The interval of x in which the inequality 5^{ | vedclass

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(A) Given inequality: $5^{\frac{1}{4}(\log_5 x)^2} \geq 5 \cdot x^{\frac{1}{5}(\log_5 x)}$Taking $\log_5$ on both sides:$\frac{1}{4}(\log_5 x)^2 \geq

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$(0, 5^{-2\sqrt{5}}] \cup [5^{2\sqrt{5}}, \infty)$

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(A) Given inequality: $5^{\frac{1}{4}(\log_5 x)^2} \geq 5 \cdot x^{\frac{1}{5}(\log_5 x)}$Taking $\log_5$ on both sides:$\frac{1}{4}(\log_5 x)^2 \geq \log_5(5 \cdot x^{\frac{1}{5}\log_5 x})$$\frac{1}{4}(\log_5 x)^2 \geq 1 + \frac{1}{5}(\log_5 x)^2$Let $u = \log_5 x$. Then $\frac{1}{4}u^2 \geq 1 + \frac{1}{5}u^2$$(\frac{1}{4} - \frac{1}{5})u^2 \geq 1$$\frac{1}{20}u^2 \geq 1 \implies u^2 \geq 20$$u \geq 2\sqrt{5}$ or $u \leq -2\sqrt{5}$$\log_5 x \geq 2\sqrt{5} \implies x \geq 5^{2\sqrt{5}}$$\log_5 x \leq -2\sqrt{5} \implies x \leq 5^{-2\sqrt{5}}$Since $x > 0$,the solution is $x \in (0, 5^{-2\sqrt{5}}] \cup [5^{2\sqrt{5}}, \infty)$.

The interval of $x$ in which the inequality $5^{\frac{1}{4}(\log_5 x)^2} \geq 5x^{\frac{1}{5}(\log_5 x)}$ holds is:

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