Understanding the Logarithmic Expression

The problem asks us to find a value that the expression \(k = \log_x\left(\frac{x}{y}\right) + \log_y\left(\frac{y}{x}\right)\) can never attain, given the condition that \(x \ge y > 1\). We need to analyze the properties of logarithms and the given constraints to determine the possible range of \(k\).

Simplifying the Expression for k

We can simplify the expression for \(k\) using the fundamental properties of logarithms:

• The quotient rule states that \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\).
• The logarithm of a base to itself is 1, meaning \(\log_b(b) = 1\).

Applying these rules to the terms in the expression for \(k\):

\(\log_x\left(\frac{x}{y}\right) = \log_x(x) - \log_x(y) = 1 - \log_x(y)\)

\(\log_y\left(\frac{y}{x}\right) = \log_y(y) - \log_y(x) = 1 - \log_y(x)\)

Now, substitute these simplified terms back into the equation for \(k\):

\(k = \left(1 - \log_x(y)\right) + \left(1 - \log_y(x)\right)\)

Combining the constants and the logarithmic terms:

\(k = 2 - \left(\log_x(y) + \log_y(x)\right)\)

Analyzing the Term \(\log_x(y) + \log_y(x)\)

To simplify further, let's make a substitution. Let \(a = \log_x(y)\).

We know that \(\log_y(x)\) is the reciprocal of \(\log_x(y)\), according to the change of base property (\(\log_b(c) = \frac{1}{\log_c(b)}\)). So, \(\log_y(x) = \frac{1}{\log_x(y)} = \frac{1}{a}\).

Substituting \(a\) into the expression for \(k\):

\(k = 2 - \left(a + \frac{1}{a}\right)\)

Now, we need to understand the possible values of \(a\) based on the given constraint \(x \ge y > 1\).

• Case 1: \(x = y\). If \(x\) equals \(y\), then \(a = \log_x(x) = 1\).
• Case 2: \(x > y > 1\). If \(x\) is strictly greater than \(y\) (and both are greater than 1), then \(\log_x(y)\) will be a positive number less than 1. That is, \(0 < a < 1\).

Consider the term \(a + \frac{1}{a}\). For any positive real number \(a\), the AM-GM inequality tells us that \(\frac{a + \frac{1}{a}}{2} \ge \sqrt{a \cdot \frac{1}{a}}\).

This simplifies to:

\(\frac{a + \frac{1}{a}}{2} \ge \sqrt{1}\)

\(\frac{a + \frac{1}{a}}{2} \ge 1\)

\(a + \frac{1}{a} \ge 2\)

Equality holds only when \(a = \frac{1}{a}\), which means \(a^2 = 1\). Since \(a\) must be positive in our case (\(a = \log_x(y)\) where \(x, y > 1\)), equality holds only when \(a = 1\).

Therefore, for any valid \(a\) derived from \(x \ge y > 1\), we have \(a + \frac{1}{a} \ge 2\).

Determining the Range of k

Let's use the inequality \(a + \frac{1}{a} \ge 2\) in our expression for \(k\):

\(k = 2 - \left(a + \frac{1}{a}\right)\)

Since \(a + \frac{1}{a}\) is always greater than or equal to 2, the term \(-\left(a + \frac{1}{a}\right)\) will always be less than or equal to -2.

So, \(k \le 2 - 2\)

\(k \le 0\)

This inequality shows that the value of \(k\) must always be less than or equal to zero. It can be zero (when \(x=y\), because then \(a=1\) and \(k=2-(1+1)=0\)) and it can be negative (when \(x>y\), because then \(a<1\) and \(a + 1/a > 2\), making \(k = 2 - (\text{number greater than 2}) < 0\)).

Identifying the Impossible Value

We need to find which of the given options \(k\) can never be equal to. Our analysis concluded that \(k \le 0\). Let's check the options:

• Option 1: -1. Since \(-1 \le 0\), this value is possible.
• Option 2: \(-\frac{1}{2}\). Since \(-\frac{1}{2} \le 0\), this value is possible.
• Option 3: 0. Since \(0 \le 0\), this value is possible (occurs when \(x=y\)).
• Option 4: 1. Since \(1 > 0\), this value contradicts our finding that \(k \le 0\).

Therefore, the value of \(k\) can never be equal to 1.