While working on a homework assignment, I came across the following problem:
Problem: Give the values of $x$ for which $x>500\lg{x}$.
Solutions: $0<x<1.00139$, $x>6311.93$
In spite of having the solutions to this problem, I am having a hard time making progress towards solving it myself. I started by substituting $\lg{x}=\log_{2}{x}$ to obtain
$$x>500\log_{2}{x}.$$
The next logical step seemed to be getting both $x$s on the same side of the inequality so I substituted $\log_{2}{x}=\frac{\ln{x}}{\ln{2}}$
$$x>500\frac{\ln{x}}{\ln{2}}$$
and then multiplied both sides by $\frac{\ln{2}}{500x}$ to obtain
$$\frac{\ln{x}}{x}<\frac{\ln{2}}{500}.$$
This is pretty much where my progress ends. I've tried applying the various logarithmic identities to get rid of $\ln{x}$ but to no avail. A push in the right direction would be much appreciated.
Multiply both sides of the inequality be $500x$ and take the exponent of both sides, giving $$e^{500\ln x}<e^{\ln2\cdot x}$$ Then use the identity $a^{b\cdot c}=\left(a^b\right)^c$, giving $$\left(e^{\ln x}\right)^{500}<\left(e^{\ln2}\right)^x$$ now since $e^{\ln x}=x$, we can simplify to $$x^{500}<2^x$$ You can't find the answer numerically, but you can use the Lambert W function to continue. The Lambert W function is the inverse of $x\cdot e^x$ or finding $x$ if $z=x\cdot e^x$ and $z$ is known. This explains how to manipulate the algebra then apply the Lambert W fanction