Finding the inverse of a function

[anubis01]

1. The problem statement, all variables and given/known data
1)Find the inverse of a f(x)=1+e^x/1-e^x

2)solve for x when e^ax=ce^bx where a doesn't equal b.


2. Relevant equations
1)ln(e^x)=x

2)ln(e^x)=x


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

sqrt(ln2/ln2)=x

I think this is what x comes out to but I'm not sure.

2)
ln(e^ax)=ln(ce^bx)
ax=bxC
a-b=x-x

Again I'm not sure if this is right, any help is much appreciated.

[Pacopag]

I'm thrown off by the division by 1.
i.e. e^x/1 = e^x?

[HallsofIvy]

1. The problem statement, all variables and given/known data
1)Find the inverse of a f(x)=1+e^x/1-e^x

2)solve for x when e^ax=ce^bx where a doesn't equal b.


2. Relevant equations
1)ln(e^x)=x

2)ln(e^x)=x

You realize those are the same equation, don't you?:smile:


3. The attempt at a solution
1)
(1+e^x/1-e^x)(1+e^x/1+e^x)=(2+e^x^2/2)
No, that's just bad algebra. [(1+e^x)(1+e^x)= 1+ 2ex+ e2x is the numerator and the denominator is 1- e2x, not "2".0

ln(2+e^x^2/2)= ln2+x^2/ln2
And even if it were correct, ln(A+ B) is not ln(A)+ ln(B).

sqrt(ln2/ln2)=x

I think this is what x comes out to but I'm not sure.
No, it doesn't. The if y= f(x), then x= f-1 of y so the standard way to find the inverse of y= f(x) is to solve the equation for x. If y= (1+ex)/(1- ex) then (1- ex)y= 1+ ex or y- yex= 1+ ex. y(1+ ex)= y- 1, 1+ ex= (y- 1)/y, ex= (y-1)/y- 1= (y-1)/y- y/y= -1/y. Can you solve that for x?

2)
ln(e^ax)=ln(ce^bx)
ax=bxC
No, ln(AB) is not A ln(B) ln(ce^(bx))= ln(c)+ ln(bx)= ln(c)+ bx.
a-b=x-x

Again I'm not sure if this is right, any help is much appreciated.
Since the "x"s cancel out in what you have, you should be sure it is not right!

[anubis01]

1) IF e^x=-1/y then ln(e^x)=ln(-1/y)
x=ln(-1/y)
y=ln(-1/x)

[hancyu]

is the answer to the problem y= ln(x/2) ?