Solve this simple equation without trial and error.

$\large { 2 }^{ x }-4\quad =\quad 4x+8\\ \therefore2^x=4(x+3)=2^2*(x+3).\\$
Since RHS is a power of 2, LHS also must be a power of 2.
So (x+3) also must be power of two.
However we already have power 2 consumed in $2^2$ . So x must be >2.
Under these conditions, minimum x=5 makes (x+3)=8= $2^3$ .
So $for\ \boxed{x=5}\ 2^2*(x+3)=2^2*8=2^5$ satisfies our equation.