- Home
- Standard 11
- Mathematics
અસમતા ${5^{(1/4)(\log _5^2x)}}\, \geqslant \,5{x^{(1/5)(\log _5^x)}}$ નો ઉકેલ ગણ મેળવો
$\left( {0,{5^{ - 2\sqrt 5 }}} \right]$
$\left[ {{5^{2\sqrt 5 }},\infty } \right)$
બંને $(A)$ $\&$ $(B)$
$(0, \infty )$
Solution
${5^{\frac{1}{4}\left( {\log _2^5x} \right)}} \ge 5{x^{\frac{1}{5}\left( {{{\log }_5}x} \right)}}$
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 $ $x \in\left(0,5^{-2 \sqrt{5}}\right] \cup\left[5^{2 \sqrt{5}}, \infty\right)$
