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Question Number 105301 by Don08q last updated on 27/Jul/20

$$\mathrm{Without}\mathrm{using}\mathrm{tables}\mathrm{or}\mathrm{calculator},$$$$compare{6}^7\mathrm{and}{7}^6$$

Answered by 1549442205PVT last updated on 28/Jul/20

$$\frac{6^7}{7^6}=6\times {\left(\frac{6}{7}\right)}^6>6\times {\left(\frac{4}{5}\right)}^6=6\times {\left(\frac{64}{125}\right)}^2$$$$>6\times {\left(\frac{64}{128}\right)}^2=6\times {\left(\frac{1}{2}\right)}^2=\frac{6}{4}>1$$$$\Rightarrow 6^7>7^6$$

Commented by Don08q last updated on 28/Jul/20

$$ThankyouSir!$$

Answered by JDamian last updated on 27/Jul/20

$$7^6={(6+1)}^6=$$$$=6^6+6\cdot 6^5+15\cdot 6^4+20\cdot 6^3+15\cdot 6^2+6\cdot 6^1+1=$$$$=2\cdot 6^6+15\cdot 6^4+20\cdot 6^3+15\cdot 6^2+6\cdot 6^1+1=$$$$=2\cdot 6^6+15\cdot 6^4+20\cdot 6^3+16\cdot 6^2+1$$$$$$$$Ifwedividebothquantitiesundertestby{6}^6:$$$$\frac{6^7}{6^6}=6$$$$\frac{7^6}{6^6}=2+\frac{15}{6^2}+\frac{20}{6^3}+\frac{16}{6^4}+\frac{1}{6^6}$$$$$$$$6>2+\frac{15}{6^2}+\frac{20}{6^3}+\frac{16}{6^4}+\frac{1}{6^6}$$$$4>\frac{15}{6^2}+\frac{20}{6^3}+\frac{16}{6^4}+\frac{1}{6^6}$$$$Therefore{6}^7>7^6$$

Commented by Don08q last updated on 27/Jul/20

$$Excellent.ThankyouSir.$$

Commented by malwaan last updated on 27/Jul/20

$$\frac{15}{6^2}<\frac{1}{2}<1;\frac{20}{6^3}<\frac{1}{10}<1$$$$\frac{16}{6^4}=\frac{1}{81}\ll 1;\frac{1}{6^6}\ll 1$$$$thankyousirJDamian$$

Answered by mathmax by abdo last updated on 27/Jul/20

$$\mathrm{for}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{naturels}\mathrm{and}\mathrm{x}>\mathrm{y}\mathrm{let}\mathrm{compare}{\mathrm{x}}^{\mathrm{y}}\mathrm{and}{\mathrm{y}}^{\mathrm{x}}$$$$\Rightarrow \mathrm{let}\mathrm{compare}\ln \left({\mathrm{x}}^{\mathrm{y}}\right)\mathrm{and}\ln \left({\mathrm{y}}^{\mathrm{x}}\right)(\ln \mathrm{is}\mathrm{increazing}\mathrm{function}..)$$$$\mathrm{let}\phi \left(\mathrm{x}\right)=\mathrm{ylnx}-\mathrm{xlny}\mathrm{for}\mathrm{y}\mathrm{fixed}\mathrm{we}\mathrm{have}\mathrm{y}⩾1\mathrm{and}\mathrm{x}\in ]\mathrm{y},+\mathrm{\infty }[$$$$\phi \left(\mathrm{y}\right)=0\mathrm{and}{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\phi \left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{x}(\mathrm{y}\frac{\mathrm{lnx}}{\mathrm{x}}-\mathrm{lny})=-\mathrm{\infty }$$$${\phi }^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{lny}\Rightarrow {\phi }^{{}^{″}}\left(\mathrm{x}\right)=-\frac{\mathrm{y}}{{\mathrm{x}}^2}<0\Rightarrow {\phi }^{{}^{\prime }}\mathrm{is}\mathrm{decreazing}\mathrm{and}$$$${\phi }^{{}^{\prime }}\left(\mathrm{y}\right)=1-\mathrm{lny}<0\Rightarrow \phi \mathrm{is}\mathrm{decreazing}\Rightarrow \phi \left(\mathrm{x}\right)<0\Rightarrow$$$$\mathrm{ylnx}-\mathrm{xlny}<0\Rightarrow \mathrm{ylnx}<\mathrm{xlny}\Rightarrow {\mathrm{x}}^{\mathrm{y}}<{\mathrm{y}}^{\mathrm{x}}$$$$\mathrm{x}=7\mathrm{and}\mathrm{y}=6\mathrm{x}>\mathrm{y}\Rightarrow {\mathrm{x}}^{\mathrm{y}}<{\mathrm{y}}^{\mathrm{x}}\Rightarrow 7^6<6^7$$$$$$$$$$

Commented by Don08q last updated on 28/Jul/20

$$ThankyouSir!$$

Commented by abdomsup last updated on 28/Jul/20

$$youarewelcome$$

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