what's infinity?

$\begin{aligned} \ln (2x^2) - 2x^2 + 1 & = 0 & \small \color{#3D99F6} \text{Let }u = 2x^2 \\ \ln u & = u - 1 & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }u \\ \frac 1u & = 1 \\ \implies u & = 1 \\ 2x^2 & = 1 \\ x & = \frac 1{\sqrt 2} \approx \boxed{0.707} & \small \color{#3D99F6} \text{For }x > 0 \end{aligned}$

Let $\ f(x) = Ln(2x^2) \ and \ g(x) = 2x^2 - 1$ . Then: $\ f'(x) = \frac{2}{x} \ and \ g'(x) = 4x$ $\ f'(x) = g'(x) \Rightarrow\ x = \frac{1}{\sqrt{2}}$ Now plugging this value into both functions gives $\ f( \frac{1}{\sqrt{2}} ) = g( \frac{1}{\sqrt{2}} )$ Finally, $\ f(x) < g(x) \iff |x| > \frac{1}{\sqrt{2}}$ and $\ f(x) < g(x) \iff |x| < \frac{1}{\sqrt{2}}$ . Hence, $\ x = \frac{1}{\sqrt{2}} = 0.707..$ is a unique $\ positive$ solution. ( You may also want to refer to a sketch where the two functions are mutually tangent at $\ x = \pm \frac{1}{\sqrt{2}}$ )

Letting $\alpha = 2x^{2},$ a root of $f(x)$ must satisfy the equation

$\ln(\alpha) - \alpha + 1 = 0 \Longrightarrow \alpha - \ln(\alpha) = 1.$

By observation we see that $\alpha = 1$ is a solution, which in turn gives us

$x = \sqrt{\dfrac{\alpha}{2}} = \dfrac{1}{\sqrt{2}}.$

where we took the positive root as specified. Is this solution unique? We look at the function $g(\alpha) = \alpha - \ln(\alpha).$ The critical points will be such that

$\dfrac{dg}{d\alpha} = 1 - \dfrac{1}{\alpha} = 0 \Longrightarrow \alpha = 1.$

Now since $\dfrac{d^{2}g}{d\alpha^{2}} = \dfrac{1}{\alpha^{2}} \gt 0$ for $\alpha = 1,$ (for all $\alpha$ in fact), we see that, by the second derivative test, $g(\alpha)$ has a minimum value of $g(1) = 1 - \ln(1) = 1.$ From this we can conclude that $x = \dfrac{1}{\sqrt{2}} = \boxed{0.707....}$ is the only positive real root of the given function $f(x).$