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Question Number 97412 by mr W last updated on 07/Jun/20

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

Answered by bobhans last updated on 08/Jun/20

$$\sqrt{8}<\sqrt{9}\Rightarrow \log {}_2\left(\sqrt{8}\right)<\log {}_2\left(\sqrt{9}\right)$$$$\frac{3}{2}<\log {}_2\left(3\right).\Rightarrow \sqrt{2}<\frac{3}{2}\iff \sqrt{2}<\log {}_2\left(3\right)$$$$3<2^{\sqrt{3}}\Rightarrow \log {}_2\left(3\right)<\log {}_2\left(2^{\sqrt{3}}\right)$$$$\log {}_2\left(3\right)<\sqrt{3}$$$$\therefore \sqrt{2}<\log {}_2\left(3\right)<\sqrt{3}$$

Answered by 1549442205 last updated on 08/Jun/20

$$\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\sqrt{2}<{\log }_{2}3<\sqrt{3}(\ast )$$$$\iff 2^{\sqrt{2}}<3<2^{\sqrt{3}}.\mathrm{Since}{2}^{\sqrt{2}}<2^{\frac{3}{2}}=\sqrt{2^3}=\sqrt{8}<\sqrt{9}=3\mathrm{and}$$$$2^{\sqrt{3}}>2^{1.6}=2^{\frac{8}{5}}{=}^5\sqrt{2^8}{=}^5\sqrt{256}{>}^5\sqrt{243}=3,$$$$\mathrm{it}\mathrm{follows}\mathrm{that}\mathrm{the}\mathrm{inequality}(\ast )\mathrm{true}$$

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