Solution

Taking log on both sides:
\( \log(x^x) = \log(2) \)
=> \( x \log(x) = \log(2) \) as \( \log(x^n) = n \log(x) \)
Differentiating with respect to \( x \):
\( \frac{d}{dx}(x \log(x)) = \frac{d}{dx}(\log(2)) \)
According to differentiation product rule:
\( x \cdot \frac{d}{dx}(\log(x)) + \log(x) \cdot \frac{d}{dx}(x) = 0 \) as \( \frac{d}{dx}(\text{constant}) = 0 \) and \( \log(2) \) is a constant.
\( x \cdot \frac{1}{x} + \log(x) \cdot 1 = 0 \) as \( \frac{d}{dx}(\log(x)) = \frac{1}{x} \) and \( \frac{d}{dx}(x) = 1 \)
\( 1 + \log(x) = 0 \)
=> \( \log(x) = -1 \)
=> \( x = 10^{-1} \) This is also a formula that is if \( \log(a) = b \) => \( a = 10^b \)
So, \( x = \frac{1}{10} \)
=> \( x = \frac{1}{10} = 0.1 \)