Question

e^2x-2e^x=3 how do I solve this?

Answer 1

I'd suggest reducing it to something more familiar, perhaps let
\[\large u = e^x\]
Thus making it
\[\large u^2 - 2u = 3\]

Answer 2

Let u=e^x

Answer 3

Dammit @terenzreignz

Answer 4

I've heard that many a time, @agent0smith :)
(in real life too XD )

Answer 5

ok then what

Answer 6

Then solve it the same way you'd solve a quadratic formula, find u. Then go back and use u=e^x to find x.

Answer 7

Then, you solve for the value of u...\[\large u^2 -2u -3=0\]
Using any method that you feel comfy with :)

Answer 8

dammit @agent0smith
:D

Answer 9

haha :D

Answer 10

all this is to get you to factor this beast
with a little practice you can factor
\[e^{2x}-e^x-3=0\] with just a mental substitution

Answer 11

well not really, but you can for
\[e^{2x}-2e^x-3=0\]

Answer 12

(2e^x-1)(e^x+2)

Answer 13

I mean 3

Answer 14

careful

Answer 15

?

Answer 16

Try factoring this first...
\[\large u^2 - 2u - 3\]

Answer 17

oh you pull the 2 out first? so 2(u-1)(u+3)

Answer 18

There is no "2" to pull out...

Answer 19

i think im confusing myself

Answer 20

Don't give up, though.
I'm not really good at explaining how to factor, so you're going to have to get that from someone else...

Answer 21

I think I got it

Answer 22

What did you get?

Answer 23

e^x+1=0 so e^x=-1 and then e^x=3 for the other one

Answer 24

Well, can \[\Large e^x=-1\] have a solution?
To which exponent do you raise e so that it becomes negative?

That's not entirely infeasible, but it's probably beyond the scope of this question....



btw for future reference...
\[\huge e^{i\pi}= -1\]

Answer 25

So the only thing you have to worry about is
\[\huge e^x = 3\]
so what's x?

Answer 26

so you just take the ln of 3 right?1.0986

Answer 27

Yes. Precisely :)