$ 2^{\frac{x-1}{3}} \cdot 2^{-\frac{2x+2}{2}} = 2\sqrt{2} $
$ 2^{\frac{x-1}{3}} \cdot 2^{-x-1} = 2^{\frac{3}{2}} $
stessa base
$ \frac {x-1}{3} -x-1 = \frac{3}{2}$
$ 2x-2-6x-6 = 9$
$ -4x = 17$
$ x = -\frac{17}{4} $
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