Question Number 146680 by mathdanisur last updated on 14/Jul/21
$${Compare}:\:\:\mathrm{100}^{\mathrm{101}} \:\:{and}\:\:\:\mathrm{101}^{\mathrm{100}} \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
$$\mathrm{log}_{\mathrm{10}} \mathrm{101}\:=\:\mathrm{2}+\mathrm{log}_{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:=\:\mathrm{200}+\mathrm{100log}_{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{200}+\mathrm{100}×\frac{\mathrm{1}}{\mathrm{100}} \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{201}\:<\:\mathrm{202}\:=\:\mathrm{101}×\mathrm{2}\:=\:\mathrm{101log}_{\mathrm{10}} \mathrm{100} \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{101log}_{\mathrm{10}} \mathrm{100} \\ $$$$\mathrm{101}^{\mathrm{100}} \:<\:\mathrm{100}^{\mathrm{101}} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thankyou}\:{Ser}\:{cool} \\ $$
Answered by mr W last updated on 15/Jul/21
$${y}={x}^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{y}=\frac{\mathrm{ln}\:{x}}{{x}} \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{1}−\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\ $$$${y}'=\frac{{x}^{\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{ln}\:{x}\right)<\mathrm{0}\:{for}\:{x}>{e} \\ $$$${that}\:{means}\:{if}\:{x}>{e},\:{y}={x}^{\frac{\mathrm{1}}{{x}}} \:{is} \\ $$$${strictly}\:{decreasing}. \\ $$$$\Rightarrow\mathrm{100}^{\frac{\mathrm{1}}{\mathrm{100}}} >\mathrm{101}^{\frac{\mathrm{1}}{\mathrm{101}}} \\ $$$$\Rightarrow\mathrm{100}^{\frac{\mathrm{101}}{\mathrm{100}}} >\mathrm{101} \\ $$$$\Rightarrow\mathrm{100}^{\mathrm{101}} >\mathrm{101}^{\mathrm{100}} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thankyou}\:{Ser}\:{cool} \\ $$
Commented by mr W last updated on 15/Jul/21
$${applying}\:{this}\:{we}\:{can}\:{get} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt[{\mathrm{4}}]{\mathrm{4}}>\sqrt[{\mathrm{5}}]{\mathrm{5}}>… \\ $$$${since}\:\sqrt[{\mathrm{4}}]{\mathrm{4}}=\sqrt{\mathrm{2}},\:{so}\:{we}\:{get}\:{also} \\ $$$$\sqrt{\mathrm{2}}<\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$