Question

If x>y>1 then the maximum value of $\log_x (\frac{x}{y}) + \log_y (\frac{y}{x})$ is equal to

• A. -2
• B. 0
• C. 2
• D. 4

Answer 1

Answer: (B)

0

Answer 2

Let

$S = \log_{x}\left(\frac{x}{y}\right) + \log_{y}\left(\frac{y}{x}\right)$

Express in terms of natural logarithms:

$\log_{x}\left(\frac{x}{y}\right) = \frac{\ln(x/y)}{\ln x} = \frac{\ln x - \ln y}{\ln x} = 1 - \frac{\ln y}{\ln x}$

$\log_{y}\left(\frac{y}{x}\right) = \frac{\ln(y/x)}{\ln y} = \frac{\ln y - \ln x}{\ln y} = 1 - \frac{\ln x}{\ln y}$

Thus,

$S = 2 - \left(\frac{\ln y}{\ln x} + \frac{\ln x}{\ln y}\right)$

Let $t = \frac{\ln x}{\ln y}$. Note that since $x > y > 1$, both $\ln x$ and $\ln y$ are positive and $t > 1$. Then:

$\frac{\ln y}{\ln x} = \frac{1}{t}$

and

$S = 2 - \left(t + \frac{1}{t}\right)$

The minimum value of $t + \frac{1}{t}$ for $t > 0$ is $2$ when $t = 1$ (by AM-GM inequality). Since $t > 1$, the closest we can get is as $t \to 1^{+}$, which gives:

$S_{\text{max}} = 2 - 2 = 0$