I feel the underlying root object is the exact solution:

$\displaystyle \text{myexp}=\text{Root}[\#{}^{\wedge}t-1-u \#+u\&,k]$

I know that's cryptic but that's 21'st century for ya'. Once I have that in Mathematica, I can access any of the solutions I want for any value of t, k, and u. For example, suppose the exponent is 10, the factor z/x=2.5, and I want all the solutions (roots) of the expression. I'd write:

Code:

myexp = Root[#^t - 1 - u # + u &, k]
Table[N[myexp /. {t -> 10, u -> 2.5, k -> kval}], {kval, 1, 10}]
{0.602522, 1., -1.0932 - 0.382235 I, -1.0932 +
0.382235 I, -0.608924 - 0.969707 I, -0.608924 + 0.969707 I,
0.130353 - 1.10888 I, 0.130353 + 1.10888 I, 0.770506 - 0.736209 I,
0.770506 + 0.736209 I}

. . . well, I guess it's only nine roots since y not equal to one. Suppose I should have coded it as:

$\displaystyle \text{Solve}\left[u==\frac{y^{10}-1}{y-1},y\right]$

in which case Mathematica would have returned only nine root objects. You work out all the kinks ok.