Calculate Rate constant, k, for a consecutive reaction where k1<<k2
In a consecutive reaction A-B-C, A decays as $A=A_0\exp(-k_1t)$ with $A_0$ the initial amount (initial B and C are assumed to be zero) and B rate expression $\displaystyle \frac{dB}{dt}=k_1A-k_2B$  which when substituting for $A$ and integrating gives $\displaystyle B=A_0\frac{k_1}{k_2-k_1}\left(\exp(-k_1t) -\exp(-k_2t  \right)$. C is found as $A_0-A-B$

then $\displaystyle C= A_0\left(1+\frac{k_2\exp(-k_1t)-k_1\exp(-k_2t)}{k_1-k_2}  \right) $

In your case $k_1&lt;&lt;k_2$ then $\displaystyle C= A_0\left( 1 - \exp(-k_1t)  \right) $ which is almost your equation.

Taking logs gives $\displaystyle\ln\left(\frac{C}{A_0}\right)=\ln\left(1-\exp(-k_1t)\right)$ and this is what you should use.

[To go further you will have to expand the series for log and exponential but this will lead to a polynomial and will only be an approximation anyway.  The first few terms are $\log(k_1t)  - k_1t/2 + (k_1t)^2/24\cdots$; very messy.]