show-that-2-lt-log-2-3-lt-3-

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

s h o w t h a t \sqrt{2} < \log_{2} 3 < \sqrt{3}

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

2^{\sqrt{2}} < 3 < 2^{\sqrt{3}}

{l o g}_{2} 2^{\sqrt{2}} < {l o g}_{2} 3 < {l o g}_{2} 2^{\sqrt{3}}

\sqrt{2} < {l o g}_{2} 3 < \sqrt{3}

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

I s m y w a y c o r r e c t s i r ?

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

w e m a y n o t u s e c a l c u l a t o r . h o w i s i t

o b v i o u r t h a t 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}

{(\log_{2} 3)}^{2} > {(\frac{3}{2})}^{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}

{(\log_{2} 3)}^{2} < {(\frac{5}{3})}^{2} = \frac{25}{9} < \frac{27}{9} = 3

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

Commented by som(math1967) last updated on 27/Feb/21

o k s i r