Solve the equation $\displaystyle 2^{4-x} - 2^{3-x} = \frac{1}{8}$
$\displaystyle \left(\frac{2^4}{2^x}\right) - \left(\frac{2^3}{2^x}\right) = \frac{1}{8}$
$\displaystyle \frac{1}{2^x}\left(16 - 8 \right) = \frac{1}{8}$
$\displaystyle \frac{8}{2^x} = \frac{1}{8}$
$\displaystyle 64 = 2^x$
$\displaystyle 2^6 = 2 ^x$
$\displaystyle x = 6$
am i right?