CAT 2017 Slot 2 — QA Question 26

Arithmetic mean of log(2² × 3³ × 5), (log 2⁶ × 3 × 5⁷) and (log 2 × 3² × 5⁴)

= $\frac{1}{3}$[log(2² × 3³ × 5) + log(2⁶ × 3 × 5⁷) + log(log 2 × 3² × 5⁴)]

= â€‹â€‹â€‹â€‹â€‹â€‹â€‹$\frac{1}{3}$[log(2² × 3³ × 5 × 2⁶ × 3 × 5⁷ × 2 × 3² × 5⁴)]

= $\frac{1}{3}$[log(2^{2 + 6 + 1}) × (3^{3 + 1 + 2}) × (5^{1 + 7 + 4})]

= ${\log }_{}\left[2^{9\times \frac{1}{3}}\times 3^{6\times \frac{1}{3}}\times 5^{12\times \frac{1}{3}}\right]$

= log [2³ × 3² × 5⁴]

Now log(2^a × 3^b × 5^c) = log(2³ × 3² × 5⁴)

(Note: In the given question it should have been specified that a, b, c are integers, as there can be multiple answers for the same. Since it is a TITA question, we have to assume the same and equate power of 2, 3 and 5)

∴ Equating powers of 2, 3 and 5 we get

a = 3, b = 2 and c = 4

Hence, 3.