For a positive number $a$, the function $f(x)$ defined on $x \geq -1$ is
$$f(x) = \begin{cases} -x^2 + 6x & (-1 \leq x < 6) \\ a\log_4(x-5) & (x \geq 6) \end{cases}$$
For a real number $t \geq 0$, let $g(t)$ denote the maximum value of $f(x)$ on the closed interval $[t-1, t+1]$. If the minimum value of the function $g(t)$ on the interval $[0, \infty)$ is 5, find the minimum value of the positive number $a$. [4 points]