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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    
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re(2): possible solution--correction | Comment 4 of 9 |
(In reply to re: possible solution--correction by Charlie)

Try log(x) = -1; x=1/e, so x = 0.36787944117144232159552377016146087...

Then, e.g. (b)^0.36787944117144232159552377016146087= 

-0.99999999999999999999999999999999999/(log(b))

gives  b = 0.065988035845312537076790187596846423 plus or minus a tiny complex part of the order of i*10^-19, according to Mathematica.

I suspect the rest of it has mostly to deal with how the computer handles approximations. Mathematica gives 1 solution for b={(1/2),(1/4),(1/pi),(1/8)} and 3 for b={(1/16),(1/32),(1/64), etc}.* 

Algebraically, b^x=log(x)/log(b), for x=1/e is b^(1/e)=log((1/e))/log(b), so b=e^(-e)

and that is 0.06598803584531253707679018759..., as was to be expected.

Accordingly the answer to the original puzzle is e^(-e) < b < 1.  Not sure why this was D2.

* By setting x =(i*y) Mathematica can be forced to admit the existence of complex solutions in other cases and it is worth trying to see what happens when this is done.

 

Edited on December 13, 2012, 7:27 am
  Posted by broll on 2012-12-13 04:21:06

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