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

Prove that 3>(log_2 3)^2 >2.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{3}>\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} >\mathrm{2}. \\ $$

Commented byFilupS last updated on 06/Mar/17

for l=log_n x if n>1 and x≥1, l≥0 ⇒ if 1<x<n, 0<l<1 (∗) (log_2 3)^2 =(log_2 3)(log_2 3) ∴ (3/(log_2 3))>log_2 3>(2/(log_2 3)) 3log_3 2>log_2 3>2log_3 2 log_2 3>1 ⇒ 1>log_3 2 from (∗) ⇒ 3>3log_3 2 ∴ 3>log_2 3>2log_3 2 log_2 3>2log_3 2 log_3 2<1 from (∗) 2log_3 2<2 ⇒ 2>2log_2 3 ∴3>3log_3 2>log_2 3>2>2log_2 3 3log_3 2>log_2 3>2log_2 3 working

$$\mathrm{for}\:\:{l}=\mathrm{log}_{{n}} {x} \\ $$ $$\mathrm{if}\:{n}>\mathrm{1}\:\mathrm{and}\:{x}\geqslant\mathrm{1},\:{l}\geqslant\mathrm{0} \\ $$ $$\:\:\:\Rightarrow\:\mathrm{if}\:\mathrm{1}<{x}<{n},\:\mathrm{0}<{l}<\mathrm{1}\:\:\:\:\left(\ast\right) \\ $$ $$\: \\ $$ $$\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right)\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right) \\ $$ $$\therefore\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}}>\mathrm{log}_{\mathrm{2}} \mathrm{3}>\frac{\mathrm{2}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}} \\ $$ $$\mathrm{3log}_{\mathrm{3}} \mathrm{2}>\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{2log}_{\mathrm{3}} \mathrm{2} \\ $$ $$\: \\ $$ $$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{1}\:\:\Rightarrow\:\:\mathrm{1}>\mathrm{log}_{\mathrm{3}} \mathrm{2}\:\:\:\:\:\mathrm{from}\:\left(\ast\right) \\ $$ $$\Rightarrow\:\mathrm{3}>\mathrm{3log}_{\mathrm{3}} \mathrm{2} \\ $$ $$\: \\ $$ $$\therefore\:\mathrm{3}>\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{2log}_{\mathrm{3}} \mathrm{2} \\ $$ $$\: \\ $$ $$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{2log}_{\mathrm{3}} \mathrm{2} \\ $$ $$\mathrm{log}_{\mathrm{3}} \mathrm{2}<\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{from}\:\left(\ast\right) \\ $$ $$\mathrm{2log}_{\mathrm{3}} \mathrm{2}<\mathrm{2}\:\:\Rightarrow\:\:\mathrm{2}>\mathrm{2log}_{\mathrm{2}} \mathrm{3} \\ $$ $$\: \\ $$ $$\therefore\mathrm{3}>\mathrm{3log}_{\mathrm{3}} \mathrm{2}>\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{2}>\mathrm{2log}_{\mathrm{2}} \mathrm{3} \\ $$ $$\mathrm{3log}_{\mathrm{3}} \mathrm{2}>\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{2log}_{\mathrm{2}} \mathrm{3} \\ $$ $$\: \\ $$ $$\mathrm{working} \\ $$

Answered by mrW1 last updated on 07/Mar/17

2log_2 3=log_2 3^2 =log_2 9>log_2 8=log_2 2^3 =3 ⇒log_2 3>(3/2) ⇒(log_2 3)^2 >((3/2))^2 =(9/4)>(8/4)=2 ...(i) 3log_2 3=log_2 3^3 =log_2 27<log_2 32=log_2 2^5 =5 ⇒log_2 3<(5/3) ⇒(log_2 3)^2 <((5/3))^2 =((25)/9)<((27)/9)=3 ...(ii) (i) and (ii): 2<(log_2 3)^2 <3 or (√2)<log_2 3<(√3)

$$\mathrm{2log}_{\mathrm{2}} \:\mathrm{3}=\mathrm{log}_{\mathrm{2}} \:\mathrm{3}^{\mathrm{2}} =\mathrm{log}_{\mathrm{2}} \:\mathrm{9}>\mathrm{log}_{\mathrm{2}} \:\mathrm{8}=\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{3}} =\mathrm{3} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\Rightarrow\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} >\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}>\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{2}\:\:\:\:\:...\left({i}\right) \\ $$ $$ \\ $$ $$\mathrm{3log}_{\mathrm{2}} \:\mathrm{3}=\mathrm{log}_{\mathrm{2}} \:\mathrm{3}^{\mathrm{3}} =\mathrm{log}_{\mathrm{2}} \:\mathrm{27}<\mathrm{log}_{\mathrm{2}} \:\mathrm{32}=\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} =\mathrm{5} \\ $$ $$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\Rightarrow\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} <\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{9}}<\frac{\mathrm{27}}{\mathrm{9}}=\mathrm{3}\:\:\:\:...\left({ii}\right) \\ $$ $$ \\ $$ $$\left({i}\right)\:{and}\:\left({ii}\right): \\ $$ $$\mathrm{2}<\left(\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\right)^{\mathrm{2}} <\mathrm{3} \\ $$ $${or} \\ $$ $$\sqrt{\mathrm{2}}<\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\sqrt{\mathrm{3}} \\ $$

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