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Question Number 1100 by malwaan last updated on 13/Jun/15

compare log_2 3 and log_3 5

$${compare}\:{log}_{\mathrm{2}} \mathrm{3}\:{and}\:{log}_{\mathrm{3}} \mathrm{5} \\ $$

Answered by 123456 last updated on 13/Jun/15

log_2 3=x⇔2^x =3 log_3 5=y⇔3^y =5 3<5 2^x <3^y log_2 3=((log 3)/(log 2))>1 log_3 5=((log 5)/(log 3))>1 log 5>log 3>log 2 log^2 5>log 3log 5>log 2log 5 log 3log 5>log^2 3>log 2log 3 log 2log 5>log 2log 3>log^2 2 log 3log 5>log^2 3>log 2log 3 log 3log 5>log 2log 5>log 2log 3 ........... 2^x <3^y x<log_2 3^y =ylog_2 3=log_3 5log_2 3 =((log 5)/(log 3))∙((log 3)/(log 2))=((log 5)/(log 2))=log_2 5 y>log_3 2^x =xlog_3 2=log_2 3log_3 2 =((log 3)/(log 2))∙((log 2)/(log 3))=1 1<log_2 3<log_2 5 1<log_3 5 log_2 5>log_3 5>1

$$\mathrm{log}_{\mathrm{2}} \mathrm{3}={x}\Leftrightarrow\mathrm{2}^{{x}} =\mathrm{3}\: \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}={y}\Leftrightarrow\mathrm{3}^{{y}} =\mathrm{5} \\ $$$$\mathrm{3}<\mathrm{5} \\ $$$$\mathrm{2}^{{x}} <\mathrm{3}^{{y}} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}=\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}>\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{3}}>\mathrm{1} \\ $$$$\mathrm{log}\:\mathrm{5}>\mathrm{log}\:\mathrm{3}>\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{log}^{\mathrm{2}} \mathrm{5}>\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{5} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}^{\mathrm{2}} \mathrm{3}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{2log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3}>\mathrm{log}^{\mathrm{2}} \mathrm{2} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}^{\mathrm{2}} \mathrm{3}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$........... \\ $$$$\mathrm{2}^{{x}} <\mathrm{3}^{{y}} \\ $$$${x}<\mathrm{log}_{\mathrm{2}} \mathrm{3}^{{y}} ={y}\mathrm{log}_{\mathrm{2}} \mathrm{3}=\mathrm{log}_{\mathrm{3}} \mathrm{5log}_{\mathrm{2}} \mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{3}}\centerdot\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{2}}=\mathrm{log}_{\mathrm{2}} \mathrm{5} \\ $$$${y}>\mathrm{log}_{\mathrm{3}} \mathrm{2}^{{x}} ={x}\mathrm{log}_{\mathrm{3}} \mathrm{2}=\mathrm{log}_{\mathrm{2}} \mathrm{3log}_{\mathrm{3}} \mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}\centerdot\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}}=\mathrm{1} \\ $$$$\mathrm{1}<\mathrm{log}_{\mathrm{2}} \mathrm{3}<\mathrm{log}_{\mathrm{2}} \mathrm{5} \\ $$$$\mathrm{1}<\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{5}>\mathrm{log}_{\mathrm{3}} \mathrm{5}>\mathrm{1} \\ $$

Commented by malwaan last updated on 14/Jun/15

where is the answer ? Is log_2 3 > or < log_3 5 ?

$${where}\:{is}\:{the}\:{answer}\:? \\ $$$${Is}\:{log}_{\mathrm{2}} \mathrm{3}\:>\:{or}\:<\:{log}_{\mathrm{3}} \mathrm{5}\:? \\ $$

Answered by prakash jain last updated on 14/Jun/15

9>8 3>2(√2)=2^(3/2) log_2 3>1.5 ...(a) 25<27 5<3(√3)=3^(3/2) log_3 5<1.5 ...(b) From (a) and (b) log_2 3>log_3 5

$$\mathrm{9}>\mathrm{8} \\ $$$$\mathrm{3}>\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{2}^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{1}.\mathrm{5}\:\:\:...\left({a}\right) \\ $$$$\mathrm{25}<\mathrm{27} \\ $$$$\mathrm{5}<\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{3}^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}<\mathrm{1}.\mathrm{5}\:\:\:\:...\left({b}\right) \\ $$$$\mathrm{From}\:\left({a}\right)\:\mathrm{and}\:\left({b}\right) \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$

