Solving $x^2+ \ln^2(x)=1$ Solve $$x^2+ \ln^2(x)=1$$
Graph of the given equation

From the graph or simply putting $x=1$ satisfy the equation one of the solution is $1$ . How I can find the other solution of the given equation $?$
 A: First of all, this isn't a quadratic. Secondly, this is a type of an equation that cannot be solved explicitly, that is, to find a closed form expression for $x$. The only way to work such equations around is using a Numerical Analysis approach with a preffered method.
Apart from that, indeed one can see that $x=1$ trully satisfies the equation $x^2 + \ln^2 x - 1 = 0$. One could study the behavior of the function $f(x)$ yielded by that expression, as to know where the other solution is around. 
Let $f(x) = x^2 + \ln^2 x - 1$. Then, $f'(x) = 2x - \frac{2\ln x}{x}$. Now, investigate :
$$f'(x) = 0 \Rightarrow 2x - \frac{2 \ln x}{x} = 0 \Rightarrow 2x^2 - 2\ln x =0$$
We've again come across a similar, non-explicitly solvable equation. Working now, as :
$$f''(x) = 0 \Rightarrow 4x - \frac{2}{x} = 0 \Rightarrow 4x^2 - 2 = 0 \Leftrightarrow x = \frac{\sqrt{2}}{2}$$
Note that we excluded the negative part of the solution, as $x >0$ since we are working with a logarithm expression. That, now, can give you a hint on where to look for the other solution and thus how to manipulate/start your preffered numerical method.
A: First of all, $x=1$ is a precise answer. Since this equation can't be solved analytically, instead I considered solving $$x_{n+1}=e^{-\sqrt{1-x_n^2}}$$recursively using Python and by having $x_0=0.1$ I obtained $$x\approx 0.4$$
A: Well, it's simple
Take $\ln x =k$
So $e^k=x$
Now plugin these equations in ur equation
You'll get $e^{2k}+k^2-1=0$
Now use maclaurin's expansion for $e^{2k}$
So the equation becomes 
$$2x+3x^2+\frac{8x^3}{3!} + \frac{16x^4}{4!}+\frac{32x^5}{5!}+\frac{64x^6}{6!}+\frac{128x^7}{7!}=0$$
Used upto degree 7 to get more precise answer
Solve this equation you'll get $k=0,-0.918$
Now $x=e^0=1$ and $x=e^{-0.918}=3.993 \approx 0.4$
