Question 1000276
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Find x and y given that e^x + 3e^y=3 and e^2x - 9e^2y=6, expressing your answers as a logarithm to base e.
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Introduce two new variables u = {{{e^x}}} and v = {{{e^y}}}.

Then your system will be reduced to the system

u   +  3v  = 3,     (1)
u^2 - 9v^2 = 6.     (2)

Factor the second equation

(u-3v)*(u+3v) = 6

and substitute 3 instead of u+3x in the last equation. You will get the system

u   +  3v  = 3,         (1')
u   -  3v  = {{{6/3}}} = 2.    (2')

Add the two last equations. You will get

2u = 5. Hence, u = {{{5/2}}}.

Now, distract (2) from (1'). You will get 

6v = 1. Hence, v = {{{1/6}}}.

Thus we get 

u = {{{e^x}}} = {{{5/2}}} -----> x = {{{ln(5/2)}}} = {{{ln(5) - ln((2))}}}.

v = {{{e^y}}} = {{{1/6}}} -----> y = {{{ln(1/6)}}} = {{{-ln(6)}}}.
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