\[\begin{cases} z^x = y^{2x} \\ 2^z = 2(4^x) \\ x+y+z = 16 \end{cases} \]
Find the integral values of \(x\), \(y\), and \(z\) satisfying the system of equations above. Submit \(xyz\).
+128473 z^x = y^(2x) take the log of both sides
log z^x = log y^(2x)
x log z = x log y^2
log z = log y^2 implies that z = y^2 ⇒ sqrt (z) = y (1)
And
2^z = 2 (4^x) divide both sides by 2
2^(z - 1) = 4^x
2^(z -1) = (2^2)^x
2^(z -1) = 2^(2x) implies that
z -1 = 2x
(z -1) / 2 = x (2)
x + y + z =16 (3)
Sub (1) and (2) into (3)
(z -1)/2 + sqrt (z) + z = 16
z - 1 + 2sqrt (z) + 2z = 32
3z + 2sqrt (z) = 33
Note that this is true when z = 9
(9 -1) / 2 = x = 8/2 = 4
sqrt (9) = y = 3
So
xyz = 4(3)(9) = 108
