If x, y, z in R^+ are such that z y x 1 | vedclass

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(C) Let $u = \log_{x}y$. Then $\log_{y}x = \frac{1}{u}$.Given $\frac{1}{u} + u = \frac{5}{2}$,which implies $2u^2 - 5u + 2 = 0$.Solving for $u$,we get

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$6$

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(C) Let $u = \log_{x}y$. Then $\log_{y}x = \frac{1}{u}$.Given $\frac{1}{u} + u = \frac{5}{2}$,which implies $2u^2 - 5u + 2 = 0$.Solving for $u$,we get $(2u - 1)(u - 2) = 0$,so $u = 2$ or $u = \frac{1}{2}$.Since $y > x > 1$,$\log_{x}y > 1$,so $u = 2$,which means $y = x^2$.Let $v = \log_{y}z$. Then $\log_{z}y = \frac{1}{v}$.Given $\frac{1}{v} + v = \frac{10}{3}$,which implies $3v^2 - 10v + 3 = 0$.Solving for $v$,we get $(3v - 1)(v - 3) = 0$,so $v = 3$ or $v = \frac{1}{3}$.Since $z > y > 1$,$\log_{y}z > 1$,so $v = 3$,which means $z = y^3$.Substituting $y = x^2$ into $z = y^3$,we get $z = (x^2)^3 = x^6$.Therefore,$\log_{x}z = \log_{x}(x^6) = 6$.

If $x, y, z \in R^+$ are such that $z > y > x > 1$,$\log_{y}x + \log_{x}y = \frac{5}{2}$ and $\log_{z}y + \log_{y}z = \frac{10}{3}$,then $\log_{x}z$ is equal to

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