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.
$\endgroup$ 73 Answers
$\begingroup$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 $x$.
$\endgroup$ 2 $\begingroup$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 $x=\lim_{k \to \infty} x_k$
$\endgroup$ $\begingroup$Taking the easy route here:
There is no analytic solution. (Or else, Wolfram Alpha would get it)
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