What is the solution for x ; such that x.Log(1 + a/x) = b ?
Currently, I am calculating the value of x by substituting a range of values for x in the LHS and find the minimum deviation from b.
Anybody can help me to express x as an algebraic solution? i.e., how to express the solution for the equation x.Log(1 + a/x) = b as x = function (a,b) ?
You can't get an analytic solution in terms of elementary functions. However, there is a solution:
x = -b/W(-b/a) where W is the Lambert W function. The Lambert function isn't elementary so you cannot express it in terms of "things you know", but it's a well-studied function and in MATLAB you can use lambertw, so
x= -b/lambertw(-b/a)
will approximate the solution using a computational approximation to W.
(Note: as pointed out, this is wrong and solves x log(a/x) = b)
Hi Chris Rackauckas, Thanks for the reply.
I have a doubt.
The solution is given by you, x= -b/lambertw(-b/a) , will work for all values in natural number?
I have a practical problem from my research : x.Log(1+25/x) = 21.3767. Expected value of x=70.
but; x= -b/lambertw(-b/a) is not giving me the expected answer. Should I considor anything more in the expression?
Hi Chris Rackauckas,
I think the solution given by you is for the expression: x.Log( a/x) = b ; but I need the expression for x.Log(1 + a/x) = b
My bad, missed the 1. With the 1 in there I don't know if you can get an algebraic expression. You can get an approximate one by expanding at x=infty to give
log(1+a/x) = a - a^2 / 2x + a^3 / 3x^2 -a^4 / 4x^3 + O(1/x^4)
You can take this series out further (all the way to 1/x^5), and then multiply by the highest value in the denominator. At 1/x^5 this will give a quartic equation which you can solve (with Mathematica), whereas anything higher is probably not solvable. This is an approximation, but a pretty good one if x is large.
If you cannot use a large x approximation, you can take a series at a value much closer to the range of values you're considering, say x=100 Taylor series. However, this formula will be very messy and will require Mathematica or some other CAS to solve it. And in the end it will still be an approximation.
If it's a good enough approximation, then you can use it as an initial condition to a Newton's method or some other nonlinear solver to refine it closer to the true value.
Again, sorry for the mistake above!
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