Thread: Rearranging exponential equation

Hey mpylon.

What I will do is use implicit differentiation to see if you can separate the variables so that they can be calculated independent of each other.

d/dx(Ax - xe^(By)) = d/dx(y)

dy/dx
= A - d/dx(xe^(By))
= A - [e^(By) + Bx*e^(By)dy/dx] So this implies:

dy/dx[1 + Bxe^(By)] = A - e^(By) or

dy/dx = [A - e^(By)]/[1 + Bxe^(By)]

This looks like all derivatives will involve x's and y's which means that you can't get y as a function purely of x (or the other way around unless you use specific restrictions or find a parameterization.

Originally Posted by chiro

Hey mpylon.

What I will do is use implicit differentiation to see if you can separate the variables so that they can be calculated independent of each other.

d/dx(Ax - xe^(By)) = d/dx(y)

dy/dx
= A - d/dx(xe^(By))
= A - [e^(By) + Bx*e^(By)dy/dx] So this implies:

dy/dx[1 + Bxe^(By)] = A - e^(By) or

dy/dx = [A - e^(By)]/[1 + Bxe^(By)]

This looks like all derivatives will involve x's and y's which means that you can't get y as a function purely of x (or the other way around unless you use specific restrictions or find a parameterization.

Why are you differentiating at all? It should be obvious that y can't be isolated because it appears as both an exponent and a polynomial...

The point is to show that the functions change depends on its current value which shows its implicit.

Recall that a function is uniquely determined by its Taylor series and one way to show implicit behavior is through its derivative.

There's a similar problem here.

Solution of equation of form ln(a+bx) = cx

Thanks "a tutor", the Lambert's W function was the way to go, although below a certain number the Lambert's W function gives complex numbers. Is there any way round this?

No worries, it was a quirk of the Excel function I downloaded, but have solved it. Thanks for your help.