Let $$x = x(t)$$ and $$y = y(t)$$ be solutions of the differential equations $$\frac{dx}{dt} + ax = 0$$ and $$\frac{dy}{dt} + by = 0$$ respectively, $$a, b \in \mathbb{R}$$. Given that $$x(0) = 2$$; $$y(0) = 1$$ and $$3y(1) = 2x(1)$$, the value of $$t$$, for which $$x(t) = y(t)$$, is :

$$\frac{dx}{dt} + ax = 0 \Rightarrow x(t) = x(0)e^{-at} = 2e^{-at}$$

$$\frac{dy}{dt} + by = 0 \Rightarrow y(t) = y(0)e^{-bt} = e^{-bt}$$

Given $$3y(1) = 2x(1)$$:

$$3e^{-b} = 2 \cdot 2e^{-a} = 4e^{-a}$$

$$3e^{-b} = 4e^{-a}$$

$$e^{a-b} = \frac{4}{3}$$, so $$a - b = \ln\frac{4}{3}$$.

We need $$x(t) = y(t)$$:

$$2e^{-at} = e^{-bt}$$

$$2 = e^{(a-b)t}$$

$$(a-b)t = \ln 2$$

$$t = \frac{\ln 2}{\ln(4/3)} = \log_{4/3} 2$$

The answer is $$\log_{4/3} 2$$, which corresponds to Option (4).