# Rearranging exponential equation

• Jan 22nd 2013, 01:20 AM
mpylon
Rearranging exponential equation
Hi,

I've tried this on a Maths teacher friend of mine, who is currently stumped, so I hope it classes as 'advanced'. Essentially, I'm trying to rearrange the following equation:

x = y / [A - e^(By)]

to express y in terms of x. Is this possible?

Thanks.
• Jan 22nd 2013, 05:20 PM
chiro
Re: 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.
• Jan 22nd 2013, 10:31 PM
Prove It
Re: Rearranging exponential equation
Quote:

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...
• Jan 22nd 2013, 10:35 PM
chiro
Re: Rearranging exponential equation
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.
• Jan 22nd 2013, 10:42 PM
a tutor
Re: Rearranging exponential equation
There's a similar problem here.