Determine all positive roots of the equation xx=1√2.
Problem
Source: IMO LongList 1967, Sweden 1
Tags: algebra, equation, roots, IMO Shortlist, IMO Longlist, Exponential equation
16.12.2004 22:00
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions
08.01.2006 18:55
Taking ln both sides, we have that xlnx=12⋅ln12⇒x12=ln12lnx but since a≥b⇒lna≥lnb, we must have one of max, so we must have x=\frac{1}{2}
08.01.2006 23:57
The solution of campos is not sufficient ! It's easy to see that x_1=\frac 14,\ x_2=\frac 12 are solutions . There are not other solutions because of the function f(x)=x^x is decreasing on ]0,\frac 1e] and increasing on [\frac 1e,+\infty[
09.01.2006 00:12
i thought this can't be so easy..., isn't f(x)=\ln x strictly increasing on \mathbb{R^+}?
03.02.2006 20:42
i think i've found a solution... hope it's correct this time... as in the previous post we have x^x=\frac{1}{\sqrt 2} \Leftrightarrow x \ln x= \frac{1}{2}\ln \frac{1}{2} now, let x=\left(\frac{1}{2}\right)^k, then, x \ln x= \frac{1}{2}\ln \frac{1}{2} \Rightarrow \left(\frac{1}{2}\right)^k\ln \left(\frac{1}{2}\right)^k=\frac{1}{2}\ln \frac{1}{2} \\\\\Rightarrow 2^k=2k but, since f(x)=2^x is an exponential function and g(x)=2x is a linear function, we must have f(x) and g(x) intersect in at most two points, namely k=1 and k=2... so we must have x=\frac{1}{2} and x=\frac{1}{4} are the solutions to the equation.
27.02.2006 01:32
How do you know that x must be a power of 1/2?
27.02.2006 01:36
As I see it, k is not necessary an integer...
06.04.2020 07:22
Derivating, we have that (x^x)’ = x^x(\ln x + 1). Therefore, x^x is decreasing when 0 < x < \frac{1}{e} and increasing when x > \frac{1}{e}. It follows that there can be at most 2 solutions. After some trial and error, we see that \frac{1}{4} and \frac{1}{2} are solutions and by the aforementioned reasoning, they are the only solutions. \blacksquare
15.08.2023 00:32
We can see that x^x is decreasing when 0 < x < \frac{1}{e}, and increasing when x> \frac{1}{e}, I used trial and error, to get that \frac{1}{4} and \frac{1}{2}, are the only solutions. \square