ab+log⁡ba)\begin{aligned} &\log_a\!\left(\tfrac{a}{b}\right) + \log_b\!\left(\tfrac{b}{a}\right) = (1-\log_a b) + (1-\log_b a)\\ &\Rightarrow\ = 2 - \left(\log_a b + \log_b a\right) \end{aligned}
2

Bound the bracket. Since $10.Writing. Writingt=\log_a b,wehave, we have\log_b a=\tfrac1t,andapositivenumberplusitsreciprocalisatleast, and a positive number plus its reciprocal is at least2$.

log⁡ab+log⁡ba=t+1t≥2(AM–GM)\begin{aligned} &\log_a b + \log_b a = t + \tfrac{1}{t} \ge 2 \quad\text{(AM–GM)} \end{aligned}
3

Conclude the range. Subtract the bound from 22.

2−(log⁡ab+log⁡ba)≤2−2=0\begin{aligned} &2 - \left(\log_a b + \log_b a\right) \le 2 - 2 = 0 \end{aligned}

So the expression is always ≤0\le 0. Values 00, −1-1, −0.5-0.5 are attainable (e.g. a=ba=b gives 00), but a positive value like 11 is impossible.

It cannot equal 1\text{It cannot equal } 1
CAT 2020 Slot 2 QA Q16: The value of⁡ log a ⁡ a b + log b ⁡ b a , for 1 < a &le; b cannot be equal to — Solution | TheCATExam