Logarithmic aptitude

Question

Given that $y>1$ and $x\ge y$, then $\log_{x}(x/y)+\log_{y}(x/y)$ can never be:

1. $-1$
2. $-0.5$
3. $0$
4. $1$

Please provide the answer with the solution and brief explanation.

Answer 1

$$\log_x\frac xy+\log_y\frac xy=$$

$$\log_xx-\log_xy+\log_yx-\log_yy=$$

$$1-\frac1{\log_yx}+\log_yx-1=$$

$$\log_yx-\frac1{\log_yx}\tag{1}$$

Introduce $a=\log_yx$. The expression (1) becomes:

$$a-\frac1a=\frac{a^2-1}a\tag{2}$$

...with $a$ having the following propery.

$$y>1,\space x\ge y\implies a\ge1$$

Obviously expression (2) cannot be negative for $a\ge1$ so you should drop the first two options. For $a=1$ expression (2) is equal to zero. I will leave it up to you to prove that the fourth option is possible. Just show that equation $a-\frac1a=1$ has a solution greater than 1.