The problem of x = ln(x)

Question

I am trying to find x values for points along the normal distribution curve, and I ended up with a problem that goes back to the method of solving $x = \ln x$. Right now, I have $\ln(a \mu) - \ln(10) = \frac {-1} {2a^2} $. This is not something I know how to solve. I tried Googling this type of problem and nothing came up. How does one solve for a variable both inside and outside of a natural logorithm, specifically in the example problem above? Could anyone explain how this could be done?

Answer 1

This kind of problem requires the use of the Lambert W function, which cannot be expressed with regular algebra but can be approximated with something like Newton's Method. However, since $x$ is linear and $\ln(x)$ is logarithmic (and at $x=1$, $\ln(x)$ is only at $0$ while $x$ is at $1$, and linear functions increase faster than logarithmic functions), there are no solutions for $x$.

Answer 2

There are no real numbers $x$ such that $x=\ln(x)$.

First of all, note that any such $x$ would need to be positive or $\ln(x)$ is not even defined. Since $x$ would need to be positive, so would $\ln(x)$, so that means $x$ would have to be greater than $1$.

Now consider two functions $f$ with $f(x)=x$ and $g$ with $g(x)=\ln(x)$. At input $1$, $f$ has a larger output: $f(1)=1$ and $g(1)=0$.

Forever onward (for all $x>1$) $f$ has a steeper slope: $f'(x)$ is constantly $1$, but $g'(x)=\frac1x$, is less than $1$. So the function $g$ that is already giving a lower output at input $1$ can never catch up with $f$ so that they have equal outputs.

You can get a great deal of clarity just by looking at the graphs of $y=x$ and $y=\ln(x)$ and observing that they never touch.

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The equation $\ln(a\mu)-\ln(10)=\frac{-1}{2a^2}$ can be solved for $a$ numerically (to arbitrarily high decimal precision) using Newton's Method. Define $$f(a)=\ln(a\mu)-\ln(10)+\frac{1}{2a^2}$$ and we have $$f'(a)=\frac1a-\frac{1}{a^3}$$

Start with a decent guess $a_0$ for a solution, and then keep iterating $$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$$ and the $\{a_n\}$ will converge to a solution. It generally converges fast, so you do not need too many iterations to get high decimal precision as long as you choose $a_0$ with a grain of care. Of course, you have to have a numerical value for $\mu$ to do this.