How Can You Solve These Variable Equations?

[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

 

[HallsofIvy]

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

 

[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.

 

[Swapnil]

I don't know! Someone else, help!

 

[leon1127]

I don't know! Someone else, help!

Can you factor out the y?

 

[^_^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.

 

[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?

 

[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