Complex Inequality Expression (Independent Study)

[IWhitematter]

While this is not technically an assignment for any particular class (that I'm aware of, at least), I think the nature of this problem makes it suitable for this forum. Please, inform me if I should direct my question elsewhere.

Find x>3 such that ln(x)<x^0.1 (hint: The number is "huge")

At first, I disregarded the hint and picked 3.01, which is a solution, as are an infinite amount of values approaching 3.01. I then rewrote the given relation as 1<e^x^.1/x. After spending a page worth of paper on other algebraic manipulations with no direction, I started to half-guess and came up with 4^26. There are obviously infinite solution possibilities and although I did technically solve this problem, I would like to know how to create an expression that produces the solutions for x to the given inequalities.

I have searched google for the problem and found a variety of responses (and solutions) similar to both of my answers (that is, either very large or near 3.01 to several hundredths or less), however there do not appear to be any sufficiently thorough explanations.

Also note: ln(x) = 2coth^(-1)((1+x)/(x-1)), which may or may not be useful in this scenario.

 

[LCKurtz]

Well, you know that any positive power of x will eventually become larger than ln(x). If you call f(x) = ln(x) and g(x) = x^1/10, consider h(x) = f(x) - g(x).

Then

$$h'(x) = \frac 1 x - \frac 1 {10x^{\frac {9} {10}}}$$

This is obviously 0 when x = 10¹⁰. That is where h(x) is as large as it gets and it will decrease thereafter. You need some numerical method to actually find where the curves cross after that. Maple gives 3.430631121*10¹⁵.