Question

what is the inverse function of y=e^-x-2e^-2x ?

Answer 1

\[e^{-x}-2e^{-2x}\]?

Answer 2

yep

Answer 3

hmmmm....

Answer 4

its a hard one :/

Answer 5

yeah

Answer 6

http://www.wolframalpha.com/input/?i=find+the+inverse+y%3De^%28-x%29-2e^%28-2x%29

Answer 7

I think they must use methods I dont know about
\[y = e^{-x}-2e^{-2x}=e^{-x}(1-2e^{-x})\\ln(y) = ln(e^{-x}(1-2e^{-x}))\\so\\ln(y) = ln(e^{-x})+ln(1-2e^{-x})\\ln(y) = -x+ln(1-2e^{-x})\]

Answer 8

and im stuck

Answer 9

are those special rules??

Answer 10

which part?

Answer 11

i get what you have done i dont get the link but

Answer 12

\[ln(y) = -x+ln(1-\frac{2}{e^x})=-x+ln(\frac{e^x-2}{e^x})=-x+ln(e^x-2)-1\]

Answer 13

that god damn 2...

Answer 14

@KingGeorge got any ideas?

Answer 15

I posted this wolfram link to the inverse, but no idea how they got it....there is an 8 somehow....lol

Answer 16

I'm also stuck right now, but I'm trying to go backwards from WA's solution.

Answer 17

i love little complicated things like this:)

Answer 18

I'm definitely not getting the same thing :/

Answer 19

same here. If something special was going on I would think that wolfram would make a note or something...

Answer 20

Well, I really need to go to bed now, but it's possible the wolfram made a mistake. If I plug in the equation for y, I don't get x back if I do it by hand, but WA still does. http://www.wolframalpha.com/input/?i=log%28%281-sqrt%281-8+%28e%5E%28-x%29-2e%5E%28-2x%29%29%29%29%2F%282+%28e%5E%28-x%29-2e%5E%28-2x%29%29%29%29

Answer 21

crazy

Answer 22

Is there even an answer? Because I graphed the function and it is not one-one.

Answer 23

y = e^(-x) - 2e^(-2x)
Let y = f(x) and z = f^-1(x).
f(z) = ff^-1(x) = x
e^(-z) - 2e^(-2z) = x
1 / e^z - 2 / e^(2z) = x
Let e^z = u.
1 / u - 2 / u^2 = x
u - 2 = xu^2
xu^2 - u + 2 = 0
u = (1 +- sqrt (1 - 4(x)(2))) / (2x) = (1 +- sqrt (1 - 8x)) / (2x)
e^z = (1 +- sqrt (1 - 8x)) / (2x)
z = ln (1 +- sqrt (1 - 8x)) - ln (2x)
f^-1(x) = ln (1 +- sqrt (1 - 8x)) - ln (2x)

Answer 24

The domain of y needs to be restricted for the inverse to work.

Answer 25

I was afraid something like that was the case. Good job.