Finding the inverse of a function

[anubis01]

Homework Statement

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.

Homework Equations

1)ln(e^x)=x

2)ln(e^x)=x

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]

Homework Statement

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.

Homework Equations

1)ln(e^x)=x

2)ln(e^x)=x

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

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+ 2e^x+ e^2x is the numerator and the denominator is 1- e^2x, 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+e^x)/(1- e^x) then (1- e^x)y= 1+ e^x or y- ye^x= 1+ e^x. y(1+ e^x)= y- 1, 1+ e^x= (y- 1)/y, e^x= (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) ?