Question

For positive non-zero real variables \( x \) and \( y \), if \[ \ln\left(\frac{x + y}{2}\right) = \frac{1}{2}\left[\ln(x) + \ln(y)\right], \] then the value of \( \frac{x}{y} + \frac{y}{x} \) is:

• A. 1
• B. \( \frac{1}{2} \)
• C. 2
• D. 4

Hint:

When dealing with logarithmic equations, always simplify using properties such as \( \ln(ab) = \ln(a) + \ln(b) \) and exponentiate to eliminate the logarithm.

Answer 1

The Correct Option is C

The given equation is: \[ \ln\left(\frac{x + y}{2}\right) = \frac{1}{2}\left[\ln(x) + \ln(y)\right]. \] Step 1: Simplify using logarithmic properties.
Using the property of logarithms, \( \ln(ab) = \ln(a) + \ln(b) \), the right-hand side can be rewritten as: \[ \frac{1}{2}\left[\ln(x) + \ln(y)\right] = \ln\left(\sqrt{xy}\right). \]
Equating both sides, we get: \[ \ln\left(\frac{x + y}{2}\right) = \ln\left(\sqrt{xy}\right). \] Step 2: Exponentiate both sides.
By removing the logarithm, we have: \[ \frac{x + y}{2} = \sqrt{xy}. \] Step 3: Solve for \( \frac{x}{y} + \frac{y}{x} \).
Square both sides: \[ \left(\frac{x + y}{2}\right)^2 = xy \implies \frac{(x + y)^2}{4} = xy \implies x^2 + y^2 + 2xy = 4xy. \]
Simplify to find \( x^2 + y^2 \): \[ x^2 + y^2 = 2xy. \]
Thus: \[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = \frac{2xy}{xy} = 2. \] Final Answer: \[ \boxed{2} \]