Question

how would you find the solutions for:
e^ { 2 x } - 6 e^ { x } +8 = 0

Answer 1

let u=e^x
then factor

Answer 2

what do you mean by u

Answer 3

let u=e^x

Answer 4

e^x is u
replace e^x with u
replace e^(2x) with u^2

Answer 5

then factor the quadratic

Answer 6

then replace u with e^x

Answer 7

i dont think im deriving the solutions

Answer 8

\[u^2-6u+8=0\]
so you didn't replace e^x with u?
you just factor now
(u-2)(u-4)=0
u=2 or u=4
but u=e^x
so replace u with e^x
e^x=2 or e^x=4
then solve for x