Question

Find the inverse of the function: f(x)=1+e^x/1-e^x

Answer 1

\[\frac{ 1+e ^{x} }{ 1-e ^{x} }\]
I know the opposite of e is ln, but thats about it

Answer 2

is it\[f(x)=\frac{1+e^x}{1-e^x}\]

Answer 3

Yep

Answer 4

\[x=\frac{1+e^y}{1-e^y}\] find e^y

Answer 5

here we swapped x with y

Answer 6

I get that, but do I just multiply by ln on the top and bottom?

Answer 7

before lets multiply both sides by 1-e^y

Answer 8

oh to get it off the bottom?

Answer 9

yes

Answer 10

do i get:

Answer 11

do i get:\[x(1-e ^{x})=(1+e ^{x})(1-e ^{x})\]

Answer 12

i mean those exponents to be y's

Answer 13

right hand side does not have the 1-e^x part\[\frac{ 1+e^x }{ \cancel{1-e^x }}\cancel{1-e^x}\]

Answer 14

oh gotcha!

Answer 15

now i mutiply by ln on both sides?

Answer 16

so we have\[x(1-e^y)=1-e^y\]

Answer 17

open bracket then take take e^y to the same side

Answer 18

huh?

Answer 19

I got as far as your last equation and then I multiplied by ln on both sides and get confused on what can cancel

Answer 20

do you get this\[x(1-e^y)=1-e^y\]\[x-xe^y=1-e^y\]

Answer 21

hey we never multiplied by ln both sides

Answer 22

oh gotcha, distribute the x through, ok

Answer 23

so can you take xe^y to the right and 1 to the left

Answer 24

alrighty

Answer 25

\[x+1=-xe ^{y}-e ^{y}\]

Answer 26

notice that its now +xe^y not -xe^y
change of sign

Answer 27

oh ok, mistake on just copying it to here

Answer 28

ok lol so
\[x-1=e^y(x-1)\]
amazingly canceling x-1 both sides we have \[1=e^y\]

Answer 29

how can we just cancel both sides?

Answer 30

oh by dividing, nevermind

Answer 31

yes so\[\ln e^y=\ln 1\]

Answer 32

so y=ln1 , which is y=0. thanks!

Answer 33

wait somthing went wrong

Answer 34

Its no biggie. I think i get the basic idea now. Thanks though

Answer 35

i can always just ask my teacher for sure on monday

Answer 36

I think its because its actually - on top and + on the bottom of the main funciton. No sense doing it over though

Answer 37

I miscopied it from the book i guess

Answer 38

\[x=\frac{ 1+e^y }{ 1-e^y }\]
\[x-e^y=1\color{red}{+}e^y\] we wrote - earlier so its +
\[x-1=xe^y+e^y\]
\[1-x=e^y(1+x)\]
\[e^y=\frac{ 1-x }{ 1+x}\]

Answer 39

yep! that makes the books answer now if i just ln both sides. thanks so much!

Answer 40

\[y=\ln \left(\frac{ x-1 }{ x+1 }\right)\]

Answer 41

thanks so much. i fanned you and gave you best response

Answer 42

yw