# Solving for a variable.

1. Nov 27, 2006

### FizX

How would you go about solving these?

solving for x, x ln(x)=A

solving for y, y Exp[y] =B

and also what would happen if you turned A into A+x and B into B+y.

Thanks

2. Nov 28, 2006

### HallsofIvy

Use Lambert's W function which is DEFINED as the inverse function to f(x)= xex.

3. Nov 29, 2006

### Swapnil

$$x \ln(x) = A \Rightarrow x = \frac{A}{W\{A\}}$$
$$ye^{y} = B \Rightarrow y = W\{B\}$$
$$x \ln(x) = A + x \Rightarrow x = \frac{A}{W\{\frac{A}{e}\}}$$
$$y e^{x} = A + y \Rightarrow y = ?$$

I am still working on the fourth one...looks hard!

edit: sorry, the last one is suppose to have a y in the argument of the exponential instead of an x.

Last edited: Nov 30, 2006
4. Nov 29, 2006

### Swapnil

I don't know! Someone else, help!!

5. Nov 30, 2006

### leon1127

Can you factor out the y?

6. Nov 30, 2006

### ^_^physicist

Leon22's logic is right, subtract y from the righthand side and place it with the terms y*e^x to get an expression that looks like:

y*(e^(x))-y=A

You can then factor out the y, and get an expression:

y* ((e^(x))-1)=A.

Then just solve for y.

Sorry I don't have much experance with the function that was referanced eariler and didn't bother trying to put into that format, but the solution gained from leon22's logic should be vaild.

Good Luck.

7. Nov 30, 2006

### Swapnil

Oops... There wasn't suppose to be any x in that equation. All the variable were suppose to be y. That's why I had to resort to using the Lambert W funcition. It was suppose to be:

$$y e^{y} = A + y \Rightarrow y = ?$$

Sorry about that. So does anyone now know how to solve for y?

8. Nov 30, 2006

### Swapnil

I tried mathematica too but no luck. I am guessing that it is probably not possible to solve for y in closed-form.

BTW, here is another problem that I came across a while back which doesn't seem to have a closed-form solution either:

$$x^x = Ax$$

Last edited: Nov 30, 2006