2018/003) For the set of equations $z^x=y^{2x}, 2^z= 2.4^x,x+y+z = 16$ find integral values of x,y,z

we are given
$z^x=y^{2x}\cdots(1)$
$2^z= 2.4^x\cdots(2)$
$x+y+z = 16\cdots(3)$

$z^x=y^{2x}$ is true when $z=y^2$ or $x = 0$
we deal both cases
case 1
$z=y^2\cdots(4)$
the equation (2) gives
$2^z = 2^{2x+1}$
or $z=2x+1$
or $2x = z-1=y^2-1\cdots(5)$
from (3), (2), (5) we get
$y^2-1 +2y + 2y^2 = 32$
or $3y^2 +2y-33=0$ or $(y-3)(3y+11)=0$ so y =3 as y is integer
hence z=9 and x= 4
so solution $x=4,y=3,z=9$
case 2
x = 0
so from (2) $2^z =2=>z=1$
hence y = 15
so solution $x=0,y=15,z=1$