Prove-that-3-gt-log-2-3-2-gt-2-

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

Prove that 3 > {(\log_{2} 3)}^{2} > 2 .

Commented by FilupS last updated on 06/Mar/17

for l = \log_{n} x

if n > 1 and x ⩾ 1, l ⩾ 0

\Rightarrow if 1 < x < n, 0 < l < 1 (*)

{(\log_{2} 3)}^{2} = (\log_{2} 3) (\log_{2} 3)

∴ \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 from (*)

\Rightarrow 3 > {3 \log}_{3} 2

∴ 3 > \log_{2} 3 > {2 \log}_{3} 2

\log_{2} 3 > {2 \log}_{3} 2

\log_{3} 2 < 1 from (*)

{2 \log}_{3} 2 < 2 \Rightarrow 2 > {2 \log}_{2} 3

∴ 3 > {3 \log}_{3} 2 > \log_{2} 3 > 2 > {2 \log}_{2} 3

{3 \log}_{3} 2 > \log_{2} 3 > {2 \log}_{2} 3

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 {(\log_{2} 3)}^{2} > {(\frac{3}{2})}^{2} = \frac{9}{4} > \frac{8}{4} = 2 \dots (i)

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

(i) a n d (i i) :

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

o r

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

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