3x3 non-linear system of equations (Known and unkown isotope decay) TLDR version: is it possible to solve the following system for X, Y, Z?
$C_1=X+Y$
$C_2=C_3 X + Y Z^{C_4} $
$C_5=C_6 X + Y Z^{C_7} $
Background/Context: I have a mixture with a known isotope A (known decay constant $λ_Α$) and an unknown one B (unknown decay constant $λ_B$) in unknown proportions. When taking three activity measurements at different times $t_0$, $t_1$ and $t_2$ we get the following system of equations:
$Measurement(t_0)=A_0+B_0$
$Measurement(t_1)=A_0e^{-λ_At_1}+B_0e^{-λ_Bt_1} $
$Measurement(t_2)=A_0e^{-λ_At_2}+B_0e^{-λ_Bt_2} $
where $A_0$, $B_0$ and $λ_B$ are unknowns.
After bunching some known constants together, I reached the TLDR form of the system where X is $A_0$, Y is $B_0$ and Z is $e^{λ_B}$.
Thanks
 A: I don't know if you can solve it, but the first steps seems pretty easy, Use the first equation to remove $X$ from the others:
\begin{align}
C_1&=X+Y\\
C_2&=C_3 X + Y Z^{C_4} \\
C_5&=C_6 X + Y Z^{C_7} 
\end{align}
becomes
\begin{align}
C_2&=C_3 (C_1 - Y) + Y Z^{C_4} \\
C_5&=C_6 (C_1 - Y) + Y Z^{C_7} 
\end{align}
and now, to save ugliness, I'm going to change to mostly lowercase, and combine some constants into new ones. 
\begin{align}
u&=a - by + y Z^{C_4} \\
v&=c - dy + y Z^{C_7} 
\end{align}
Letting $z = Z^{C_4}$, and $s = C_7/C_4$
\begin{align}
u&=a - by + y z \\
v&=c - dy + y z^s 
\end{align}
\begin{align}
u-a&=y(z - b)\\
v-c&=y(z^s - d) 
\end{align}
Now taking a quotient we get
$$
\frac{u-a}{v-c} = \frac{z-b}{z^s - d}
$$
which I'll write
$$
K(z^s- 1) = z - b
$$
so 
$$
Kz^s - z + b - K = 0
$$
Sadly, for various powers of $s$, such an equation can have many roots; any one of these will produce a value for $y$, and hence for $x$. 
So...unless your $s$ value is nice, like $s = 1$, (or you have some bounded range for $z$, which might help), chances are good there's no unique solution. 
A: Make sure to choose $t_2=2t_1$.
Then the system writes
$$\begin{cases}A+B&=M_0,
\\Ar+Bs&=M_1,
\\Ar^2+Bs^2&=M_2,\end{cases}$$
where $r$ and $M_k$ are known.
Solving the first two,
$$A=\frac{M_0s-M_1}{s-r},\\B=\frac{M_1-M_0r}{s-r}$$ and plugging in the third,
$$(M_0s-M_1)r^2+(M_1-M_0r)s^2=M_2(s-r)$$ is quadratic in $s$.

Of course you can also work with other $\dfrac{t_2}{t_1}=\alpha$ ratios, but the equation gets a little harder, of the form
$$s^\alpha+ps+q=0.$$
A: Let's simplify a bit the notation and write
$$
\left\{ \matrix{
  C_{\,0}  = A + B \hfill \cr 
  C_{\,1}  = Ae^{ - \alpha \,t}  + Be^{ - \beta \,t}  \hfill \cr 
  C_{\,2}  = Ae^{ - \alpha \,u}  + Be^{ - \beta \,u}  \hfill \cr}  \right.
$$
After that, let's continue and semplify further, by putting
$$
\left\{ \matrix{
  1 = {A \over {C_{\,0} }} + {B \over {C_{\,0} }} = a + b \hfill \cr 
  {{C_{\,1} } \over {C_{\,0} }} = c = ae^{ - \alpha \,t}  + be^{ - \beta \,t}  \hfill \cr 
  {{C_{\,2} } \over {C_{\,0} }} = d = ae^{ - \alpha \,u}  + be^{ - \beta \,u}  \hfill \cr}  \right.
$$
and
$$
\left\{ \matrix{
  1 = a + b \hfill \cr 
  c = e^{ - \alpha \,t}  + b\left( {e^{ - \beta \,t}  - e^{ - \alpha \,t} } \right) = e^{ - \alpha \,t} \left( {1 + b\left( {e^{\left( {\alpha  - \beta } \right)\,t}  - 1} \right)} \right) \hfill \cr 
  d = e^{ - \alpha \,u}  + b\left( {e^{ - \beta \,u}  - e^{ - \alpha \,u} } \right)
 = e^{ - \alpha \,u} \left( {1 + b\left( {e^{\left( {\alpha  - \beta } \right)\,u}  - 1} \right)} \right) \hfill \cr}  \right.
$$
and also
$$
\left\{ \matrix{
  1 = a + b \hfill \cr 
  \gamma  = ce^{\alpha \,t}  \hfill \cr 
  \delta  = de^{\alpha \,u}  \hfill \cr 
  \gamma  - 1 = b\left( {e^{\left( {\alpha  - \beta } \right)\,t}  - 1} \right) \hfill \cr 
  \delta  - 1 = b\left( {e^{\left( {\alpha  - \beta } \right)\,u}  - 1} \right) \hfill \cr}  \right.
$$
and finally reach to
$$
\left\{ \matrix{
  1 = a + b \hfill \cr 
  \gamma  = ce^{\alpha \,t}  \hfill \cr 
  \delta  = de^{\alpha \,u}  \hfill \cr 
  \gamma  - 1 = b\left( {e^{\left( {\alpha  - \beta } \right)\,t}  - 1} \right) \hfill \cr 
  {{\delta  - 1} \over {\gamma  - 1}} = \left( {{{e^{\left( {\alpha  - \beta } \right)\,u}  - 1} \over {e^{\left( {\alpha  - \beta } \right)\,t}  - 1}}} \right) \hfill \cr}  \right.
$$
where $\gamma, \, \delta, u, t$ are known, so that:
 - the 5th equation will give $\alpha -\beta$;
 - the 4th will give $b$;
 - the remaining will permit to go back to the original quantities $A,B$. 
In the last equation , the function 
$$
{{e^{\,u\,x}  - 1} \over {e^{\,t\,x}  - 1}}
$$
is steadily increasing and convex , for $t < u$, and therefore it allows
to be easily solved numerically.
