Find all values of b such that the equation
bx = logbx
has exactly one real solution.
*PLEASE SEE THE LATER CORRECTION
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