Question

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

Answer 1

$2^x=4^y=8^z\Rightarrow 2^x=2^{2y}=2^{32}$

$\Rightarrow x=2y=3z$

As $xyz=288\Rightarrow x\frac{x}{2}\times \frac{x}{3}=288$

$\Rightarrow x=12;y=6$ & $z=4$

$\therefore x+y+z=22$