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Question Number 133996 by mr W last updated on 26/Feb/21

$$showthat\sqrt{2}<{\log }_{2}3<\sqrt{3}$$

Answered by som(math1967) last updated on 26/Feb/21

$$2^{\sqrt{2}}<3<2^{\sqrt{3}}$$ $${log}_2{2}^{\sqrt{2}}<{log}_{2}3<{log}_2{2}^{\sqrt{3}}$$ $$\sqrt{2}<{log}_{2}3<\sqrt{3}$$

Commented bysom(math1967) last updated on 26/Feb/21

$$Ismywaycorrectsir?$$

Commented bymr W last updated on 26/Feb/21

$$wemaynotusecalculator.howisit$$ $$obviourthat{2}^{\sqrt{2}}<3<2^{\sqrt{3}}?$$

Answered by mr W last updated on 26/Feb/21

$$9>8$$ $$3^2>2^3$$ $$2{\log }_{2}3>3$$ $${\log }_{2}3>\frac{3}{2}$$ $${\left({\log }_{2}3\right)}^2>{\left(\frac{3}{2}\right)}^2=\frac{9}{4}>\frac{8}{4}=2$$ $$\Rightarrow {\log }_{2}3>\sqrt{2}$$ $$$$ $$27<32$$ $$3^3<2^5$$ $${3\log }_{2}3<5$$ $${\log }_{2}3<\frac{5}{3}$$ $${\left({\log }_{2}3\right)}^2<{\left(\frac{5}{3}\right)}^2=\frac{25}{9}<\frac{27}{9}=3$$ $$\Rightarrow {\log }_{2}3<\sqrt{3}$$

Commented bysom(math1967) last updated on 27/Feb/21

$$oksir$$

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