I need to solve e^(kx)=kx-c for x where k and c are constants

Question

I need to solve the expression e^(kx)=kx-c for x. Is it possible?

Answer 1

Given:

exp( kx ) = e^kx = kx - c

Find:

x ( k, c ) = ?

Assume:

The student have knowledge or will search the internet for knowledge on the topic of:

The Lambert W function (a.k.a. the product log)

Solution:

To avoid writing kx a bunch of times, I will make this substitution:

u = kx

Thus, the problem becomes:

exp( u ) = u - c

So I'm going to utilize an algebra "trick" to do something to the power of the e:

exp( c ) exp( u - c ) = u - c

That trick was "adding zero" which will leave the function ultimately unchanged. You can check it will rules of exponents

You will then want to get the exp( u - c ) to the other side of the equation like so:

exp( c ) = ( u - c ) exp[ - ( u - c ) ]

Now, for the necessity of what will be needed to solve for u, I will multiply both sides by -1 to get:

- exp( c ) = - ( u - c ) exp[ - ( u - c ) ]

Now to introduce the Lambert W function/product log:

To solve:

y e^y = x for y: W( y e^y ) = W( x ) y = W( x )

It should be noted the lambert W function has 2 branches because it can only be implicitly graphed since for some x-values, 2 y-values can exist. So be careful if you use the Lambert W function going forward.

So, taking the Lambert W function of both sides leads to:

W[ - ( u - c ) e^{-( u - c )} ] = W( -e^c ) = - ( u - c )

u - c = - W( -e^c )

u = c - W( -e^c )

Finally, re-inputting kx = u

kx = c - W( -e^c )

∴ x = [ c - W( -e^c ) ] / k

An interesting fact of the Lambert W function is that its inputs can NOT be less than - e^-1

y = W( x ) where x ≥ - e^-1

This restricts the solution for x in this problem that c can NOT be greater than -1.

So, real solutions will exist so long as:

c and k are within the set of all real numbers and c ≤ -1 and k can be any value within the set of all real numbers:

- ∞ < c ≤ -1

- ∞ < k < ∞

I hope this helps! Message me in the comments with any questions, comments, or concerns about how I did the above work! Let it be known that my knowledge of the Lamber W function/product log is pretty limited since I've taught myself most of the mechanics of this function.