Solving $X x^t+ Y y^t=1$ for a specific case with constraints Is there an analytical solution to the equation below?
$$\begin{align*}
A\frac{\alpha}{k}e^{(k-\alpha)t}+B\frac{\beta}{k}e^{(k-\beta)t}=1
\end{align*}$$
where $\alpha$ and $\beta $ are roots of
$$\begin{align*}
z^2 - (a+b+c)z + ac=0
\end{align*}$$
I tried, but in vain, to use AM-GM inequality and obtain $something+something \ge 1$. By then the two things have to be equal if those two things look like the original expressions. But not very successful.
I have been told that this somewhat looks like
$u^t+v^t=w^t$. Fermat is not happy with this equation.
Any thoughts are appreciated.
Edit: $A$, $B$, $\alpha$, $\beta$, $k$, $a$, $b$, $c$ are all positive real numbers.
 A: If we ignore the title then there is no actual constraint on $\alpha$ and $\beta.$ For arbitrary $\alpha$ and $\beta$ you can set $a=\alpha,$ $b=0$ and $c=\beta$ and the quadratic equation is satisfied.
The equation can be re-written as
$$A\alpha\exp(-\alpha t)+B\beta\exp(-\beta t)=k\exp(-kt).$$
The left hand side is a linear combination of exponentials. In the special case where $\alpha, \beta$ and $k$ are positive we are looking at the intersections between, on the one hand, a linear combination of density functions of two exponential distributions, and on the other hand, the density function of a third exponential distribution. If $A+B=1$ (this is not given) then the LHS is a classical example of a hyperexponential distribution. Telecom engineers are not happy with such a distribution.
This already illustrates the kind of difficulty we are up against: depending on clever choices of the parameters there may be zero, one or two solutions - and not in the obvious way of quadratic equations and discriminants.
A: As already said in comments and answers, finding the zero of the function$$f(t)=A\,\alpha\, e^{-\alpha t}+B\,\beta \,e^{-\beta t}-k\, e^{-kt}$$ (as Justpassingby wrote it) will require, for the most general case, numerical methods (such as Newton). 
Assuming that $A$, $B$, $\alpha$, $\beta$, $k$ are all positive real numbers, it will probably be a good idea to search for the zero of $$g(x)=\log\Big(A\,\alpha\, e^{-\alpha t}+B\,\beta \,e^{-\beta t}\Big)-\log\Big(k\, e^{-kt} \Big)$$ which should look almost as a straight line.
Starting iterations at $t_0=0$ would provide as a first iterate $$t_1=(\alpha  A+\beta  B)\frac{ \log (k)-\log (\alpha  A+\beta  B)}{\alpha  A (k-\alpha
   )+\beta  B (k-\beta )}$$
