The only horizontal force acting on the boat is the drag force, which opposes the motion. Applying Newton's second law:

$$\sum F_x = -F_d = ma$$

Substituting the given drag force and $a = dv/dt$:

$$-bv = m \frac{dv}{dt}$$

This is a separable differential equation. We rearrange the terms to integrate:

$$\frac{dv}{v} = -\frac{b}{m} dt$$

Integrate from the initial state ($t=0$, $v=v_i$) to the final state ($t$, $v_f = v_i/2$):

$$\int_{v_i}^{v_i/2} \frac{dv}{v} = \int_{0}^{t} -\frac{b}{m} dt$$ $$[\ln v]_{v_i}^{v_i/2} = -\frac{b}{m} [t]_{0}^{t}$$ $$\ln\left(\frac{v_i}{2}\right) - \ln(v_i) = -\frac{b}{m} t$$

Using the logarithm property $\ln(A/B) = \ln(A) - \ln(B)$:

$$\ln\left(\frac{v_i/2}{v_i}\right) = \ln\left(\frac{1}{2}\right) = -\ln(2) = -\frac{b}{m} t$$

Solving for the time $t$:

$$t = \frac{m}{b}\ln(2)$$