# How do you isolate for y when 0 = 2y + e^y

1. Sep 12, 2013

### MathewsMD

How do you isolate for y when you have the equation 0 = 2y + e^y - 4x + 3?

Any tips, useful links or solutions and an explanation would be greatly appreciated!

Thanks!

2. Sep 12, 2013

### johnqwertyful

Numerics seems to be the only way. Usually when you have to isolate a variable that's acted on by different types of functions, it's very difficult or impossible to do analytically.

3. Sep 12, 2013

### johnqwertyful

4. Sep 12, 2013

### MathewsMD

:(

10char

5. Sep 12, 2013

### johnqwertyful

In a lot of situations, it's not necessary to get analytic solutions. What is this for?

6. Sep 13, 2013

### HallsofIvy

Staff Emeritus
The "Wolfram Alpha" solution that johnqwertyful links to use the "Lambert W function" which is defined as the inverse function to $f(x)= xe^x$. It cannot be written in terms of any simpler function.

7. Sep 13, 2013

### jackmell

Well, when we have these mixed exponential equations, we try to put it in Lambert-W form. That is, in the form:

$$g(x,y)e^{g(x,y)}=h(x)$$

Then by definition of the Lambert W function which you can look up, we take the W function of both sides and obtain:

$$g(x,y)=W(h)$$

Now, doing a little moving around of your equation:

$$1/2 e^y=2x-3/2-y$$
$$1/2 e^{2x-3/2}=e^{-y} e^{2x-3/2}(2x-3/2-y)$$

or:

$$(2x-3/2-y)e^{2x-3/2-y}=1/2 e^{2/x-3/2}$$

I'll let you finish it to isolate y in terms of the perfectly valid (multi-valued) function of x in terms of the Lambert W-function.

Last edited: Sep 13, 2013