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

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Question

${5^{\frac{1}{4}\left( {\log _2^5x} \right)}} \ge 5{x^{\frac{1}{5}\left( {{{\log }_5}x} \right)}}$

Taking logarithm on base $5,$ then

$\frac{1}{

Vedclass · Accepted Answer

Both $(A)$ $\&$ $(B)$

Vedclass · Answer

${5^{\frac{1}{4}\left( {\log _2^5x} ight)}} \ge 5{x^{\frac{1}{5}\left( {{{\log }_5}x} ight)}}$

Taking logarithm on base $5,$ then

$\frac{1}{4}\left(\log _{5}^{2} xight) \geq 1+\frac{1}{5}\left(\log _{5} xight)\left(\log _{5} xight)$

$\Rightarrow \frac{1}{20} \log _{5}^{2} x \geq 1$

or $\left(\log _{5}^{2} xight) \geq 20$

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 $x>0$

$	herefore $ $x \in\left(0,5^{-2 \sqrt{5}}ight] \cup\left[5^{2 \sqrt{5}}, \inftyight)$

The interval of $x$ in which the inequality ${5^{(1/4)(\log _5^2x)}}\, \geqslant \,5{x^{(1/5)(\log _5^x)}}$

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