The number of solutions of the equation $$\log_4(x - 1) = \log_2(x - 3)$$ is ______.

We need to solve $$\log_4(x - 1) = \log_2(x - 3)$$.

The domain requires $$x - 1 > 0$$ and $$x - 3 > 0$$, so $$x > 3$$.

Converting to the same base using $$\log_4(x-1) = \frac{\log_2(x-1)}{2}$$, the equation becomes:

$$\frac{\log_2(x-1)}{2} = \log_2(x-3)$$

$$\log_2(x-1) = 2\log_2(x-3) = \log_2(x-3)^2$$

Therefore $$x - 1 = (x-3)^2 = x^2 - 6x + 9$$.

Rearranging: $$x^2 - 7x + 10 = 0$$, which factors as $$(x-2)(x-5) = 0$$.

So $$x = 2$$ or $$x = 5$$. Since we need $$x > 3$$, only $$x = 5$$ is valid.

Verification: $$\log_4(5-1) = \log_4(4) = 1$$ and $$\log_2(5-3) = \log_2(2) = 1$$. Both sides are equal.

The number of solutions is $$1$$.