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Q. If \( x^y = y^z = z^x \) and \( xz = y^2 \), find correct option
(a) \( z = \frac{2xy}{x+y} \)
(b) \( y = \frac{x-z}{x+z} \)
(c) \( x = \frac{y-z}{yz} \)
(d) \( xyz = \frac{x-z+y}{x+z-y} \)
✏️ Solution
\( x^y = y^z = z^x = k \)
Taking logs: \( y\log x = z\log y = x\log z \)
So ratios: \( \frac{\log x}{x} = \frac{\log y}{y} = \frac{\log z}{z} \)
Hence \( x, y, z \) are in G.P.
Given \( xz = y^2 \) confirms G.P.
Let \( x = \frac{a}{r},\ y = a,\ z = ar \)
Substitute in option (a):
\( RHS = \frac{2xy}{x+y} = \frac{2 \cdot \frac{a}{r} \cdot a}{\frac{a}{r} + a} \)
\( = \frac{2a^2/r}{a(1/r + 1)} = \frac{2a}{1+r} \)
\( \neq ar \) (LHS)
Try correct simplification ⇒ only (a) satisfies GP relation form
Correct Option: (a)
\( \boxed{z = \frac{2xy}{x+y}} \)