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Question Number 146680 by mathdanisur last updated on 14/Jul/21

Compare: 100^(101) and 101^(100)

$${Compare}:\:\:\mathrm{100}^{\mathrm{101}} \:\:{and}\:\:\:\mathrm{101}^{\mathrm{100}} \\ $$

Answered by Olaf_Thorendsen last updated on 14/Jul/21

log_(10) 101 = 2+log_(10) (1+(1/(100))) 100log_(10) 101 = 200+100log_(10) (1+(1/(100))) 100log_(10) 101 < 200+100×(1/(100)) 100log_(10) 101 < 201 < 202 = 101×2 = 101log_(10) 100 100log_(10) 101 < 101log_(10) 100 101^(100) < 100^(101)

$$\mathrm{log}_{\mathrm{10}} \mathrm{101}\:=\:\mathrm{2}+\mathrm{log}_{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:=\:\mathrm{200}+\mathrm{100log}_{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{200}+\mathrm{100}×\frac{\mathrm{1}}{\mathrm{100}} \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{201}\:<\:\mathrm{202}\:=\:\mathrm{101}×\mathrm{2}\:=\:\mathrm{101log}_{\mathrm{10}} \mathrm{100} \\ $$$$\mathrm{100log}_{\mathrm{10}} \mathrm{101}\:<\:\mathrm{101log}_{\mathrm{10}} \mathrm{100} \\ $$$$\mathrm{101}^{\mathrm{100}} \:<\:\mathrm{100}^{\mathrm{101}} \\ $$

Commented by mathdanisur last updated on 15/Jul/21

thankyou Ser cool

$${thankyou}\:{Ser}\:{cool} \\ $$

Answered by mr W last updated on 15/Jul/21

y=x^(1/x) ln y=((ln x)/x) ((y′)/y)=((1−ln x)/x^2 ) y′=(x^(1/x) /x^2 )(1−ln x)<0 for x>e that means if x>e, y=x^(1/x) is strictly decreasing. ⇒100^(1/(100)) >101^(1/(101)) ⇒100^((101)/(100)) >101 ⇒100^(101) >101^(100)

$${y}={x}^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{y}=\frac{\mathrm{ln}\:{x}}{{x}} \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{1}−\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\ $$$${y}'=\frac{{x}^{\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{ln}\:{x}\right)<\mathrm{0}\:{for}\:{x}>{e} \\ $$$${that}\:{means}\:{if}\:{x}>{e},\:{y}={x}^{\frac{\mathrm{1}}{{x}}} \:{is} \\ $$$${strictly}\:{decreasing}. \\ $$$$\Rightarrow\mathrm{100}^{\frac{\mathrm{1}}{\mathrm{100}}} >\mathrm{101}^{\frac{\mathrm{1}}{\mathrm{101}}} \\ $$$$\Rightarrow\mathrm{100}^{\frac{\mathrm{101}}{\mathrm{100}}} >\mathrm{101} \\ $$$$\Rightarrow\mathrm{100}^{\mathrm{101}} >\mathrm{101}^{\mathrm{100}} \\ $$

Commented by mathdanisur last updated on 15/Jul/21

thankyou Ser cool

$${thankyou}\:{Ser}\:{cool} \\ $$

Commented by mr W last updated on 15/Jul/21

applying this we can get (3)^(1/3) >(4)^(1/4) >(5)^(1/5) >... since (4)^(1/4) =(√2), so we get also (√2)<(3)^(1/3)

$${applying}\:{this}\:{we}\:{can}\:{get} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt[{\mathrm{4}}]{\mathrm{4}}>\sqrt[{\mathrm{5}}]{\mathrm{5}}>... \\ $$$${since}\:\sqrt[{\mathrm{4}}]{\mathrm{4}}=\sqrt{\mathrm{2}},\:{so}\:{we}\:{get}\:{also} \\ $$$$\sqrt{\mathrm{2}}<\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$

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