(3+e^y)/(3-e^y); solve for y?

Question

You're right puggy; I goofed; the problem I'm stuck on is x = (3+e^y)/(3-e^y)

Answer 1

x/1 = (3+e^y)/(3-e^y)

Cross multiply to get 3x - xe^y = 3 + e^y

Add xe^y to both sides and subtract 3 from both sides:

3x - 3 = xe^y + e^y

Factor out the e^y

3x - 3 = e^y(x + 1)

Divide both sides by (x+1)

e^y = (3x-3)/(x+1)

ln (e^y) = ln ((3x-3)(x+1))

y = ln ((3x-3)(x+1))

Answer 2

Here's the problem; you can't solve for a variable unless it's an equation, and I don't see an equal sign anywhere.

However, if you meant

x = (3 + e^y) / (3 - e^y)

Then your first step would be to multiply both sides by (3 - e^y).

x(3 - e^y) = (3 + e^y)

Expand the left hand side,

3x - xe^y = 3 + e^y

Now, move everything with (e^y) in it to the left hand side; everything else goes to the right hand side.

-xe^y - e^y = 3 - 3x

Factor e^y on the left hand side.

e^y(-x - 1) = 3 - 3x

Divide both sides by (-x - 1)

e^y = (3 - 3x) / (-x - 1)

To clean up the right hand side, I'm going to factor (-1) out of the top and bottom.

e^y = [(-1)(3x - 3)] / [(-1) (x + 1)]

Cancel the (-1) out

e^y = (3x - 3) / (x + 1)

At this point, we have to convert this to logarithmic form. Note that

a^b = c converted to logarithmic form is log[base b](c) = a

In this case, our base is "e", and log[base e] is the same as ln. Therefore, we have

ln[(3x - 3) / (x + 1)] = y

Therefore

y = ln[(3x - 3) / (x + 1)]

We can reduce that using log properties if we like, but I'll leave that up to you to figure out. This is a sufficient answer.