Why the approximate solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = c$ is $\sqrt {2\log n} $ when $n$ is large

Question

I find in a book that when $n$ is large, the approximate solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = c$, denoted by $x(n,c)$, is about

$x(n,c) \approx\sqrt {2\log n} $

And further the book states the solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = 1$ is

$x(n,1) = \sqrt {2\log n} (1 - \frac{{\log \log n}}{{4\log n}} - \frac{{\log (2\sqrt \pi )}}{{2\log n}} + o(\frac{1}{{\sqrt {\log n} }}))$

Need help with why this is true. Following is the book contents related to this, and I highlight the part.

Answer 1

The initial equation can be written

$$e^{-x^2/2-\log x}=\frac{\sqrt{2\pi}c}n,$$

and taking the cologarithm,

$$\frac{x^2}2+\log x=\log n-\log\sqrt{2\pi}c.$$

For very large $n$, this can be approximated by

$$\frac{x^2}2=\log n.$$

Below, a plot of $\dfrac{x^2}2+\log x$ vs. $\dfrac{x^2}2$.