How do you solve x^2 = log^2(x)

Question

I read a page that said that the limit as $x$ approaches infinity of (polynomial

function)/(logarithmic function) = infinity and that the limit as $x$ approaches

infinity of (logarithmic function)/(polynomial function) = $0$.

So I happened to plug in $y=x/ln(x)$ and $y=ln(x)/x$ in my calculator and the

intersection is a $.5671...$, and when I take the intersection of the ratios

$y$=$(x/ln(x))$/$(ln(x)/x)$ and $y=(ln(x)/x)$/$(x/ln(x))$ I get the same value.

So I

tried to work it out by hand and I got the equations $x=e^x$ (which evaluates to

$1$) and $e^x$=$e/x$ (which also evaluates to $1$).

I was wondering how do I go about dealing with this equation to solve for $x$ and manage to get the value of the point of intersection?

To note is that the value that I keep getting in the Lambert W function, this value is denoted W($1$).

Answer 1

You are looking for the point where $x=-\log x$ or $e^x=\frac 1x$. If you don't want to use the Lambert W function, you will have to solve it numerically. This is a nice root, any method should have no trouble at all. The simplest is bisection. You know (from graphing, say) that the root of $x+\log x=0$ is between $0.5$ (where the left side is negative) and $0.6$ (where the left side is positive. Try $0.55$ and find it is negative, so the root is in $(0.55,0.5)$ Try $0.575$ and so on. Newton-Raphson converges more quickly, but is more work to program. Or you can just ask Alpha.

Answer 2

I assume you're asking how you can see that the solution to $\frac{\log x}{x}=\frac{x}{\log x}$ is $W(1)$.

As you've seen, this simplifies to $x^2=(\log x)^2$, which is the case if either $x=\log x$ or $x=-\log x$. The former of these is impossible, so we must have $$x+\log x = 0$$ Taking the exponential on both sides of this gives $$ e^x x = 1 $$ whose solution for $x$ is the definition of $W(1)$.

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Note, however, that this intersection point does not really relate to your initial paragraph -- the behavior at $x=0.5671$ sheds essentially no light on the behavior for $x\to \infty$, since $\frac{x}{\log x}$ has a singularity at $x=1$ which is between $0.561$ and $\infty$...