fskilnik@GMATH wrote:
$${x^{1 - x}} = 4\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^{ - \left( {1 - x} \right)}} = {4^{ - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^{x - 1}} = {1 \over 4}\,\,\,\,\,\,\,\left( * \right)$$
$$?\,\, = \,\,{x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\,\, = \,\,{x^A}\,\,\,\,\,\,\,\,\,\,\,\,\left( {A = {x^x}\,\, - \,\,{{{x^x}} \over x}} \right)$$
$$A\,\, = \,\,{x^x} - {x^{x - 1}}\,\, = \,\,{x^{x - 1}}\left( {x - 1} \right)\,\,\mathop = \limits^{\left( * \right)} \,\,{{x - 1} \over 4}$$
$$?\,\, = \,\,{x^{{{x - 1} \over 4}}}\,\, = \,\,\root 4 \of {{x^{x - 1}}} \,\,\mathop = \limits^{\left( * \right)} \,\,\root 4 \of {{1 \over 4}} \,\, = \,\,\root {2 \cdot 2} \of {{{\left( {{1 \over 2}} \right)}^2}} \,\, = \,\,\sqrt {0.5} $$
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
