urgen!!A-MATHS~~y=e^2x+e^-2x

1)y=e^2x+e^-2x
for y=3
3=e^2x+(1/e^2x)
3=[ (e^2x)^2+1]/e^2x
3e^2x=(e^2x)^2+1
(e^2x)^2-3e^2x+1=0
therefore
e^2x=(3+- root(3^2-4))/2
e^2x=(3 +- root(5))/2
therefore(take loge on both sides)
we have
2x=loge(1/2(3 +- root5))
x= 1/2loge(1/2(3 +- root5))

2007-11-12 21:59:41 補充：
for y=11=e^2x e^--2x(e^2x)^2-e^2x 1=0obviously, it is a quadratic equationsince the discrimnant=1-4(1)(1)=-3therefore there is no real solution.

2007-11-12 22:01:04 補充：
1=e^2x＋ e^-2x(e^2x)^2－e^2x＋1=0

(1) 3 = e^(2x) + e^(-2x)
=> 3e^(2x) = e^(4x) + 1
=> e^(4x) - 3e^(2x) + 1 = 0
Let z = e^(2x).
Then, z^2-3z+1=0
=> z = {3 + sqrt[(-3)^2 - 4(1)(1)]}/2 or {3 - sqrt[(-3)^2 - 4(1)(1)]}/2
= [3+sqrt(5)]/2 or [3-sqrt(5)]/2
Since z =e^(2x), ln z = 2x or x = (1/2)ln z
Then, x = (1/2) ln {[3+sqrt(5)]/2} or 1/2) ln {[3-sqrt(5)]/2}

(2) 1 = e^(2x) + e^(-2x)
=> e^(4x) - e^(2x) + 1 = 0
Let z = e^(2x).
Then, z^2 - z + 1 = 0 -------------(1)
Since discriminant of quadratic equation (1) = (-1)^2 - 4(1)(1) = -3 < 0, equation (1) has no real roots.
Then, y=e^2x+e^-2x has no solution for y = 1.