- #1

- 96

- 4

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

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

I thank you in advance

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Kairos
- Start date

- #1

- 96

- 4

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

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

I thank you in advance

- #2

Simon Bridge

Science Advisor

Homework Helper

- 17,857

- 1,655

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

member 587159

Solving for what? What are A and B?

- #4

- 96

- 4

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

- #5

member 587159

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

I get

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

Well, I also get a Ath degree equation.

- #6

Simon Bridge

Science Advisor

Homework Helper

- 17,857

- 1,655

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

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

- 10,706

- 1,728

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.

Share: