Missing $x$

$\large{64^{4x}=512^{3x+8}}$ $\large{2^{6(4x)}=2^{9(3x+8)}}$ $\large{6(4x)=9(3x+8)}$ $\large{24x=27x+72}$ $\large{-72=3x}$ $\boxed{\color{#D61F06}\large{x=-24}}$

$\large \begin{aligned} 64^{4x} & = 512^{3x+8} \\ \left(2^6\right)^{4x} & = \left(2^9\right)^{3x+8} \\ 2^{6(4x)} & = 2^{9(3x+8)} \\ \implies 6(4x) & = 9(3x+8) \\ 2(4x) & = 3(3x+8) \\ 8x & = 9x + 24 \\ \implies x & = \boxed{-24}\end{aligned}$

$\begin{aligned} 64^{4x} & = 512^{3x+8} \\(8^2)^{4x} & = (8^3)^{3x+8} \\8^{8x} & = 8^{9x+24} ~~~~~~~~~~~~~~~~ [(a^n)^m = a^{nm}] \\8x & = 9x + 24 \\8x-9x & = 24 \\-x & = 24 \\ \implies x & = \dfrac{24}{-1} = \boxed{-24}.\\ \end{aligned}$

Relevant wiki: Rules of Exponents

First, we can factor out 512 to get: $64^{4x}=64^{3x+8}\times8^{3x+8}$ If we divide both sides by $64^{3x+8}$ , we get: $64^{x-8}=8^{3x+8}$ If we realize that $64=8^2$ , then we can distribute the 2 to the $x-8$ : $8^{2x-16}=8^{3x+8}$ Using logic, we can get rid of the 8 and solve for the equation: $2x-16=3x+8$ When we do that, we get the answer of $\boxed{-24}$ .