Solving an equation with logarithms

  • Thread starter Kairos
  • Start date
  • Tags
    Logarithms
In summary, Simon tried to solve an equation involving A and B, but he wasn't able to get an analytical solution. He found an approximation using the Lambert W-function.f
  • #1
166
14
Member warned about posting with no effort and without the template
can somebody help me solving:

A*ln(1+A*x)-2*ln(x)-B=0

I thank you in advance
 
  • #2
Have you tried applying the rules for logs:
If ln(a)=b then a=exp(b).
ln(a*b)=ln(a)+ln(b)
ln(a^b)=b*ln(a)
exp(ln(a))=a
 
  • #3
Solving for what? What are A and B?
 
  • #4
for solving a physical problem (about entropy). A and B are constants

Simon, the exponential gives a Ath degree equation.. not more solvable to me
 
  • #5
for solving a physical problem (about entropy). A and B are constants

Simon, the exponential gives a Ath degree equation.. not more solvable to me

I get
for solving a physical problem (about entropy). A and B are constants

Simon, the exponential gives a Ath degree equation.. not more solvable to me

Well, I also get a Ath degree equation.
 
  • #6
This is why you are supposed to say what you've tried already.

OK. You want to find an equation for x in terms of A and B?
You got as far as:

##(1+Ax)^A=e^Bx^2## ... which, just at a glance, I suspect does not have an analytic solution.
ie. taking the Ath root of both sides: ##1+Ax-e^{B/A}x^{2/A}=0##

What was the original problem? Do you have any reason to believe the equation can be solved?
Where does A and B come from?
i.e. Ax<<1:
$$x\in \frac{-A^2\pm\sqrt{A^4-4e^B}}{2e^B}$$
 
  • #7
can somebody help me solving:

A*ln(1+A*x)-2*ln(x)-B=0

I thank you in advance

I doubt that the equation can be solved in "closed-from", via some type of formula involving A and B. If you are given numerical values for A and B you can solve the equation numerically, using one of the many reasonably effective methods available. If ##|A|## is small we have the approximation that ##A \ln(1+Ax) \approx A^2 x##, and using that approximation in the equation leads to the (approximate) solution
[tex] x \approx \exp \left( -W\left( -\frac{1}{2} A^2 e^{B/2} - \frac{1}{2}B \right) \right), [/tex]
where ##W## is the so-called Lambert W-function. It is a non-elementary function: ##W(x)## is the solution of ##x = W(x) e^{W(x)}##, and is implemented in several computer algebra systems such as Maple or Mathematica.
 

Suggested for: Solving an equation with logarithms

Replies
18
Views
656
Replies
8
Views
527
Replies
4
Views
782
Replies
3
Views
681
Replies
9
Views
993
Replies
10
Views
656
Replies
2
Views
326
Replies
42
Views
2K
Back
Top