Question

solve e^2x-3e^x+2=0 for x

Answer 1

\[e^{2x}-3e^x+2=0\]\[(e^x)^2-3e^x+2=0\]Quadratic equation. Let u = e^x, you'll get
u^2 - 3u +2 =0
Solve u, then solve x

Answer 2

where did u come from tho?

Answer 3

where did the t come from as well?

Answer 4

You let u (or t) be e^x, just easier for you to see it is a quadratic equation. Of course, you don't have to.

Answer 5

is there a simplier way to solve it?

Answer 6

This is very simple.

Answer 7

Using factorization:
\[e^{2x}-3e^x+2=0\]\[(e^{x}-2)(e^x-1)=0\]Almost done.

Answer 8

oh okay.

Answer 9

do i find the zeros?

Answer 10

Hmm.. Solve for e^x.

Answer 11

well see, im not too sure how to solve for that

Answer 12

by putting each factor =0

Answer 13

is that where e equals to something?

Answer 14

e^x equal to something.
Then, take natural log for both sides.

Answer 15

Actually, you are solving for x in e^x. e is a constant, so you arr trying to find the value x so that when e is raised to the power, you get the value 1 or 2. In other words,

\[e^{x}=2 \] or \[e^{x}=1\]

Answer 16

And then just do as Callisto suggested, take the natural log of both sides.

Answer 17

oh okay, im a bit confused but i think i understand

Answer 18

Where is the confusion?

Answer 19

why is raised to 1 or 2? @calmat01