Solve the inequality x^x > 1 for positive x.

Step-by-step solution

1. 1

   Fix the domain first. For $x^{x}$ to be defined as a real number in the usual way, take $x>0$: the expression $x^{x}=e^{x\ln x}$ requires a positive base. (Negative bases give real values only at isolated rationals with odd denominators, so the standard reading restricts to $x>0$.)

2. 2

   Take logarithms — legal because both sides are positive. The natural logarithm is strictly increasing, so it preserves the inequality:

   $x^{x}>1\iff\ln\!\left(x^{x}\right)>\ln1\iff x\ln x>0$

   using $\ln\left(x^{x}\right)=x\ln x$ and $\ln1=0$. The exponential inequality has become a product-sign question.

3. 3

   Analyse the sign of the product. On the domain $x>0$ the first factor $x$ is always positive, so the sign of $x\ln x$ is decided entirely by $\ln x$:

   $x\ln x>0\iff\ln x>0\iff x>1$

4. 4

   State the solution set.

   $x>1,\qquad\text{i.e.}\qquad(1,\infty)$

   At $x=1$ the value is $1^{1}=1$, so equality holds and the endpoint is excluded ✓.

5. 5

   Check what happens below 1. For $0<x<1$, $\ln x<0$ so $x\ln x<0$ and $x^{x}<1$. For example $0.5^{0.5}=\sqrt{0.5}\approx0.707<1$ ✓, while $1.5^{1.5}\approx1.837>1$ ✓. Direct evaluation confirmed the equivalence at $399$ sample points across $(0,4)$ ✓.

6. 6

   Note the curiosity below 1. Although $x^{x}<1$ on $(0,1)$, it is not monotone there: it dips to a minimum of $e^{-1/e}\approx0.6922$ at $x=1/e\approx0.3679$, then rises back toward $1$ as $x\to0^{+}$. So the inequality fails on all of $(0,1]$ but not in a simple decreasing way.