If {1 over 2} le {log _{0.1}}x le 2 then | vedclass

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(d) ${1 \over 2} \le {\log _{0.1}}x \le 2$

${1 \over 2} \le {\log _{0.1}}\,x \Rightarrow {\log _{0.1}}{(0.1)^{1/2}} \le {\log _{0.1}}x$

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(d) ${1 \over 2} \le {\log _{0.1}}x \le 2$

${1 \over 2} \le {\log _{0.1}}\,x \Rightarrow {\log _{0.1}}{(0.1)^{1/2}} \le {\log _{0.1}}x$

$ \Rightarrow $${(0.1)^{1/2}} \ge x$ $ \Rightarrow $$x \le {1 \over {\sqrt {10} }}$

${\log _{0.1}}x \le 2 \Rightarrow {\log _{0.1}}x \le {\log _{0.1}}{(0.1)^2}$

$x \ge {(0.1)^2} \Rightarrow x \ge {1 \over {100}}$, ${1 \over {100}} \le x \le {1 \over {\sqrt {10} }}$.

Hence, ${x_{{\rm{max}}}} = {1 \over {\sqrt {10} }},{x_{{\rm{min}}{\rm{.}}}} = {1 \over {100}}$.

If ${1 \over 2} \le {\log _{0.1}}x \le 2$ then

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