The value of the logarithmic expression can never be $\ldots$ The value of the logarithmic expression  $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x},\text{where}\quad x\geq y>1\quad$  can never be 
$\bf\text{options}$ a.) $-1\quad$   b.)$\quad0.5\quad$ c.) $\quad0\quad$ d.) $\quad1$
$$\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\\
&=\frac{2\log_x y}{\log_x y}-\frac{(\log_x y)^2}{\log_x y} -\frac{1}{\log_x y}\\
&=-\frac{ (\log _{x}y-1)^{2} }{ \log _{x}y}\end{align}$$
here if i input $x=y=2$ then option $0$ is removed , not sure how to deduce the rest.
Wolfram does gives the answer as option $1$

\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\end{align} 
Let $y^p = x$, therefore $\log_y x =p$ and $ \log_x y = \frac1p $ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$ So, $1-\log_x y +1-\log_y x = 2 - (p + \frac1p)$
For the RHS of above expression to equate to -1, 0.5, 0, 1, we must have p + $\frac1p$ = 3, 1.5, 2, 1 respectively.
So, we have to solve  for $p$ the corresponding quadratic equations which work out to be $\quad$ $p^2-3p+1=0$, $\quad$ $2p^2-3p+2=0$, $\quad$ $p^2-2p+1=0$, $\quad$ $p^2-p+1=0$. $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ On solution we obtain $p=(3+\sqrt{5})/2 $ & $p=(3-\sqrt{5})/2$ for the first, $\quad$$\quad$ $p=1$ & $p=\frac12$ for the second, $\quad$$\quad$ $p=1$ for the third, $\quad$$\quad$ and $p=(1+\sqrt3i)/2$ & $p=(1-\sqrt3i)/2$ for the fourth.
Since $y^p = x$, imaginary $y$ would be out of domain of the problem and also make $x$ outside the domain of the problem (domain is x,y > 1, i.e. x, y real) and hence solution $p=1$ which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=1$ is not possible. So, answer is option d.) $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ Also notice that for second equation, $p=1$ is a false root, and $p=\frac12$ would make $y \gt x$ which is also outside the domain of the problem. So, this equation which corresponds to $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}=\frac12$ is also not possible
