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Question Number 105301 by Don08q last updated on 27/Jul/20
Without using tables or calculator, compare 6^7 and 7^6
$$\mathrm{Without}\:\mathrm{using}\:\mathrm{tables}\:\mathrm{or}\:\mathrm{calculator}, \\ $$$$\:{compare}\:\:\:\mathrm{6}^{\mathrm{7}} \:\mathrm{and}\:\:\:\mathrm{7}^{\mathrm{6}} \\ $$
Answered by 1549442205PVT last updated on 28/Jul/20
(6^7 /7^6 )=6×((6/7))^6 >6×((4/5))^6 =6×(((64)/(125)))^2 >6×(((64)/(128)))^2 =6×((1/2))^2 =(6/4)>1 ⇒6^7 >7^6
$$\frac{\mathrm{6}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{6}} }=\mathrm{6}×\left(\frac{\mathrm{6}}{\mathrm{7}}\right)^{\mathrm{6}} >\mathrm{6}×\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{6}} =\mathrm{6}×\left(\frac{\mathrm{64}}{\mathrm{125}}\right)^{\mathrm{2}} \\ $$$$>\mathrm{6}×\left(\frac{\mathrm{64}}{\mathrm{128}}\right)^{\mathrm{2}} =\mathrm{6}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{4}}>\mathrm{1} \\ $$$$\Rightarrow\mathrm{6}^{\mathrm{7}} >\mathrm{7}^{\mathrm{6}} \\ $$
Commented by Don08q last updated on 28/Jul/20
Thank you Sir!
$${Thank}\:{you}\:{Sir}! \\ $$
Answered by JDamian last updated on 27/Jul/20
7^6 =(6+1)^6 = = 6^6 +6∙6^5 +15∙6^4 +20∙6^3 +15∙6^2 +6∙6^1 +1= = 2∙6^6 +15∙6^4 +20∙6^3 +15∙6^2 +6∙6^1 +1= = 2∙6^6 +15∙6^4 +20∙6^3 +16∙6^2 +1 If we divide both quantities under test by 6^6 : (6^7 /6^6 ) = 6 (7^6 /6^6 ) = 2 + ((15)/6^2 ) + ((20)/6^3 ) + ((16)/6^4 ) + (1/6^6 ) 6 > 2 + ((15)/6^2 ) + ((20)/6^3 ) + ((16)/6^4 ) + (1/6^6 ) 4 > ((15)/6^2 ) + ((20)/6^3 ) + ((16)/6^4 ) + (1/6^6 ) Therefore 6^7 >7^6
$$\mathrm{7}^{\mathrm{6}} =\left(\mathrm{6}+\mathrm{1}\right)^{\mathrm{6}} = \\ $$$$\:\:\:\:\:=\:\mathrm{6}^{\mathrm{6}} +\mathrm{6}\centerdot\mathrm{6}^{\mathrm{5}} +\mathrm{15}\centerdot\mathrm{6}^{\mathrm{4}} +\mathrm{20}\centerdot\mathrm{6}^{\mathrm{3}} +\mathrm{15}\centerdot\mathrm{6}^{\mathrm{2}} +\mathrm{6}\centerdot\mathrm{6}^{\mathrm{1}} +\mathrm{1}= \\ $$$$\:\:\:\:\:=\:\mathrm{2}\centerdot\mathrm{6}^{\mathrm{6}} +\mathrm{15}\centerdot\mathrm{6}^{\mathrm{4}} +\mathrm{20}\centerdot\mathrm{6}^{\mathrm{3}} +\mathrm{15}\centerdot\mathrm{6}^{\mathrm{2}} +\mathrm{6}\centerdot\mathrm{6}^{\mathrm{1}} +\mathrm{1}= \\ $$$$\:\:\:\:\:=\:\mathrm{2}\centerdot\mathrm{6}^{\mathrm{6}} +\mathrm{15}\centerdot\mathrm{6}^{\mathrm{4}} +\mathrm{20}\centerdot\mathrm{6}^{\mathrm{3}} +\mathrm{16}\centerdot\mathrm{6}^{\mathrm{2}} +\mathrm{1} \\ $$$$ \\ $$$${If}\:{we}\:{divide}\:{both}\:{quantities}\:{under}\:{test}\:{by}\:\mathrm{6}^{\mathrm{6}} : \\ $$$$\frac{\mathrm{6}^{\mathrm{7}} }{\mathrm{6}^{\mathrm{6}} }\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{7}^{\mathrm{6}} }{\mathrm{6}^{\mathrm{6}} }\:=\:\mathrm{2}\:+\:\frac{\mathrm{15}}{\mathrm{6}^{\mathrm{2}} }\:+\:\frac{\mathrm{20}}{\mathrm{6}^{\mathrm{3}} }\:+\:\frac{\mathrm{16}}{\mathrm{6}^{\mathrm{4}} }\:+\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} } \\ $$$$ \\ $$$$\mathrm{6}\:>\:\mathrm{2}\:+\:\frac{\mathrm{15}}{\mathrm{6}^{\mathrm{2}} }\:+\:\frac{\mathrm{20}}{\mathrm{6}^{\mathrm{3}} }\:+\:\frac{\mathrm{16}}{\mathrm{6}^{\mathrm{4}} }\:+\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} } \\ $$$$\mathrm{4}\:>\:\frac{\mathrm{15}}{\mathrm{6}^{\mathrm{2}} }\:+\:\frac{\mathrm{20}}{\mathrm{6}^{\mathrm{3}} }\:+\:\frac{\mathrm{16}}{\mathrm{6}^{\mathrm{4}} }\:+\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} } \\ $$$${Therefore}\:\:\:\:\:\mathrm{6}^{\mathrm{7}} >\mathrm{7}^{\mathrm{6}} \\ $$
Commented by Don08q last updated on 27/Jul/20
Excellent. Thank you Sir.
