Hi,
I would write the first equation as
e^{2y} = x  2
and then take the log of both sides
ln(e^{2y}) = ln(x  2 ) or
2y = ln(x  2)
Substitute this into the second equation to get
ln(x + 3)  ln(x  2)  1 = 0
which is
ln((x + 3)/(x  2)) = 1.
Now take the exponential of both sides to get
(x + 3)/(x  2) = e^{1} = e
Solve for x.
Penny
