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

1 Answer
Jun 13, 2016

Answer:

#x = 1/2 log_e 2, y = 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 :

#{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#