without-using-calculator-what-is-larger-log-2-3-or-log-3-5-

Question Number 199432 by mr W last updated on 03/Nov/23

$${without}\:{using}\:{calculator}: \\ $$$${what}\:{is}\:{larger}?\:\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\:{or}\:\mathrm{log}_{\mathrm{3}} \:\mathrm{5}? \\ $$

Answered by witcher3 last updated on 03/Nov/23

((ln(3))/(ln(2))),((ln(5))/(ln(3))) ((ln(2x−1))/(ln(x)))=g(x),x∈[2,3] g′(x)=((((2ln(x))/(2x−1))−((ln(2x−1))/x))/(ln^2 (x))) =((2xln(x)−(2x−1)ln(2x−1))/(x(2x−1)ln^2 (x))) sign g′ is same as x→^h 2xln(x)−(2x−1)ln(2x−1) h′=2ln(x)+2−ln(2x−1)−2 =ln((x^2 /(2x−1))) x^2 ≥2x−1∀x∈R⇒h′(x)>0 h(3)=6ln(3)−5ln(5)=ln((3^6 /5^5 ))=ln(((729)/(5.625)))<0 ⇒g′(x)<0,∀x∈[2,3]⇒g decrese g(2)>g(3)⇔((ln(3))/(ln(2)))>((ln(5))/(ln(3)))

$$\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{ln}\left(\mathrm{2}\right)},\frac{\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{ln}\left(\mathrm{3}\right)} \\ $$$$\frac{\mathrm{ln}\left(\mathrm{2x}−\mathrm{1}\right)}{\mathrm{ln}\left(\mathrm{x}\right)}=\mathrm{g}\left(\mathrm{x}\right),\mathrm{x}\in\left[\mathrm{2},\mathrm{3}\right] \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\frac{\mathrm{2ln}\left(\mathrm{x}\right)}{\mathrm{2x}−\mathrm{1}}−\frac{\mathrm{ln}\left(\mathrm{2x}−\mathrm{1}\right)}{\mathrm{x}}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$$$=\frac{\mathrm{2xln}\left(\mathrm{x}\right)−\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2x}−\mathrm{1}\right)}{\mathrm{x}\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$$$\mathrm{sign}\:\mathrm{g}'\:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:\mathrm{x}\overset{\mathrm{h}} {\rightarrow}\mathrm{2xln}\left(\mathrm{x}\right)−\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2x}−\mathrm{1}\right) \\ $$$$\mathrm{h}'=\mathrm{2ln}\left(\mathrm{x}\right)+\mathrm{2}−\mathrm{ln}\left(\mathrm{2x}−\mathrm{1}\right)−\mathrm{2} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2x}−\mathrm{1}}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} \geqslant\mathrm{2x}−\mathrm{1}\forall\mathrm{x}\in\mathbb{R}\Rightarrow\mathrm{h}'\left(\mathrm{x}\right)>\mathrm{0} \\ $$$$\mathrm{h}\left(\mathrm{3}\right)=\mathrm{6ln}\left(\mathrm{3}\right)−\mathrm{5ln}\left(\mathrm{5}\right)=\mathrm{ln}\left(\frac{\mathrm{3}^{\mathrm{6}} }{\mathrm{5}^{\mathrm{5}} }\right)=\mathrm{ln}\left(\frac{\mathrm{729}}{\mathrm{5}.\mathrm{625}}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{g}'\left(\mathrm{x}\right)<\mathrm{0},\forall\mathrm{x}\in\left[\mathrm{2},\mathrm{3}\right]\Rightarrow\mathrm{g}\:\mathrm{decrese} \\ $$$$\mathrm{g}\left(\mathrm{2}\right)>\mathrm{g}\left(\mathrm{3}\right)\Leftrightarrow\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}>\frac{\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{ln}\left(\mathrm{3}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by York12 last updated on 04/Nov/23

$$\mathrm{That}'\mathrm{s}\:\mathrm{impressive} \\ $$

Commented by witcher3 last updated on 04/Nov/23

no just basic calculus not smart ideas thank You

$$\mathrm{no}\:\mathrm{just}\:\mathrm{basic}\:\mathrm{calculus}\:\mathrm{not}\:\mathrm{smart}\:\mathrm{ideas} \\ $$$$\mathrm{thank}\:\mathrm{You} \\ $$

Commented by York12 last updated on 04/Nov/23

IT is impressive because of the way you have used to translate the problem

$$\mathrm{IT}\:\mathrm{is}\:\mathrm{impressive}\:\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{way}\:\mathrm{you}\:\mathrm{have}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{translate}\:\mathrm{the}\:\mathrm{problem} \\ $$

Answered by mr W last updated on 03/Nov/23

log_2 3=((log 3)/(log 2))=((3 log 3)/(3 log 2))=((log 27)/(log 8))> ((log 25)/(log 8)) >((log 25)/(log 9))=((2 log 5)/(2 log 3))=((log 5)/(log 3))=log_3 5 ⇒log_2 3 is larger than log_3 5.

$$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}=\frac{\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{3}\:\mathrm{log}\:\mathrm{3}}{\mathrm{3}\:\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{log}\:\mathrm{27}}{\mathrm{log}\:\mathrm{8}}>\:\frac{\mathrm{log}\:\mathrm{25}}{\mathrm{log}\:\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:>\frac{\mathrm{log}\:\mathrm{25}}{\mathrm{log}\:\mathrm{9}}=\frac{\mathrm{2}\:\mathrm{log}\:\mathrm{5}}{\mathrm{2}\:\mathrm{log}\:\mathrm{3}}=\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{3}}=\mathrm{log}_{\mathrm{3}} \:\mathrm{5} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\:{is}\:{larger}\:{than}\:\mathrm{log}_{\mathrm{3}} \:\mathrm{5}. \\ $$

Commented by witcher3 last updated on 04/Nov/23