Given that $\frac{1}{2^{a-1}} = 2^3 + 2^2 + 2^3 + 2^3$, find the value of a.

1. Simplify the numerical expression on the right-hand side.

$2^3+2^2+2^3+2^3=8+4+8+8=28$
So,
$\frac{1}{2^{a-1}}=28.$

2. Use index laws to rewrite the left-hand side and solve for $a$.

$\frac{1}{2^{a-1}}=2^{-(a-1)}=2^{1-a}$
Hence,
$2^{1-a}=28.$
Taking logarithm base $2$,
$1-a=\log_2 28$
$a=1-\log_2 28$
$a\approx -3.807\ldots$