Commented by prakash jain last updated on 15/Jun/15

log_(20) 80=((log 80)/(log 20))=((1+3log 2)/(1+log 2)) log _(80) 640=((log 640)/(log 80))=((1+6log 2)/(1+3log 2)) log 2=x log_(20) 80−log_(80) 640 [(1+3x)^2 −(1+x)(1+6x)]÷[(1+x)(1+3x)] Ignoring Denominator since it does contribute to sign 2>1⇒log 2>0 =1+6x+9x^2 −1−7x−6x^2 =3x^2 −x =x(3x−1) Sign will be >0 if x>(1/3) x=log 2 10>8 10^(1/3) >2 (1/3)>log 2=x Since x=log 2<(1/3)⇒x(3x−1)<0 ⇒log_(20) 80<log_(80) 640

$$\mathrm{log}_{\mathrm{20}} \mathrm{80}=\frac{\mathrm{log}\:\mathrm{80}}{\mathrm{log}\:\mathrm{20}}=\frac{\mathrm{1}+\mathrm{3log}\:\mathrm{2}}{\mathrm{1}+\mathrm{log}\:\mathrm{2}} \\ $$$$\mathrm{log}\:_{\mathrm{80}} \mathrm{640}=\frac{\mathrm{log}\:\mathrm{640}}{\mathrm{log}\:\mathrm{80}}=\frac{\mathrm{1}+\mathrm{6log}\:\mathrm{2}}{\mathrm{1}+\mathrm{3log}\:\mathrm{2}} \\ $$$$\mathrm{log}\:\mathrm{2}={x} \\ $$$$\:\mathrm{log}_{\mathrm{20}} \mathrm{80}−\mathrm{log}_{\mathrm{80}} \mathrm{640} \\ $$$$\left[\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{2}} −\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{6}{x}\right)\right]\boldsymbol{\div}\left[\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)\right] \\ $$$$\mathrm{Ignoring}\:\mathrm{Denominator}\:\mathrm{since}\:\mathrm{it}\:\mathrm{does}\:\mathrm{contribute}\:\mathrm{to}\:\mathrm{sign} \\ $$$$\mathrm{2}>\mathrm{1}\Rightarrow\mathrm{log}\:\mathrm{2}>\mathrm{0} \\ $$$$=\mathrm{1}+\mathrm{6}{x}+\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{7}{x}−\mathrm{6}{x}^{\mathrm{2}} \\ $$$$=\mathrm{3}{x}^{\mathrm{2}} −{x} \\ $$$$={x}\left(\mathrm{3}{x}−\mathrm{1}\right) \\ $$$$\mathrm{Sign}\:\mathrm{will}\:\mathrm{be}\:>\mathrm{0}\:\mathrm{if}\:{x}>\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{10}>\mathrm{8} \\ $$$$\mathrm{10}^{\mathrm{1}/\mathrm{3}} >\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}>\mathrm{log}\:\mathrm{2}={x} \\ $$$$\mathrm{Since}\:{x}=\mathrm{log}\:\mathrm{2}<\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}\left(\mathrm{3}{x}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{20}} \mathrm{80}<\mathrm{log}_{\mathrm{80}} \mathrm{640} \\ $$

Commented by malwaan last updated on 14/Jun/15

thank you very much Mr Prakash

$${thank}\:{you}\:{very}\:{much} \\ $$$${Mr}\:{Prakash} \\ $$

Commented by malwaan last updated on 14/Jun/15

what about log_(20) 80 and log_(80) 640 ?

$${what}\:{about}\: \\ $$$${log}_{\mathrm{20}} \mathrm{80}\:{and}\:{log}_{\mathrm{80}} \mathrm{640}\:? \\ $$

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