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CAT 2024 Slot 2 — QA Question 3

LogarithmsEasy

If a, b and c are postive real numbers such that a > 10 ≥ b ≥ c and log8(a+b)log2c + log27(a-b)log3c = 23, then the greatest possible integer value of a is

Answer & solution

Answer: 14

Solution

Medium

Convert every logarithm to a common base using log⁡23=13log⁡2\log_{2^3}=\tfrac13\log_2 and log⁡33=13log⁡3\log_{3^3}=\tfrac13\log_3, then apply the change-of-base ratio to collapse the whole left side into one log⁡c\log_c. The constraint becomes a2−b2=c2a^2-b^2=c^2; maximise integer aa.

1

Simplify each term. Using log⁡8N=log⁡2N3\log_8 N=\tfrac{\log_2 N}{3}, the first term is log⁡8(a+b)log⁡2c=13log⁡2(a+b)log⁡2c=13log⁡c(a+b)\dfrac{\log_8(a+b)}{\log_2 c}=\dfrac{\tfrac13\log_2(a+b)}{\log_2 c}=\tfrac13\log_c(a+b). Likewise the second is 13log⁡c(a−b)\tfrac13\log_c(a-b).

13log⁡c(a+b)+13log⁡c(a−b)=23⇒ log⁡c[(a+b)(a−b)]=2(multiply by 3, combine logs)⇒ a2−b2=c2\begin{aligned} &\tfrac13\log_c(a+b)+\tfrac13\log_c(a-b)=\tfrac23\\ &\Rightarrow\ \log_c\big[(a+b)(a-b)\big]=2 \quad\text{(multiply by 3, combine logs)}\\ &\Rightarrow\ a^2-b^2=c^2 \end{aligned}
2

Maximise aa. So a2=b2+c2a^2=b^2+c^2 with the constraints a>10≥b≥c>0a>10\ge b\ge c>0. The largest b,cb,c can be is 1010, giving the bound:

a2=b2+c2≤102+102=200⇒ a≤200≈14.14(take square root)⇒ amax⁡=14\begin{aligned} &a^2=b^2+c^2\le 10^2+10^2=200\\ &\Rightarrow\ a\le\sqrt{200}\approx 14.14 \quad\text{(take square root)}\\ &\Rightarrow\ a_{\max}=14 \end{aligned}
3

Check a=14a=14 is attainable. Take b=10b=10 (allowed, ≤10\le 10). Then c2=a2−b2=196−100=96c^2=a^2-b^2=196-100=96, so c=96≈9.8c=\sqrt{96}\approx 9.8, which satisfies c≤b≤10c\le b\le 10 and c>0c>0. All conditions hold, so a=14a=14 is achievable.

a=14:c=196−100=96≈9.8≤10  ✓\begin{aligned} &a=14:\quad c=\sqrt{196-100}=\sqrt{96}\approx 9.8\le 10 \;\checkmark \end{aligned}
amax⁡=14a_{\max}=14
CAT 2024 Slot 2 QA Q3: If a, b and c are postive real numbers such that a > 10 ≥ b ≥ c and log 8 ( a + b ) log 2 c + log 27 ( a — Solution | TheCATExam