This is the equation: $$A\alpha\,e^{-\alpha t}+B\beta\,e^{-\beta t}=(A+B)\gamma\,e^{-\gamma t}$$ where $A$, $B$, $\alpha$, $\beta$ and $\gamma$ are all known constants.
I would like to solving for $t$, and I don't know how to take care of the L.H.S. after I take logarithm of both sides: $$ln(A\alpha\,e^{-\alpha t}+B\beta\,e^{-\beta t})=ln[(A+B)\gamma]-\gamma t$$
I am trying to do this before I turn to numerical approximation.
Thanks!
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EDIT:
The actual constants are $k_a$, $CL$, $V_1$, $Q$ and $V_2$, which are all individual positive real numbers (for simplicity, I will denote them, respectively, as $x_1$, $x_2$, $x_3$, $x_4$ and $x_5$), such that
$$\begin{cases} \beta=\frac{1}{2}[\frac{x_4}{x_3}+\frac{x_4}{x_5}+\frac{x_2}{x_3} - \sqrt{(\frac{x_4}{x_3}+\frac{x_4}{x_5}+\frac{x_2}{x_3})^2-4\frac{x_2x_4}{x_3x_5}}] \\ \alpha=\frac{1}{\beta}\frac{x_2x_4}{x_3x_5} \\ B=\frac{x_1(\frac{x_4}{x_5}-\beta)}{x_3(x_1-\beta)(\alpha-\beta)} \\ A=\frac{x_1(\frac{x_4}{x_5}-\alpha)}{x_3(x_1-\alpha)(\beta-\alpha)} \\ \gamma=x_1 \end{cases}$$
Therefore, $\alpha$, $\beta$ and $\gamma$ should be positive, while $A$ and $B$ can be positive or negative.
\begin{align} A\alpha\,\exp(-\alpha t) +B\beta\,\exp(-\beta t) &=(A+B)\gamma\,\exp(-\gamma t) \tag{1}\label{1} \end{align}
Under given conditions equation \eqref{1} can be simplified to
\begin{align} p\,\exp(-u\,t)+q\, \exp(-v\,t)-1&=0 \tag{2}\label{2} , \end{align}
where
\begin{align} u&=\alpha-\gamma ,\quad v=\beta-\gamma ,\\ p&=\frac{\alpha A}{\gamma(A+B)} ,\quad q=\frac{\beta B}{\gamma(A+B)} . \end{align}
And expression \eqref{2} is equivalent to
\begin{align} p\,x^a+q\,x-1&=0 \tag{3}\label{3} \\ \text{for }\quad x&=\exp(-v\,t) ,\\ a&=\frac uv . \end{align}
Unfortunately, equation of the form \eqref{3} does not have a closed-form solution, not even in terms of the Lambert W function, unless $a$ happens to be a small integer less than $5$. So, as @Claude Leibovici noted in the comments, the numerical root finding is the only option in this case.