Compare-100-101-and-101-100-

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

C o m p a r e : 100^{101} a n d 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

t h a n k y o u S e r c o o l

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

y = x^{\frac{1}{x}}

\ln y = \frac{\ln x}{x}

\frac{y^{'}}{y} = \frac{1 - \ln x}{x^{2}}

y^{'} = \frac{x^{\frac{1}{x}}}{x^{2}} (1 - \ln x) < 0 f o r x > e

t h a t m e a n s i f x > e, y = x^{\frac{1}{x}} i s

s t r i c t l y d e c r e a s i n g .

\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

t h a n k y o u S e r c o o l

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

a p p l y i n g t h i s w e c a n g e t

\sqrt[3]{3} > \sqrt[4]{4} > \sqrt[5]{5} > \dots

s i n c e \sqrt[4]{4} = \sqrt{2}, s o w e g e t a l s o

\sqrt{2} < \sqrt[3]{3}

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