1. Solve the differential equations

Given: $\frac{dx}{dt}+ax=0, \qquad \frac{dy}{dt}+by=0$ with initial conditions $x(0)=2, \qquad y(0)=1.$

These are standard first-order linear differential equations.

Their solutions are: $x(t)=2e^{-at}, \qquad y(t)=e^{-bt}.$

2. Use the condition $3y(1)=2x(1)$

At $t=1$, $x(1)=2e^{-a}, \qquad y(1)=e^{-b}.$

Given: $3y(1)=2x(1).$ Substitute: $3e^{-b}=2(2e^{-a})=4e^{-a}.$

So, $3e^{-b}=4e^{-a}$ $\Rightarrow e^{a-b}=\frac{4}{3}.$

3. Find $t$ such that $x(t)=y(t)$

We need: $2e^{-at}=e^{-bt}.$

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

From step 2, $e^{a-b}=\frac{4}{3}.$ Hence, $e^{(a-b)t}=\left(e^{a-b}\right)^t=\left(\frac{4}{3}\right)^t.$

Therefore, $\left(\frac{4}{3}\right)^t=2.$

So, $t=\log_{\frac{4}{3}}2.$

4. Check the options

The required value is: $\boxed{\log_{\frac{4}{3}}2}$

This matches Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So, they agree.