All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Multi-Logarithms (Posted on 2004-12-08) Difficulty: 2 of 5
If log9(x) = log12(y) = log16(x+y), then find y/x.

See The Solution Submitted by SilverKnight    
Rating: 3.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution | Comment 8 of 10 |

Let, log_9(x) = log_12(y) = log_16(x+y) = m(say)

Then, x = 9^m; y = 12 ^m; x+y = 16^m
So, x(x+y) = 144^m = 12^(2m) = y^2
Or, y^2 - xy - x^2 = 0
Or, y/x = (1 +/- sqrt(5))/2 

However, if y/x = (1- sqrt(5))/2 = -0.618. 
If x is positive, then y must be negative, so that log_12(y) is undefined. This is a contradiction.
If y is positive, then x must be negative so that log_9(x) is undefined. This is a contradiction.

Consequently, y/x = (1+ sqrt(5))/2 is the only possible solution to the given problem. 

Edited on February 28, 2008, 5:40 am
  Posted by K Sengupta on 2007-03-19 12:01:53

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information