Solving $1-x^2 = \ln x$ How to solve $1-x^2 = \ln x$ or $x^2 -1 = \ln x$
I know it can simply be checked that 1 is the answer. Even if we say $x = e^{(x^2 -1)}$ it can be easily checked that one is answer but it is not easily solved without checking numbers. or even if we say $1- 2x^2 =\ln x$ then we can't solve this by checking numbers. How this equations can be solved?
 A: We can easily see that for $x=1$ : $1-1^2 = \ln(1) \Leftrightarrow 0=0$, thus the equation is satisfied.
Let's prove that this solution is the only one as well !
$$1-x^2 = \ln(x) \Leftrightarrow \ln(x)+x^2-1 =0$$
Let $g(x) = \ln(x) + x^2 - 1$. Then, the derivative of $g(x)$, will be :
$$g'(x) = \frac{1}{x}+2x$$
Note that $g(x)$ is well defined on $\mathbb R^+ = (0,+\infty)$, which means that for our function, it's $x>0$.
For $x>0$, $g'(x) > 0 $. This means that the function $g(x)$ is strictly increasing at its domain.
Thus, if there is a $x_0 \in D_g : g(x_0)=0$, it will be the only one as well, since the function is strictly increasing.
This means that $x=1$ is the only solution.
Note : This is the only "entry level" way to work around such problems, that are easy to be handled like this.
In general, such problems that cannot be solved in a standard way, need numerical methods, which are methods that via repeating, converge to the solution of the equation given (there are many different methods with better convergence and different characteristics that serve different kind of problems).
