Suppose not. That is, suppose $$\log_3 2$$ were rational. Then we would have $$\log_3 2=\frac{a}{b}$$ for some integers a and b, with b non-zero.
Since $$\log_3 2=\frac{a}{b}$$, we must have $$3^{\log_3 2}=3^{\frac{a}{b}}$$. Simplifying the lefthand side gives us $$2=3^\frac{a}{b}$$. So $$2^b= (3^\frac{a}{b})^b = 3^a$$. But 2 and 3 have no positive factors other than 1, so we can have $$2^b=3^a$$ only when $$a=b=0$$. This contradicts our assumption that b is non-zero.
Since our assumption that $$\log_3 2$$ was rational led to a contradiction, it must be the case that $$\log_3 2$$ is irrational. QED.