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If the sets of equations z^x=y^(2x),2^x=...

If the sets of equations `z^x=y^(2x),2^x=2.4^x,x+y+z=16,` the integral roots in the order `x,y,z`

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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),
...
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