Solving an equation with logarithms

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Homework Help Overview

The discussion revolves around solving the equation A*ln(1+A*x)-2*ln(x)-B=0, which involves logarithmic functions. The context is related to a physical problem concerning entropy, with A and B being constants.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of logarithmic rules and question the nature of the constants A and B. There are attempts to reformulate the equation and explore its solvability, with some suggesting that it may not have an analytic solution.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided insights into numerical methods for solving the equation, while others express skepticism about finding a closed-form solution.

Contextual Notes

There is mention of approximations that may apply when |A| is small, and the discussion includes references to the Lambert W-function as a potential tool for finding solutions under certain conditions.

Kairos
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can somebody help me solving:

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

I thank you in advance
 
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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
 
Solving for what? What are A and B?
 
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
 
Kairos said:
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
Kairos said:
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.
 
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:
$$x\in \frac{-A^2\pm\sqrt{A^4-4e^B}}{2e^B}$$
 
Kairos said:
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.
 

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