$${Excellent}.\:{Thank}\:{you}\:{Sir}. \\ $$
Commented by malwaan last updated on 27/Jul/20
((15)/6^2 )<(1/2) <1 ; ((20)/6^3 )<(1/(10)) <1 ((16)/6^4 ) = (1/(81)) ≪ 1; (1/6^6 ) ≪ 1 thank you sir JDamian
$$\frac{\mathrm{15}}{\mathrm{6}^{\mathrm{2}} }<\frac{\mathrm{1}}{\mathrm{2}}\:<\mathrm{1}\:;\:\:\frac{\mathrm{20}}{\mathrm{6}^{\mathrm{3}} }<\frac{\mathrm{1}}{\mathrm{10}}\:<\mathrm{1}\: \\ $$$$\frac{\mathrm{16}}{\mathrm{6}^{\mathrm{4}} }\:=\:\frac{\mathrm{1}}{\mathrm{81}}\:\ll\:\mathrm{1};\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\:\ll\:\mathrm{1} \\ $$$${thank}\:{you}\:{sir}\:{JDamian} \\ $$
Answered by mathmax by abdo last updated on 27/Jul/20
for x and y naturels and x>y let compare x^y and y^x ⇒let compare ln(x^y ) and ln(y^x )(ln is increazing function..) let ϕ(x) =ylnx−xlny for y fixed we have y≥1 and x∈]y,+∞[ ϕ(y) =0 and lim_(x→+∞) ϕ(x) =lim_(x→+∞) x(y((lnx)/x)−lny) =−∞ ϕ^′ (x) =(y/x)−lny ⇒ϕ^(′′) (x) =−(y/x^2 )<0 ⇒ϕ^′ is decreazing and ϕ^′ (y) =1−lny <0 ⇒ϕ is decreazing ⇒ϕ(x)<0 ⇒ ylnx−xlny <0 ⇒ylnx<xlny ⇒x^y <y^x x=7 and y =6 x>y ⇒x^y <y^x ⇒7^6 <6^7
$$\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{naturels}\:\mathrm{and}\:\mathrm{x}>\mathrm{y}\:\mathrm{let}\:\mathrm{compare}\:\mathrm{x}^{\mathrm{y}} \:\mathrm{and}\:\mathrm{y}^{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{let}\:\mathrm{compare}\:\mathrm{ln}\left(\mathrm{x}^{\mathrm{y}} \right)\:\mathrm{and}\:\mathrm{ln}\left(\mathrm{y}^{\mathrm{x}} \right)\left(\mathrm{ln}\:\mathrm{is}\:\mathrm{increazing}\:\mathrm{function}..\right) \\ $$$$\left.\mathrm{let}\:\varphi\left(\mathrm{x}\right)\:=\mathrm{ylnx}−\mathrm{xlny}\:\:\:\mathrm{for}\:\mathrm{y}\:\mathrm{fixed}\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{y}\geqslant\mathrm{1}\:\mathrm{and}\:\mathrm{x}\in\right]\mathrm{y},+\infty\left[\right. \\ $$$$\varphi\left(\mathrm{y}\right)\:=\mathrm{0}\:\mathrm{and}\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \varphi\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{x}\left(\mathrm{y}\frac{\mathrm{lnx}}{\mathrm{x}}−\mathrm{lny}\right)\:=−\infty \\ $$$$\varphi^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{lny}\:\Rightarrow\varphi^{''} \left(\mathrm{x}\right)\:=−\frac{\mathrm{y}}{\mathrm{x}^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow\varphi^{'} \:\mathrm{is}\:\mathrm{decreazing}\:\:\mathrm{and} \\ $$$$\varphi^{'} \left(\mathrm{y}\right)\:=\mathrm{1}−\mathrm{lny}\:<\mathrm{0}\:\Rightarrow\varphi\:\mathrm{is}\:\mathrm{decreazing}\:\Rightarrow\varphi\left(\mathrm{x}\right)<\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{ylnx}−\mathrm{xlny}\:<\mathrm{0}\:\Rightarrow\mathrm{ylnx}<\mathrm{xlny}\:\Rightarrow\mathrm{x}^{\mathrm{y}} <\mathrm{y}^{\mathrm{x}} \\ $$$$\mathrm{x}=\mathrm{7}\:\mathrm{and}\:\mathrm{y}\:=\mathrm{6}\:\:\mathrm{x}>\mathrm{y}\:\Rightarrow\mathrm{x}^{\mathrm{y}} <\mathrm{y}^{\mathrm{x}} \:\Rightarrow\mathrm{7}^{\mathrm{6}} <\mathrm{6}^{\mathrm{7}} \\ $$$$ \\ $$$$ \\ $$
Commented by Don08q last updated on 28/Jul/20
Thank you Sir!
$${Thank}\:{you}\:{Sir}! \\ $$
Commented by abdomsup last updated on 28/Jul/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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