If a, b, c are distinct positive numbers, each different from 1, such that [logb a logc a - loga a] +  [loga b logc b - logb b] +  [loga c logb c - logc c] = 0, then abc =

1) 1

2) 2

3) 3

4) 4

 

Solution:

We use the property logn m = log m/log n

Given [logb a logc a – loga a] +  [loga b logc b – logb b] +  [loga c logb c – logc c] = 0

=> [(log a/log b)(log a/log c) – 1] +  [(log b/log a)(log b/log c) – 1] +  [(log c/log a)(log c/log b) – 1] = 0

=> (log a)2/log b log c +  (log b)2/log a log c +  (log c)2/log a log b = 3

Multiply both sides by log a log b log c

=> (log a)3 + (log b)3 + (log c)3 = 3 log a log b log c

=> log a + log b + log c = 0 (Use A3 + B3 + C3 = 3ABC, then A+B+C = 0)

=> log (abc) = 0

=> abc = 1

Hence option (1) is the answer.

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

Ask
Question