Question 77

If $$2^x = 4^y = 8^z$$ and xyz = 288, then, x + y + z = 

Solution
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$$2^x = 4^y = 8^z$$

$$2^x=2^{2y}=2^{3z}\ $$

=> $$x=2y=3z\ $$

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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