Exponential equations.?

Question

Could someone show me they method used to solve this equation for x.
e^(4x) + 4e^(2x) - 21 = 0

Answer 1

Factor this into:
(e^(2x) + 7)(e^(2x) - 3) = 0
either e^(2x) + 7 = 0
or
e^(2x) - 3 = 0

e^(2x) + 7 = 0
or e^(2x) = -7
This can't happen so we can disregard this answer. (e to any power will be positive, never negative.)

e^(2x) - 3 = 0
e^(2x) = 3
take logs of both sides
2x = ln3
x = (1/2)ln3

Answer 2

e^(4x) + 4e^(2x) - 21 = 0

Let z = e^2x, then:
z^2 + 4z - 21 = 0
(z + 7)(z - 3) = 0
z + 7 = 0 or z - 3 = 0
z = -7 or z = 3

Substitute back:
e^2x = -7
2x = ln -7
x = ln -7/2 = invalid since you can't take ln of neg number

e^2x = 3
2x = ln3
x = (ln3)/2

Therefore the answer is:
x = (ln3)/2

Answer 3

make y = exp(2x).