I'm just having trouble linearizing a quick function I'm graphing. The function is of a shifted exponential decay from a capacitor : $V(t) = V_0e^{-t/\tau}+V_1$. I previously had the function without the $V_1$ addition shifting it which just simply needed a natural logarithm on both sides, but I'm not sure what to do from here.
For the graph, we have a list of known values for V and t, $V_0$ and $\tau$ are the unknowns we have to find. $V_1$ is assumed to be known, but we have to use the graph to guess as to it's value. I can't take the natural log of both sides because that gets me nowhere, there's no rule for (x + b) in a natural log, and I don't think I can just subtract $V_1$ from either side because I wouldn't have my known V values to graph anymore.
Any help to this would be greatly appreciated.
From $V(t)$ for $t=0$ we have $$ V(0)=V_0+V_1\quad\Longrightarrow\quad V_0=V(0)-V_1 $$
From the value of $V(t)$ for $t=1$ we have $$\frac{V(1)-V_1}{V_0}=\mathrm e^{-1/\tau}\quad\Longrightarrow\quad \tau=\frac{1}{\log\left(\frac{V_0}{V(1)-V_1}\right)}$$
$-V_0/\tau$ is the angular coefficient of the line tangent to the exponential function at $t=0$.
The equation of the linearized function is $$ V_l(t)=V_0-V_0\frac{t}{\tau}+V_1 $$ So the vertical axis intecerpt is for $t=0$, i.e. $V_l (0)=V_0+V_1$, and the orizontal axis intercept is when $V_l(t)=0$, i.e. for $t=\tau\frac{V_0+V_1}{V_0}$. Note that for $t=\tau $ we have $V_l(\tau)=V_1$.