1.

In the sets of equations `z^x =y^(2x)`, `2^z = 2. 4^x`, x+y+z = 16 , the integral roots in the order x, y, z are :

Answer» Here, `z^x = y^(2x)`
`=>z^x = (y^2)^x`
`=>z = y^2->(1)`
`2^z = 2.4^x`
`=>2^z = 2.(2^2)^x=2^(2x+1)`
`=>z = 2x+1`
`=>x = (z-1)/2->(2)`
From (1),
`x = (y^2-1)/2`
`x+y+z = 16`
`=>(y^2-1)/2+y+y^2 = 16`
`=>y^2-1+2y+2y^2 = 32`
`=>3y^2+2y-33=0`
`=>3y^2+11y-9y-33 = 0`
`=>(3y+11)(y-3) = 0`
`y = 3 and y = -11/3`
But, we have to find only integer root.
So, `y = 3`
`x = (9-1)/2 = 4`
`z = 3^2 = 9`


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