How to solve: $x+\ln(x^2-1)=0$

Question

I'm trying to solve the following equation for a couple of hours with no success.

$x+\ln(x^2-1)=0$

I'm trying to find x.

I tried playing with logarithmic identities to transform it to something simpler, but it didn't lead me to an answer.

I would appreciate your help.

Thanks.

There is a solution, by intermediate value theorem. but I cannot find it analytically. — mez, Feb 23 '13 at 19:45
@GitGud I updated the question to make it clearer. I'm looking for x. — Einat, Feb 23 '13 at 20:05
@Einat Ok, I deleted my comment. I don't think there's a way to find $x$ unless you do it numerically, that is, find an approximate value for $x$. — Git Gud, Feb 23 '13 at 20:06
Introducing the lambert W I arrive at $ x=W(x/(x^2-1))$ ... but still only in the form of a fixed-point formula... — Gottfried Helms, Feb 24 '13 at 02:52

DasKrümelmonster · Answer 1 · 2013-02-24T01:25:16.657

1

Taking the easy route here: http://www.wolframalpha.com/input/?i=solve+x%2Bln%28x%5E2%E2%88%921%29%3D0

There is no analytic solution. (Or else, Wolfram Alpha would get it)

edited Feb 24 '13 at 01:25

answered Feb 23 '13 at 19:58

DasKrümelmonster

214

3

Are you sure Wolfram Alpha would get it? – Git Gud Feb 23 '13 at 20:01
Yes, like 99.999% sure. Similar equations pose no problem and result in an analytic solution. – DasKrümelmonster Feb 23 '13 at 20:16
and how exactly did you rearrange it to $x^2 e^x = -1$? – J.H. Feb 24 '13 at 01:02
Ähhmm ... by mistake. I should get some sleep... – DasKrümelmonster Feb 24 '13 at 01:23

Shreya · Answer 2 · 2013-02-24T00:04:49.820

1

The above equation gives:

$$e^x=\frac{1}{x^2-1}$$

I suppose you know the graph of $f(x) =e^x$,now if you know how to draw the graph of $g(x) =\frac{1}{x^2-1}$ using its derivatives then, the value of $x$ corresponding to the point of intersection of the two curves on both sides of equality is the required value of $x$.

edited Feb 24 '13 at 00:04

answered Feb 23 '13 at 20:14

Shreya

3,035

Shouldn't it be $e^{-x}$? – Daryl Feb 23 '13 at 21:26
1

thanks for notifying my error. – Shreya Feb 24 '13 at 00:05

score 1 · Answer 3 · answered Feb 24 '13 at 00:58

Set $f(x)=-\ln(x^2-1)$ then we ask for its fixpoint $x = f(x) = f(f(x)) = ... $ . Someone proposed to use Newton/Raphson, and although I think it is the perfect one (because of its rate of convergence), here is a simpler iterative method.

a) guess some range for a possible solution: To have the log having a positive argument, a positive x must be greater than 1, but then to have the log negative (so that -log(...) is positive as x) x^2-1 must be smaller than 1, so x must be somewhere in the range $1 \ldots \sqrt2 $

b) do iteration: Now with some initial value for $x_0$ from this range, say $x_0=1.1$ iterating $x_{k+1}=f(x_k)$ diverges quickly. Thus we must find the inverse function to $f(x)$ which is $$g(x)=\sqrt{\exp(-x)+1}$$ Then we get the solution simply by initializing $x_0=1.1$ and iterating $x_{k+1}=g(x_k)$ and $x=\lim_{k \to \infty} x_k$

How to solve: $x+\ln(x^2-1)=0$

3 Answers3