Solution
$$2^x = 4^y = 8^z$$
$$2^x=2^{2y}=2^{3z}\ $$
=> $$x=2y=3z\ $$
Let $$x=2y=3z\ =k$$
xyz = 288
=> k*k/2 *k/3 = 288
=> k^3 = 1728 => k =12
x=12, y=6 and z = 4
x+y+z=22
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