If x>y>0 and 2 log(x-y)=log x + log y, then what does (x/y) equal?

Aug 12, 2016

$\frac{x}{y} = \frac{1}{2} \left(3 + \sqrt{5}\right)$

Explanation:

From $2 \log \left(x - y\right) = \log x + \log y$ we know

${\left(x - y\right)}^{2} = x y$. Now calling $x = \lambda y$ and substituting

${\left(\lambda y - y\right)}^{2} = \lambda {y}^{2}$ or supposing $y > 0$

${\left(\lambda - 1\right)}^{2} = \lambda$

Solving for $\lambda$

${\lambda}^{2} - 3 \lambda + 1 = 0$

$\lambda = \frac{1}{2} \left(3 \pm \sqrt{5}\right)$

so remembering that $x > y$ the feasible solution is $\frac{x}{y} = \frac{1}{2} \left(3 + \sqrt{5}\right)$

Aug 12, 2016

$\frac{x}{y} = \frac{3 + \sqrt{5}}{2}$.

Explanation:

Given that, $2 \log \left(x - y\right) = \log x + \log y$

$\Rightarrow \log {\left(x - y\right)}^{2} = \log \left(x y\right)$

$\Rightarrow {\left(x - y\right)}^{2} = x y$

$\Rightarrow {x}^{2} - 3 x y + {y}^{2} = 0$

Dividing by ${y}^{2} \ne 0$, we get, ${\left(\frac{x}{y}\right)}^{2} - 3 \left(\frac{x}{y}\right) + 1 = 0$

Hence, $\frac{x}{y} = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 1}}{2} = \frac{3 \pm \sqrt{5}}{2}$

But $x > y > 0 \Rightarrow \frac{x}{y} > 1 \Rightarrow \frac{x}{y} = \frac{3 - \sqrt{5}}{2} < 1$ is not admissible.

Therefore, $\frac{x}{y} = \frac{3 + \sqrt{5}}{2}$.