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

Question

1717 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-13T12:30:23

$x = 1/2 log_e 2, y = 1/4 log_e 2$

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 :

${X = sqrt[2], Y = pm 2^(1/4)}$

so

${e^x = sqrt(2), e^y = 2^(1/4)}$ and finally

$x = 1/2 log_e 2, y = 1/4 log_e 2$

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