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Given that,

$\log_{y}{x} = a\cdot \log_{z}{y} = b\cdot \log_{x}{z} = ab \longrightarrow (1)$

Let’s take first two terms.

$\log_{y}{x} = a\cdot \log_{z}{y}$

$\Rightarrow \boxed{a= \frac{\log_{y}{x}}{\log_{z}{y}}}$

And now, let’s take first and third terms.

$\log_{y}{x} = b \cdot \log_{x}{z}$

$\Rightarrow \boxed{b= \frac{\log_{y}{x}}{\log_{x}{z}}}$

Now, $ab = \left[\frac{\log_{y}{x}}{\log_{z}{y}}\right] \times \left[ \frac{\log_{y}{x}}{\log_{x}{z}} \right]$

$\Rightarrow ab = \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{y}}{\log_{c}{z}}}\right] \times \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{z}}{\log_{c}{x}}}\right] \quad \left[\because\log_{b}{a} = \frac{\log_{c}{a}}{\log_{c}{b}}\right]$

$\Rightarrow ab = \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{z}}{\log_{c}{y}}\right] \times \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{x}}{\log_{c}{z}}\right]$

$\Rightarrow ab = \left(\frac{\log_{c}{x}}{\log_{c}{y}}\right)^{3}$

$\Rightarrow ab = (\log_{y}{x})^{3}$

$\Rightarrow ab = (ab)^{3}$

$\Rightarrow ab – a^{3} b^{3} = 0$

$\Rightarrow ab (1- a^{2} b^{2}) = 0$

$ \Rightarrow \boxed{ab = 0} or 1- a^{2} b^{2} = 0$

$\Rightarrow a^{2} b^{2} = 1$

$\Rightarrow (ab)^{2} = 1$

$\Rightarrow \boxed{ab = \pm1}$

Now, we can check all of the options.

$A. (-2,\frac{1}{2}) \Rightarrow -2 \times \frac{1}{2} = -1$ (Possible)

$B. (1,1) \Rightarrow 1 \times 1 = 1$ (Possible)

$C. (\pi,\frac{1}{\pi}) \Rightarrow \pi \times \frac{1}{\pi} = 1$ (Possible)

$D. (2,2) \Rightarrow 2 \times 2 = 4$ (Not Possible)

Correct Answer $: \text{D}$

$\log_{y}{x} = a\cdot \log_{z}{y} = b\cdot \log_{x}{z} = ab \longrightarrow (1)$

Let’s take first two terms.

$\log_{y}{x} = a\cdot \log_{z}{y}$

$\Rightarrow \boxed{a= \frac{\log_{y}{x}}{\log_{z}{y}}}$

And now, let’s take first and third terms.

$\log_{y}{x} = b \cdot \log_{x}{z}$

$\Rightarrow \boxed{b= \frac{\log_{y}{x}}{\log_{x}{z}}}$

Now, $ab = \left[\frac{\log_{y}{x}}{\log_{z}{y}}\right] \times \left[ \frac{\log_{y}{x}}{\log_{x}{z}} \right]$

$\Rightarrow ab = \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{y}}{\log_{c}{z}}}\right] \times \left[\dfrac{\frac{\log_{c}{x}}{\log_{c}{y}}}{\frac{\log_{c}{z}}{\log_{c}{x}}}\right] \quad \left[\because\log_{b}{a} = \frac{\log_{c}{a}}{\log_{c}{b}}\right]$

$\Rightarrow ab = \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{z}}{\log_{c}{y}}\right] \times \left[\frac{\log_{c}{x}}{\log_{c}{y}} \times \frac{\log_{c}{x}}{\log_{c}{z}}\right]$

$\Rightarrow ab = \left(\frac{\log_{c}{x}}{\log_{c}{y}}\right)^{3}$

$\Rightarrow ab = (\log_{y}{x})^{3}$

$\Rightarrow ab = (ab)^{3}$

$\Rightarrow ab – a^{3} b^{3} = 0$

$\Rightarrow ab (1- a^{2} b^{2}) = 0$

$ \Rightarrow \boxed{ab = 0} or 1- a^{2} b^{2} = 0$

$\Rightarrow a^{2} b^{2} = 1$

$\Rightarrow (ab)^{2} = 1$

$\Rightarrow \boxed{ab = \pm1}$

Now, we can check all of the options.

$A. (-2,\frac{1}{2}) \Rightarrow -2 \times \frac{1}{2} = -1$ (Possible)

$B. (1,1) \Rightarrow 1 \times 1 = 1$ (Possible)

$C. (\pi,\frac{1}{\pi}) \Rightarrow \pi \times \frac{1}{\pi} = 1$ (Possible)

$D. (2,2) \Rightarrow 2 \times 2 = 4$ (Not Possible)

Correct Answer $: \text{D}$