Question

Not understanding this concept, please Help!
Find the inverse formula for the following function:
See the first reply for the equation I need to find an inverse for.

Answer 1

\(\ \huge y = \frac{1+e^x}{1-e^x}. Find the inverse? \)

Answer 2

The attempt at a solution

\[(1+e^x/1-e^x) (1+e^x/1+e^x) = (2+e^x^2/2)\]
\[\ln(2+e^x^2/2)= \ln2+x^2/\ln2\]

Answer 3

The attempt at a solution
1)
(1+e^x/1-e^x)(1+e^x/1+e^x)=(2+e^x^2/2)
ln(2+e^x^2/2)= ln2+x^2/ln2

Answer 4

@Study23 is that correct!

Answer 5

So did you switch the y and the x?

Answer 6

you solve for x then at the end you swap variables.

Answer 7

So how do you begin to solve for x?

Answer 8

if the question was this:\[y=\frac{1+z}{1-z}\]would you be able to solve for z?

Answer 9

because if you can do this, your problem requires one more step afterwards.

Answer 10

x = ln((-1+y)/(1+y))

now you can swap x for y

Answer 11

Well wouldn't you multiply both sides by (1-z), @joemath314159 ?

Answer 12

@timo86m how did you get to that x=ln part?

Answer 13

yes, thats correct:\[y=\frac{1+z}{1-z}\iff y(1-z)=1+z\iff y-yz=1+z\]\[\iff -z-yz=1-y\iff z(-1-y)=1-y\iff \]\[z=\frac{1-y}{-1-y}\]

Answer 14

Now switch z with e^x, and it should be clear how to proceed.

Answer 15

So, @joemath314159 basically you substituted a variable \(\ \huge \color{red}{z}\) for \(\ \huge \color{green}{e^x}?\)

Answer 16

Wouldn't you have to use \(\ \huge \color{blue}{ln}?\).

Answer 17

yes. its not necessary, but for some reason the problem looks sorta hard with e^x's floating about. yes you would have to use ln after where I left off.

Answer 18

Where would use ln? Where would you substitute the z value

Answer 19

@joemath314159

Answer 20

apply ln of both sides of jomaths final equation solved for z

Answer 21

since z= e^x

the ln of e^x is ln^(e^x) is just x
so x= ln ((1-y)/(-1-y))