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

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

Answered by Olaf_Thorendsen last updated on 14/Jul/21

$${\log }_{10}101=2+{\log }_{10}(1+\frac{1}{100})$$$${100\log }_{10}101=200+{100\log }_{10}(1+\frac{1}{100})$$$${100\log }_{10}101<200+100\times \frac{1}{100}$$$${100\log }_{10}101<201<202=101\times 2={101\log }_{10}100$$$${100\log }_{10}101<{101\log }_{10}100$$$${101}^{100}<{100}^{101}$$

Commented by mathdanisur last updated on 15/Jul/21

$$thankyouSercool$$

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

$$y=x^{\frac{1}{x}}$$$$\ln y=\frac{\ln x}{x}$$$$\frac{y^{\prime }}{y}=\frac{1-\ln x}{x^2}$$$$y^{\prime }=\frac{x^{\frac{1}{x}}}{x^2}(1-\ln x)<0forx>e$$$$thatmeansifx>e,y=x^{\frac{1}{x}}is$$$$strictlydecreasing.$$$$\Rightarrow {100}^{\frac{1}{100}}>{101}^{\frac{1}{101}}$$$$\Rightarrow {100}^{\frac{101}{100}}>101$$$$\Rightarrow {100}^{101}>{101}^{100}$$

Commented by mathdanisur last updated on 15/Jul/21

$$thankyouSercool$$

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

$$applyingthiswecanget$$$$\sqrt[3]{3}>\sqrt[4]{4}>\sqrt[5]{5}>...$$$$since\sqrt[4]{4}=\sqrt{2},sowegetalso$$$$\sqrt{2}<\sqrt[3]{3}$$

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