# How do you solve e^(x-y) = e^y and e^(x+2y) = 2?

Jun 13, 2016

$x = \frac{1}{2} {\log}_{e} 2 , y = \frac{1}{4} {\log}_{e} 2$

#### Explanation:

This is equivalent to

{(e^x/(e^y)= e^y), (e^x e^{2y] = 2) :}

Making ${e}^{x} = X$ and ${e}^{y} = Y$
we have the equivalent system

{ (X/Y= Y), (X Y^2 = 2) :}

Solving for $X , Y$ for real solutions, we get :

$\left\{X = \sqrt{2} , Y = \pm {2}^{\frac{1}{4}}\right\}$

so

$\left\{{e}^{x} = \sqrt{2} , {e}^{y} = {2}^{\frac{1}{4}}\right\}$ and finally

$x = \frac{1}{2} {\log}_{e} 2 , y = \frac{1}{4} {\log}_{e} 2$