Question

Hi
How can I inverse these two equations:

1/ f(x)=2−e^2x

2/ f(x)=−8-(4/x)

Thanks,

Answer 1

\[Let \qquad f(x)=y\]To find the inverse function, we'll swap x and y in the equation.\[y=2-e^{2x} \rightarrow \qquad x=2-e^{2y}\] And now we want to solve for y :D Hmmm.

Answer 2

\[\huge e^{2y}=2-x\]

Answer 3

Now we need to get the y out of the exponent, hmmm do you know how to do that? :D
I suppose we could take the natural log of both sides.
\[\huge \ln ( e^{2y})=\ln ( 2-x)\]

Answer 4

Next, we want to apply this rule of logarithms to get the y out of the log.
\[\huge \log(a^b)=b \log (a)\]

Answer 5

\[\huge (2y) \ln (e) = \ln (2-x)\]

Answer 6

Andddd do you remember what the natural log of e is? It's basically asking, e to what power gives us e?
e to the first power yes? :O
so ln e = 1
If you're confused about that, maybe just try punching it into your calculator ^^ it's a good one to remember though.

Answer 7

\[\huge 2y= \ln (2-x)\]

Answer 8

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

Answer 9

Confused about anything in there? :O it's quite a few steps to get through! :D

Answer 10

Thank you zepdrix :) that's really helpful. I was trying many many time and every time i get this: ln(x+2)/2

Answer 11

Ah, yah i kinda flew through that step :3 hopefully you see what's going on.
\[\large x=2-e^{2y}\]\[\large x-2=-e^{2y}\]Multiplying both sides by -1 gives:\[\large 2-x=e^{2y}\]

Answer 12

yup :D i get it now.
Can you help me with the 2nd equation plz?

Answer 13

\[\large y=-8-\frac{4}{x}\]Taking the inverse gives us:\[\large x=-8-\frac{4}{y}\]

Answer 14

\[\large x+8=-\frac{4}{y}\]Multiplying both sides by y gives us:\[\large y(x+8)=-4\]

Answer 15

Dividing both sides by (x+8) gives us:\[\large y=\frac{-4}{x+8}\]
\[\huge f^{-1}(x)=\frac{-4}{x+8}\]

Answer 16

omg thank you. I was doing it wrong every time by multiplying y in x and then 8 and getting crazy answers!
Sounds easy now, Thanks again :)

Answer 17

hah :D