prove-2-lt-log-2-3-lt-3-

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

p r o v e

\sqrt{2} < \log_{2} 3 < \sqrt{3}

Answered by bobhans last updated on 08/Jun/20

\sqrt{8} < \sqrt{9} \Rightarrow \log_{2} (\sqrt{8}) < \log_{2} (\sqrt{9})

\frac{3}{2} < \log_{2} (3) . \Rightarrow \sqrt{2} < \frac{3}{2} \Leftrightarrow \sqrt{2} < \log_{2} (3)

3 < 2^{\sqrt{3}} \Rightarrow \log_{2} (3) < \log_{2} (2^{\sqrt{3}})

\log_{2} (3) < \sqrt{3}

∴ \sqrt{2} < \log_{2} (3) < \sqrt{3}

Answered by 1549442205 last updated on 08/Jun/20

It is easy to see that \sqrt{2} < \log_{2} 3 < \sqrt{3} (*)

\Leftrightarrow 2^{\sqrt{2}} < 3 < 2^{\sqrt{3}} . Since 2^{\sqrt{2}} < 2^{\frac{3}{2}} = \sqrt{2^{3}} = \sqrt{8} < \sqrt{9} = 3 and

2^{\sqrt{3}} > 2^{1.6} = 2^{\frac{8}{5}} =^{5} \sqrt{2^{8}} =^{5} \sqrt{256} >^{5} \sqrt{243} = 3,

it follows that the inequality (*) true