It's just simple log, isn't it?

$\begin{aligned} \log_2(x+4) & = \log_{4x+16}8 \\ & = \frac {\log_2 8}{\log_2 (4(x+4))} \\ & = \frac 3{\log_2 4 + \log_2 (x+4)} \\ \implies \log_2(x+4) & = \frac 3{2 + \log_2 (x+4)} & \small \color{#3D99F6} \text{Let }y = \log_2 (x+4) \\ y & = \frac 3{2+y} \\ y^2 + 2y - 3 & = 0 \\ (y+3)(y-1) & = 0 \\ \implies y & = \log_2 (x+4) = \begin{cases} -3 \\ 1 \end{cases} \end{aligned}$

$\implies \log_2(x+4) = \begin{cases} -3 & \implies x + 4 = \dfrac 18 & \implies x = -\dfrac {31}8 \\ 1 & \implies x + 4 = 2 & \implies x = - 2 \end{cases}$

$\implies ab = \left(-\dfrac {31}8 \right) (-2) = \boxed{\dfrac {31}4}$

$\log_{4x+16}8$ can be written in the form of $\frac{\log_{2}8}{\log_{2}(4x+16)}=\frac{3}{\log_{2}4(x+4)}=\frac{3}{\log_{2}4+\log_{2}(x+4)}$ .

Now, the equation becomes:

$\log_{2}(x+4)=\frac{3}{2+\log_{2}(x+4)}$

$\log_{2}^{2}(x+4)+2\log_{2}(x+4)-3=0$ .

Setting $t=\log_{2}(x+4)$ , we get simple quadratic equation: $t^{2}+2t-3=0$

Solving for $t$ , we get:

$t_{1}=-3 , t_{2}=1$ .

Now, going back to solve for $x$ :

$\log_{2}(a+4)=-3$

$\log_{2}(b+4)=1$ , where $a$ and $b$ are two different real solution of the equation.

Solving for $a$ :

$\log_{2}(a+4)=\log_{2}2^{-3}$

$a+4=\frac{1}{8}$

$a=-\frac{31}{8}$

and then for $b$ :

$\log_{2}(b+4)=\log_{2}2$

$b+4=2$

$b=-2$ .

Hence, $a\times b=-\frac{31}{8}\times (-2)=\boxed{\frac{31}{4}}.$