x cannot be 1 since the denominator of the exponent
would be 0.
Take logs of both sides:
Use a rule of logs to write the exponent as a coefficient
of a log:
Substitute for k in
This will be true for every allowable value of x
Thus the equation
has a solution for every positive value of x except 1
That is, for all x ∈ {x|0 < x < 1 or x > 1}
(0,1) U (1, ∞)
Edwin
