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# The equation \( (x^{x})^{x} = 2 \) is satisfied when x is equal to?

## 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, \( x = \frac{1}{10} \)

=> \( x = \frac{1}{10} = 0.1 \)