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Question Number 10995 by Nadium last updated on 06/Mar/17

$$\mathrm{Prove}\mathrm{that}3>{\left({\log }_{2}3\right)}^2>2.$$

Commented by FilupS last updated on 06/Mar/17

$$\mathrm{for}l={\log }_{n}x$$$$\mathrm{if}n>1\mathrm{and}x⩾1,l⩾0$$$$\Rightarrow \mathrm{if}1<x<n,0<l<1(\ast )$$$$$$$${\left({\log }_{2}3\right)}^2=\left({\log }_{2}3\right)\left({\log }_{2}3\right)$$$$\therefore \frac{3}{{\log }_{2}3}>{\log }_{2}3>\frac{2}{{\log }_{2}3}$$$${3\log }_{3}2>{\log }_{2}3>{2\log }_{3}2$$$$$$$${\log }_{2}3>1\Rightarrow 1>{\log }_{3}2\mathrm{from}(\ast )$$$$\Rightarrow 3>{3\log }_{3}2$$$$$$$$\therefore 3>{\log }_{2}3>{2\log }_{3}2$$$$$$$${\log }_{2}3>{2\log }_{3}2$$$${\log }_{3}2<1\mathrm{from}(\ast )$$$${2\log }_{3}2<2\Rightarrow 2>{2\log }_{2}3$$$$$$$$\therefore 3>{3\log }_{3}2>{\log }_{2}3>2>{2\log }_{2}3$$$${3\log }_{3}2>{\log }_{2}3>{2\log }_{2}3$$$$$$$$\mathrm{working}$$

Answered by mrW1 last updated on 07/Mar/17

$${2\log }_{2}3={\log }_2{3}^2={\log }_{2}9>{\log }_{2}8={\log }_2{2}^3=3$$$$\Rightarrow {\log }_{2}3>\frac{3}{2}$$$$\Rightarrow {\left({\log }_{2}3\right)}^2>{\left(\frac{3}{2}\right)}^2=\frac{9}{4}>\frac{8}{4}=2...\left(i\right)$$$$$$$${3\log }_{2}3={\log }_2{3}^3={\log }_{2}27<{\log }_{2}32={\log }_2{2}^5=5$$$$\Rightarrow {\log }_{2}3<\frac{5}{3}$$$$\Rightarrow {\left({\log }_{2}3\right)}^2<{\left(\frac{5}{3}\right)}^2=\frac{25}{9}<\frac{27}{9}=3...\left(ii\right)$$$$$$$$\left(i\right)and\left(ii\right):$$$$2<{\left({\log }_{2}3\right)}^2<3$$$$or$$$$\sqrt{2}<{\log }_{2}3<\sqrt{3}$$

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