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 Exp=log (Posted on 2012-12-12)
Find all values of b such that the equation

bx = logbx

has exactly one real solution.

 See The Solution Submitted by Jer No Rating

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 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

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

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