Question Number 1100 by malwaan last updated on 13/Jun/15
$${compare}\:{log}_{\mathrm{2}} \mathrm{3}\:{and}\:{log}_{\mathrm{3}} \mathrm{5} \\ $$
Answered by 123456 last updated on 13/Jun/15
$$\mathrm{log}_{\mathrm{2}} \mathrm{3}={x}\Leftrightarrow\mathrm{2}^{{x}} =\mathrm{3}\: \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}={y}\Leftrightarrow\mathrm{3}^{{y}} =\mathrm{5} \\ $$$$\mathrm{3}<\mathrm{5} \\ $$$$\mathrm{2}^{{x}} <\mathrm{3}^{{y}} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}=\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}>\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{3}}>\mathrm{1} \\ $$$$\mathrm{log}\:\mathrm{5}>\mathrm{log}\:\mathrm{3}>\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{log}^{\mathrm{2}} \mathrm{5}>\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{5} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}^{\mathrm{2}} \mathrm{3}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{2log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3}>\mathrm{log}^{\mathrm{2}} \mathrm{2} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}^{\mathrm{2}} \mathrm{3}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{3log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{5}>\mathrm{log}\:\mathrm{2log}\:\mathrm{3} \\ $$$$……….. \\ $$$$\mathrm{2}^{{x}} <\mathrm{3}^{{y}} \\ $$$${x}<\mathrm{log}_{\mathrm{2}} \mathrm{3}^{{y}} ={y}\mathrm{log}_{\mathrm{2}} \mathrm{3}=\mathrm{log}_{\mathrm{3}} \mathrm{5log}_{\mathrm{2}} \mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{3}}\centerdot\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{2}}=\mathrm{log}_{\mathrm{2}} \mathrm{5} \\ $$$${y}>\mathrm{log}_{\mathrm{3}} \mathrm{2}^{{x}} ={x}\mathrm{log}_{\mathrm{3}} \mathrm{2}=\mathrm{log}_{\mathrm{2}} \mathrm{3log}_{\mathrm{3}} \mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}\centerdot\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}}=\mathrm{1} \\ $$$$\mathrm{1}<\mathrm{log}_{\mathrm{2}} \mathrm{3}<\mathrm{log}_{\mathrm{2}} \mathrm{5} \\ $$$$\mathrm{1}<\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{5}>\mathrm{log}_{\mathrm{3}} \mathrm{5}>\mathrm{1} \\ $$
Commented by malwaan last updated on 14/Jun/15
$${where}\:{is}\:{the}\:{answer}\:? \\ $$$${Is}\:{log}_{\mathrm{2}} \mathrm{3}\:>\:{or}\:<\:{log}_{\mathrm{3}} \mathrm{5}\:? \\ $$
Answered by prakash jain last updated on 14/Jun/15
$$\mathrm{9}>\mathrm{8} \\ $$$$\mathrm{3}>\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{2}^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{1}.\mathrm{5}\:\:\:…\left({a}\right) \\ $$$$\mathrm{25}<\mathrm{27} \\ $$$$\mathrm{5}<\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{3}^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{5}<\mathrm{1}.\mathrm{5}\:\:\:\:…\left({b}\right) \\ $$$$\mathrm{From}\:\left({a}\right)\:\mathrm{and}\:\left({b}\right) \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{3}>\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$
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](https://www.tinkutara.com/question/Q1114.png)
$$\mathrm{log}_{\mathrm{20}} \mathrm{80}=\frac{\mathrm{log}\:\mathrm{80}}{\mathrm{log}\:\mathrm{20}}=\frac{\mathrm{1}+\mathrm{3log}\:\mathrm{2}}{\mathrm{1}+\mathrm{log}\:\mathrm{2}} \\ $$$$\mathrm{log}\:_{\mathrm{80}} \mathrm{640}=\frac{\mathrm{log}\:\mathrm{640}}{\mathrm{log}\:\mathrm{80}}=\frac{\mathrm{1}+\mathrm{6log}\:\mathrm{2}}{\mathrm{1}+\mathrm{3log}\:\mathrm{2}} \\ $$$$\mathrm{log}\:\mathrm{2}={x} \\ $$$$\:\mathrm{log}_{\mathrm{20}} \mathrm{80}−\mathrm{log}_{\mathrm{80}} \mathrm{640} \\ $$$$\left[\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{2}} −\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{6}{x}\right)\right]\boldsymbol{\div}\left[\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)\right] \\ $$$$\mathrm{Ignoring}\:\mathrm{Denominator}\:\mathrm{since}\:\mathrm{it}\:\mathrm{does}\:\mathrm{contribute}\:\mathrm{to}\:\mathrm{sign} \\ $$$$\mathrm{2}>\mathrm{1}\Rightarrow\mathrm{log}\:\mathrm{2}>\mathrm{0} \\ $$$$=\mathrm{1}+\mathrm{6}{x}+\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{7}{x}−\mathrm{6}{x}^{\mathrm{2}} \\ $$$$=\mathrm{3}{x}^{\mathrm{2}} −{x} \\ $$$$={x}\left(\mathrm{3}{x}−\mathrm{1}\right) \\ $$$$\mathrm{Sign}\:\mathrm{will}\:\mathrm{be}\:>\mathrm{0}\:\mathrm{if}\:{x}>\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{10}>\mathrm{8} \\ $$$$\mathrm{10}^{\mathrm{1}/\mathrm{3}} >\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}>\mathrm{log}\:\mathrm{2}={x} \\ $$$$\mathrm{Since}\:{x}=\mathrm{log}\:\mathrm{2}<\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}\left(\mathrm{3}{x}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{20}} \mathrm{80}<\mathrm{log}_{\mathrm{80}} \mathrm{640} \\ $$
Commented by malwaan last updated on 14/Jun/15
$${thank}\:{you}\:{very}\:{much} \\ $$$${Mr}\:{Prakash} \\ $$
Commented by malwaan last updated on 14/Jun/15
$${what}\:{about}\: \\ $$$${log}_{\mathrm{20}} \mathrm{80}\:{and}\:{log}_{\mathrm{80}} \mathrm{640}\:? \\ $$