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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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Solution Solution | Comment 7 of 9 |


Let        y = bx = logbx,  so that we have two simultaneous equations:

                                    y = bx   and   x = by

Since these are mirror images of each other in the line y = x, all solutions

must lie on the line y = x. So for there to be only one solution, the graph

of y = bx must intersect the line y = x at only one point.

*When 0 < b < 1 this always happens, since y = bx is strictly decreasing.

When b = 1 there is also only one intersection (y = 1 being a straight line).

When b > 1 the graph of y = bx is strictly increasing and can cross the line

y = x at two points or touch it at one point. For a single point of intersection,

we need the gradient to be 1 when x = y.

y = bx  =>  dy/dx = ln(b)*bx, so at the point of tangency:

            ln(b)*bx = 1       and       x = bx

which give:        x = 1/ln(b) and therefore  1/ln(b) = b^(1/ln(b)) which

simplifies to b = e^(1/e)  = 1.444667861..... and x = e.

In summary: There will be one real solution when 0 < b <= 1  or b = e1/e.


Edited on December 13, 2012, 2:52 pm

Edited on December 13, 2012, 10:58 pm
  Posted by Harry on 2012-12-13 14:49:12

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