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

$$withoutusingcalculator:$$$$whatislarger?{\log }_{2}3or{\log }_{3}5?$$

Answered by witcher3 last updated on 03/Nov/23

$$\frac{\ln \left(3\right)}{\ln \left(2\right)},\frac{\ln \left(5\right)}{\ln \left(3\right)}$$$$\frac{\ln (2\mathrm{x}-1)}{\ln \left(\mathrm{x}\right)}=\mathrm{g}\left(\mathrm{x}\right),\mathrm{x}\in [2,3]$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\frac{\frac{2\ln \left(\mathrm{x}\right)}{2\mathrm{x}-1}-\frac{\ln (2\mathrm{x}-1)}{\mathrm{x}}}{{\ln }^2\left(\mathrm{x}\right)}$$$$=\frac{2\mathrm{xln}\left(\mathrm{x}\right)-(2\mathrm{x}-1)\ln (2\mathrm{x}-1)}{\mathrm{x}(2\mathrm{x}-1){\ln }^2\left(\mathrm{x}\right)}$$$$\mathrm{sign}{\mathrm{g}}^{\prime }\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{x}\stackrel{\mathrm{h}}{\to }2\mathrm{xln}\left(\mathrm{x}\right)-(2\mathrm{x}-1)\ln (2\mathrm{x}-1)$$$${\mathrm{h}}^{\prime }=2\ln \left(\mathrm{x}\right)+2-\ln (2\mathrm{x}-1)-2$$$$=\ln \left(\frac{{\mathrm{x}}^2}{2\mathrm{x}-1}\right)$$$${\mathrm{x}}^2⩾2\mathrm{x}-1\mathrm{\forall }\mathrm{x}\in \mathbb{R}\Rightarrow {\mathrm{h}}^{\prime }\left(\mathrm{x}\right)>0$$$$\mathrm{h}\left(3\right)=6\ln \left(3\right)-5\ln \left(5\right)=\ln \left(\frac{3^6}{5^5}\right)=\ln \left(\frac{729}{5.625}\right)<0$$$$\Rightarrow {\mathrm{g}}^{\prime }\left(\mathrm{x}\right)<0,\mathrm{\forall }\mathrm{x}\in [2,3]\Rightarrow \mathrm{g}\mathrm{decrese}$$$$\mathrm{g}\left(2\right)>\mathrm{g}\left(3\right)\iff \frac{\ln \left(3\right)}{\ln \left(2\right)}>\frac{\ln \left(5\right)}{\ln \left(3\right)}$$$$$$$$$$$$$$

Commented by York12 last updated on 04/Nov/23

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{impressive}$$

Commented by witcher3 last updated on 04/Nov/23

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

Commented by York12 last updated on 04/Nov/23

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

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

$${\log }_{2}3=\frac{\log 3}{\log 2}=\frac{3\log 3}{3\log 2}=\frac{\log 27}{\log 8}>\frac{\log 25}{\log 8}$$$$>\frac{\log 25}{\log 9}=\frac{2\log 5}{2\log 3}=\frac{\log 5}{\log 3}={\log }_{3}5$$$$\Rightarrow {\log }_{2}3islargerthan{\log }_{3}5.$$

Commented by witcher3 last updated on 04/Nov/23

$$\mathrm{batter}\mathrm{solution}\mathrm{than}\mathrm{mine}\mathrm{ThankYou}\mathrm{for}\mathrm{sharing}\mathrm{It}$$

Commented by mr W last updated on 04/Nov/23

$$ifoundyourmethodusingfunction$$$$agreatidea.thanksforthat!$$

Commented by York12 last updated on 04/Nov/23

$$\mathrm{ur}\mathrm{the}\mathrm{goat}\mathrm{lmao}$$

Commented by York12 last updated on 04/Nov/23

$$\mathrm{yeah}\mathrm{really}\mathrm{great}\mathrm{ngl}$$

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