Pre-Calc question (in comments)
Answer is ln4 = about 1.386

Question

kittiwitti1 · Answer

$$e^2x-3e^x-4=0$$How to solve?
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I used these steps:$$2e^x-3e^x-4=0$$$$-e^x-4=0$$But that makes for ln(-4) which is impossible...

anonymous · Answer

Have you tried -(e^x +4)  ?

Just put the negate out side and distribute after?

I havent sat down and punched it it, but it should work.

Pardon me if its wrong.

kittiwitti1 · Answer

$$-(e^x+4)=0$$$$e^x+4=0$$$$e^x=-4$$That's incorrect, sorry.

anonymous · Answer

Hmm, now I should try.

kittiwitti1 · Answer

Eh.... = w =;;

anonymous · Answer

Are you allowed to use a calculator?

anonymous · Answer

That, or it is, NO solution

kittiwitti1 · Answer

Yes, but you can't find ln/log e of a negative, can you?
It would give a nonreal answer, which we aren't allowed to have.

agent0smith · Answer

$$\large e^2x-3e^x-4=0$$there is no way to solve ^^that^^ algebraically

$$\large e^{2x}-3e^x-4=0$$^^that can be solved algebraically. use substitution, this makes it easier. Let e^x = u... then e^(2x) is u^2
$$\large u^2 -3u-4=0$$factor and then replace u with e^x

agent0smith · Answer

There will only be one solution btw

anonymous · Answer

$$\large u^2 -3u-4=0\
 (u-4)(u+1)=0\
u-4=0\
u+1=0\
u=4\
u=-1$$

anonymous · Answer

$$e^x=4\iff x=\ln(4)$$ which is your answer
$$e^x=-1$$ is not possible, since $e^x>0$ always

kittiwitti1 · Answer

Okay then... @agent0smith ? @ - @

shamil98 · Answer

u = e^x
u = 4
e^x = 4
take the ln of both sides 
ln(e^x) = ln 4
x = 1.386

kittiwitti1 · Answer

@shamil98 how does $$e^x=4$$ when the original was $$e^x+4=0$$

agent0smith · Answer

@kittiwitti1 see satellite's solution above.

kittiwitti1 · Answer

@agent0smith @shamil98 @satellite73 @mostwanting
The answer was to factor it like a quadratic with the variable being e^x:$$e^{2x}-3e^x-4=0$$$$(e^x+4)(e^x-1)$$Since only 4 works in the equation when plugged back in,$$e^x=4$$$$ln4\approx1.386$$

ranga · Answer

^^^ your signs are reversed.  It should be:$$(e^x-4)(e^x+1) = 0$$

kittiwitti1 · Answer

@ranga I asked about that but apparently it's correct.

agent0smith · Answer

@kittiwitti1 it is not correct. 

The correct solution has already been posted by satellite. 
"The answer was to factor it like a quadratic with the variable being e^x"
^That is exactly what satellite and I did, i'm not quite sure how you did not notice that. Using a substitution is perfectly valid and makes things more familiar.

Your teacher is wrong.

kittiwitti1 · Answer

Ah, yes, I forgot to MENTION that @satellite73 did it right as well... I know that the answer should be 4 and -1; however if you plug that in, you do NOT end up getting ln4 so I had to do it the other way (this happens on a lot of homework and test problems too, even last year when I was in Algebra 2).

Our teacher is correct, @agent0smith, the answer is exactly the same as in the book .-.

agent0smith · Answer

Nope, your teacher and the book are both wrong. Want even more proof? http://www.wolframalpha.com/input/?i=factor+e%5E%7B2x%7D-3e%5Ex-4%3D0 $$\large (e^x+4)(e^x-1) = 0$$is 100% incorrect. It foils to $$\large e^{2x} +3e^x -4 = 0$$and does that look like the original problem?

agent0smith · Answer

"this happens on a lot..."

the only way this happens is if something is wrong. Either the question is written wrongly, or the answer is wrong.

kittiwitti1 · Answer

Well, I didn't ask for a check after I finished rewriting the answer, so maybe I'll ask her again on Monday. @agent0smith 
We'll see after that

agent0smith · Answer

Okay, but, you do realize that the answer you gave is wrong? You seem to not be sure. 

This isn't a case of "maybe your teacher is wrong". It's definite, there is no doubt here. Either you've posted the wrong solution, or the teacher gave you the wrong solution. When a guy with a math degree who teaches precalc tells you it's incorrect, as well as ranga and satellite showing it's incorrect, AND wolfram alpha tells you it's incorrect, you can be pretty damn sure that it is indeed incorrect.

kittiwitti1 · Answer

The answer key says ln4 @agent0smith

agent0smith · Answer

Ln4 is correct.

 This equation is not correct $$\large (e^x+4)(e^x-1) = 0$$that will not give ln4 as a solution, it would give x=0 as the only solution

$$\large (e^x-4)(e^x+1) = 0$$^that is the correct factorization and will give x=ln4 as a solution.

kittiwitti1 · Answer

I'll have to ask the teacher. @agent0smith 
Of course I think that the second equation is correct, though.

kittiwitti1 · Answer

@agent0smith, turns out I've got the right answer, only I wrot down the wrong signs ^^;;
Thanks for the help all!

agent0smith · Answer

lol that's what we already told you :P the signs were wrong.