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