c
\({5^{\frac{1}{4}\left( {\log _2^5x} \right)}} \ge 5{x^{\frac{1}{5}\left( {{{\log }_5}x} \right)}}\)

Taking logarithm on base \(5,\) then

\(\frac{1}{4}\left(\log _{5}^{2} x\right) \geq 1+\frac{1}{5}\left(\log _{5} x\right)\left(\log _{5} x\right)\)

\(\Rightarrow \frac{1}{20} \log _{5}^{2} x \geq 1\)

or \(\left(\log _{5}^{2} x\right) \geq 20\)

or \(\log _{5} x \geq 2 \sqrt{5}\) and \( \log _{5} x \leq-2 \sqrt{5}\)

or \(x \geq 5^{2 \sqrt{5}} \) and \( x \leq 5^{-2 \sqrt{5}}\)

But \(x>0\)

\(\therefore \) \(x \in\left(0,5^{-2 \sqrt{5}}\right] \cup\left[5^{2 \sqrt{5}}, \infty\right)\)