$ln\left(\frac{x+y}{2}\right) = \frac{1}{2} [ln\left(x\right) + ln\left(y\right)]$
then, the value of $\frac{x}{y} + \frac{y}{x}$ is
We are given the equation for positive non-zero real variables $x$ and $y$: $ ln\left(\frac{x+y}{2}\right) = \frac{1}{2} [ln\left(x\right) + ln\left(y\right)] $
Using the logarithm property $ln(a) + ln(b) = ln(ab)$, the right side becomes:
$ \frac{1}{2} [ln\left(x\right) + ln\left(y\right)] = \frac{1}{2} ln(xy) $Using the property $c \cdot ln(a) = ln(a^c)$, this simplifies to:
$ ln((xy)^{\frac{1}{2}}) = ln(\sqrt{xy}) $The equation now is: $ ln\left(\frac{x+y}{2}\right) = ln(\sqrt{xy}) $ Since the natural logarithm function is one-to-one, we can equate the arguments:
$ \frac{x+y}{2} = \sqrt{xy} $Multiply both sides by 2:
$ x+y = 2\sqrt{xy} $Square both sides to eliminate the square root:
$ (x+y)^2 = (2\sqrt{xy})^2 $ $ x^2 + 2xy + y^2 = 4xy $Rearrange the terms to form a quadratic expression:
$ x^2 - 2xy + y^2 = 0 $Factor the expression, which is a perfect square:
$ (x-y)^2 = 0 $Taking the square root gives:
$ x-y = 0 $ $ x = y $We need to find the value of $\frac{x}{y} + \frac{y}{x}$. Since we found $x=y$, we can substitute $y$ with $x$ (or vice versa):
$ \frac{x}{x} + \frac{x}{x} = 1 + 1 $ $ = 2 $Therefore, the value of $\frac{x}{y} + \frac{y}{x}$ is 2.
Given $f(x, y) = x^2 - 2xy + y^2$
The complete contour of the equation $f(x, y) = 1$ is described by the option(s) ___.