# Solving an equation with logarithms

• Kairos
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

#### Kairos

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

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

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.

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:
$$x\in \frac{-A^2\pm\sqrt{A^4-4e^B}}{2e^B}$$

can somebody help me solving:

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

$$x \approx \exp \left( -W\left( -\frac{1}{2} A^2 e^{B/2} - \frac{1}{2}B \right) \right),$$