Need math help dont know how to right in natral log form?

Question

Solve the following exponential equeation. Express the solution in set in terms of natural logarithms.

e^4x + 2e^2x - 8 =0

Answer 1

Substitute y=e^2x into the equation:

y^2 + 2y -8 = 0

(y + 4)(y - 2) = 0

That means:

y = -4 or y = 2

e^2x = -4 is not possible in real numbers, so we know:

e^2x = 2

Take the natural log of both sides:

ln(e^2x) = ln(2)

simplify:

x = ln(2)/2

Answer 2

e^4x + 2e^2x - 8 =0
Put e^2x = u .....(1)
Then:
u^2 + 2u - 8 = 0
(u + 4)(u - 2) = 0
u = -4 or u = 2
Therefore from (1):
e^2x = -4 or e^2x = 2
Rejecting the first of these, as e^2x cannot be -ve:
2x = ln(2)
x = ln(2) / 2.

Answer 3

let u = e^2x. then you're looking at

u² + 2u - 8 = 0, ....which is quadratic.
(u + 4)(u - 2) = 0
u = -4 or u = 2

e^2x = -4
2x = ln -4
no solution

e^2x = 2
2x = ln 2
x = (ln 2)/2

Answer 4

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

e^2x + 4 = 0 and e^2x - 2 = 0
e^2x = -4 and e^2x = 2

since the first has no solution then solve e^2x = 2

ln (e^2x ) = ln 2

2x * lne = ln 2 remember lne = 1

2x = ln 2

x = ln2 / 2