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

Home > Just Math
Exp=log (Posted on 2012-12-12) Difficulty: 2 of 5
Find all values of b such that the equation

bx = logbx

has exactly one real solution.

See The Solution Submitted by Jer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution - correction Comment 9 of 9 |
Thanks Charlie, I can now see my error my assumption that all solutions must

lie on y = x is wrong. In fact, the curves y = bx and x = by can also intersect at

points off that line and, because of the symmetry about y = x, there would then

be an odd number of solutions, with one on the line and others in symmetric

pairs about y = x.

This is what is happening when 0 < b < e-e, as Broll discovered.

At (e-e, e-e) the curve y = bx crosses y = x with gradient -1 and for b < e-e

there will be three solutions. For example:

when b = 0.05,  x ~= 0.35022    (y ~= 0.35022)
                        x ~= 0.13736    (y ~= 0.66266)
                        x ~= 0.66266    (y ~= 0.13736)

So:       when     0 < b < e-e         there are 3 solutions
            when     e-e <= b <=1     there is 1 solution
            when     1 < b < e1/e        there are 2 solutions
            when     b = e1/e              there is 1 solution
            when     e1/e < b              there are no solutions



  Posted by Harry on 2012-12-13 22:54:53
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 (2)
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