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

by MathewsMD
Tags: isolate
 P: 285 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!
 P: 308 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.
 P: 308 http://www.wolframalpha.com/input/?i...5Ey+-+4x+%2B+3 Yeah, it's impossible.
P: 285

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

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10char
 P: 308 In a lot of situations, it's not necessary to get analytic solutions. What is this for?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,900 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.
P: 1,666
 Quote by 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!
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.

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