In the most general case, you cannot inverse the sum of exponential functions and you need some numerical method (Newton being the simplest.
But, instead of trying to solve for $t$
$$C(t)=\sum_{i=1}^n a_i\, e^{-k_i t}$$ solving
$$\log[C(t)]=\log\Big[\sum_{i=1}^n a_i\, e^{-k_i t}\Big]$$ would be much better since the rhs is much closer to linearity.
So, let us say that we look for the zeo of function
$$f(t)=\log\Big[\sum_{i=1}^n a_i\, e^{-k_i t}\Big]-k$$ $k$ being given.
We have
$$f(0)=\log\Big[\sum_{i=1}^n a_i\Big]-k >0\qquad \text{and}\qquad f''(0) >0$$ So, by Darboux theorem, starting iterations at $t_0=0$ ensures that we shall never face an overshoot of the solution.
But we can try to make an estimate trying to write
$$\sum_{i=1}^n a_i\, e^{-k_i t}\sim A\,e^{-K t}=C(0)\,e^{-K t}$$ and define
$$\frac {\sum_{i=1}^n a_i }K=\sum_{i=1}^n \frac {a_i}{k_i}\implies K=\frac{\sum_{i=1}^n a_i } {\sum_{i=1}^n \frac {a_i}{k_i} }$$ to match the areas under the curves from $t=0$ to $\infty$.
Trying withe random numbers
$$a_1=29.8488\qquad a_2=50.0334\qquad a_3=21.9958$$
$$k_1=0.109865\qquad k_2=0.0760897\qquad k_3=0.0650516$$ gives $A=101.878 $ and $K=0.0803852$.
By curiosity, plot the two curves (they almost overlap).
Suppose that the target is $C(t)=45.678$; this gives as an estimate $t=9.97895$.
Now, Newton iterates
$$\left(
\begin{array}{cc}
m & t_m \\
0 & 9.97895 \\
1 & 9.76084 \\
2 & 9.76091
\end{array}
\right)$$
Edit (for the biexponential decay function)
Since it is better to work on a logarithmic scale, consider that we want to approximate
$$\log\Big[ a_1\, e^{-k_1 t}+ a_2\, e^{-k_2 t}\Big]\qquad \text{by} \qquad \log\big[ a_1+ a_2\big]- K t$$ and consider two points $t_i=\frac{\log(2)}{k_i}$ corresponding to the half-times.
Minimzing the sum of the squared errors leads to
$$(t_1^2+t_2^2) K=(t_1+t_2) \log(a_1+a_2)-$$ $$\Big[t_1 \log \left(a_1 e^{-k_1 t_1}+a_2 e^{-k_2 t_1}\right)+t_2
\log \left(a_1 e^{-k_1 t_2}+a_2 e^{-k_2 t_2}\right)\Big]$$
Trying with $a_1=30$, $k_1=0.11$, $a_2=50$, $k_2=0.07$, this would give $K\sim 0.0834$. Trying for a few values of $C(t)$
$$\left(
\begin{array}{ccc}
C(t) & \text{estimate} & \text{solution} \\
70 & 1.60118 & 1.57641 \\
60 & 3.44960 & 3.40984 \\
50 & 5.63582 & 5.59716 \\
40 & 8.31154 & 8.30198 \\
30 & 11.7611 & 11.8338 \\
20 & 16.6231 & 16.8954 \\
10 & 24.9346 & 25.7633
\end{array}
\right)$$
Using this estimate, Newton method would converge very fast. For example, using the last (and worst) point, we would have
$$\left(
\begin{array}{cc}
n & t_n \\
0 & 24.9346 \\
1 & 25.7623 \\
2 & 25.7633
\end{array}
\right)$$
Best Answer
The first derivative equation is
$$k_1(x+1)e^x-k_2e^{k_2x}-k_3=0$$
and the second one,
$$k_1(x+2)e^x-k_2^2e^{k_2x}=0.$$
The solution(s) of the latter can be formulated in terms of the Lambert function $W$. https://en.wikipedia.org/wiki/Lambert_W_function#Examples
The roots of the second derivative correspond to the extrema of the first derivative, which allow to to detect all changes of sign. Then you refine the roots with a bracketing method such as regula falsi.
These give you the extrema of the initial function, the changes of sign and then the roots.
This is an indirect approach not amenable to analytical study, by it guarantees the correct enumeration of the solutions.