# Homework Help: Solving an equation with logarithms

1. May 15, 2016

### 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

2. May 15, 2016

### Simon Bridge

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. May 15, 2016

### Math_QED

Solving for what? What are A and B?

4. May 15, 2016

### Kairos

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. May 15, 2016

### Math_QED

I get
Well, I also get a Ath degree equation.

6. May 15, 2016

### Simon Bridge

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. May 15, 2016

### Ray Vickson

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