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


The next logical step seemed to be getting both $x$s on the same side of the inequality so I substituted $\log_{2}{x}=\frac{\ln{x}}{\ln{2}}$


and then multiplied both sides by $\frac{\ln{2}}{500x}$ to obtain


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.

  • $\begingroup$ I don't think you can analytically solve for those points $a\approx 1.00139, b\approx 6311.93$. Those were likely obtained numerically. However, you can structurally prove that there will be intervals $0<x<a$ and $x>b$, since you can take a derivative of the function $f(x)=\log(x)/x$ to show that $f$ increases for $0<x<1$, decreases for $x>1$, and $f'(1)=0$. $\endgroup$ – Michael Jan 16 '18 at 2:32

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


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