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CAT 2024 Slot 3 — QA Question 19

IndicesEasy

If 3a = 4, 4b = 5, 5c = 6, 6d = 7, 7e = 8 and 8f = 9, then the value of the product abcdef is

Answer & solution

Correct answer: 2

Answer: 2

Solution

Medium

Each exponent is a log: a=log⁡34a=\log_3 4, b=log⁡45b=\log_4 5, and so on. Their product telescopes via the change-of-base rule, leaving a single log that evaluates cleanly.

1

Express each variable as a log. From 3a=43^a=4, a=log⁡34=log⁡4log⁡3a=\log_3 4=\dfrac{\log 4}{\log 3}, and likewise for the rest.

a=log⁡4log⁡3, b=log⁡5log⁡4, c=log⁡6log⁡5,⇒ d=log⁡7log⁡6, e=log⁡8log⁡7, f=log⁡9log⁡8\begin{aligned} &a=\frac{\log4}{\log3},\ b=\frac{\log5}{\log4},\ c=\frac{\log6}{\log5},\\ &\Rightarrow\ d=\frac{\log7}{\log6},\ e=\frac{\log8}{\log7},\ f=\frac{\log9}{\log8} \end{aligned}
2

Telescoping product. Multiply; every numerator cancels the next denominator.

abcdef=log⁡4log⁡3⋅log⁡5log⁡4⋯log⁡9log⁡8⇒ abcdef=log⁡9log⁡3(everything cancels)⇒ abcdef=log⁡39\begin{aligned} &abcdef=\frac{\log4}{\log3}\cdot\frac{\log5}{\log4}\cdots\frac{\log9}{\log8}\\ &\Rightarrow\ abcdef=\frac{\log9}{\log3}\quad\text{(everything cancels)}\\ &\Rightarrow\ abcdef=\log_3 9 \end{aligned}
3

Evaluate. Since 9=329=3^2.

log⁡39=2\begin{aligned} &\log_3 9=2 \end{aligned}
abcdef=2abcdef=2

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CAT 2024 Slot 3 QA Q19 — Indices solution | TheCATExam