Solving a certain equation involving logarithms

**bitrex** · Jun21-09, 11:20 PM

I'm reading book called "Prime Obession" which attempts to give a layperson's introduction to the Riemann Hypothesis. In laying the groundwork in one of the early chapters the author is explaining the fact that the function log(x) increases more slowly in total than x raised to any power. For example, log(x) crosses the function x^0.1 somewhere less than e^e, and then falls below it again somewhere in the quadrillions. I'm wondering how one would go about solving the equation to find the Y axis intercepts? I want to find where log(x) = x^0.1, in trying to simplify the problem I end up with log(x)/x = e^0.1, but I'm not able to go any farther. The equation obviously has two solutions, so I imagine there's a polynomial involved somewhere? Any advice would be appreciated.

**CRGreathouse** · Jun21-09, 11:37 PM

Originally Posted by bitrex

For example, log(x) crosses the function x^0.1 somewhere less than e^e, and then falls below it again somewhere in the quadrillions. I'm wondering how one would go about solving the equation to find the Y axis intercepts? I want to find where log(x) = x^0.1, in trying to simplify the problem I end up with log(x)/x = e^0.1, but I'm not able to go any farther. The equation obviously has two solutions, so I imagine there's a polynomial involved somewhere? Any advice would be appreciated.

log(x)/x = e^0.1 is not the same as log(x) = x^0.1.

Generally, equations like this have no easy solutions (though they can be solved in terms of a special function called Lambert's W). The best way is probably by numerical methods:

Code:

gp> solve(x=1,9,log(x)-x^.1)
time = 0 ms.
%1 = 3.0597266796208088546065494702258610157
gp> solve(x=1e15,1e16,log(x)-x^.1)
time = 0 ms.
%2 = 3430631121407801.2027753365093892641824

**bitrex** · Jun22-09, 12:56 AM

Yes, you're right. I made a mistake in playing around with the equation: log(x^0.1) != log(x)^0.1 . I considered that numerical methods would probably be a way to solve it; I just wasn't sure if there were some trick that I was missing. Thanks so much for your speedy reply!

**Mentallic** · Jun22-09, 06:27 AM

This is equivalent to the idea that any function

for all values a>1 will rise faster than any polynomial of nth degree such as

n very large.

**g_edgar** · Jun24-09, 03:57 PM

log(x) = x^(0.1)
exact solution is not possible in terms of your standard schoolbook functions
modern CASs like Maple will solve this using the Lambert W function...

x = 10000000000 (W(-0.1))¹⁰