4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)
Solve for X
$\displaystyle 4^{1996} = 4^3 (4^{1993})$
$\displaystyle 4^{1995} = 4^2 (4^{1993})$
$\displaystyle 4^{1994} = 4 (4^{1993})$
Set $\displaystyle y = (4^{1993})$
Then we have:
$\displaystyle 64y - 16y - 4y + y = 45(2^x)$
$\displaystyle 45y = 45(2^x)$
$\displaystyle y = 2^x$
$\displaystyle 4^{1993} = 2^x$
$\displaystyle 2^{2 \times 1993} = 2^x$
$\displaystyle x = 3986 $