I am assuming by "time constant of the capacitor" you mean the time constant of an RC circuit $\tau=RC$. Let's assume we are looking at a discharging capacitor. Then the voltage across the capacitor as a function of time is given by
where $V_0$ is the initial potential across the capacitor.
Now, let's compare this to what you have discussed in your answer:
So, as you have said, the gradient, or slope, of $\ln(V(t))$ gives us the desired $\tau$ we want. Slopes are very easy to calculate from linear plots. Even if your data is not perfectly linear, linear models are probably the easiest models to fit data to. You could even get a good estimate of a line of best fit by just drawing one by hand and then determining the slope from there. Pretty simple.
But what about the original expression for $V(t)$? Well this is an exponential function. You would most likely need some sort of program to fit your data to the exponential function. And this is not as simple as a linear function. You could draw an exponential curve you think fits the data, but it would be harder than a line (and the curve you draw might not even be a true exponential function with base $e$). Even then, it's harder to pull the time constant from the exponential decay than it is to find the slope of a line.
Therefore, in this context the linear function and its slope are easier to work with. Lines are simple to visualize and work with. Although with today's technology, either method should be fine if you have the software to do it. In the labs I would TA for, we would fit directly to the exponential functions with very few issues.