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Question-178320

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Question Number 178320 by Shrinava last updated on 15/Oct/22
Commented by mr W last updated on 15/Oct/22
only if log 3 means log_e 3, i.e. ln 3
$${only}\:{if}\:\mathrm{log}\:\mathrm{3}\:{means}\:\mathrm{log}_{{e}} \:\mathrm{3},\:{i}.{e}.\:\mathrm{ln}\:\mathrm{3} \\ $$
Commented by Acem last updated on 15/Oct/22
Good sir
$${Good}\:{sir} \\ $$
Answered by mr W last updated on 15/Oct/22
3=((12)/4)>(9/4) ⇒(√3)>(√(9/4))=(3/2) 2.5<e<3 (proof see below) e^3 >2×2×2.5=10>9 ⇒e^(3/2) >9^(1/2) =3 e<3<e^(3/2) ln e<ln 3<ln e^(3/2) ⇒1<ln 3<(3/2) ((π(√3))/9)<((4×2)/9)=(8/9)<1<ln 3 ((π(√3))/3)>((3(√3))/3)=(√3)>(3/2)>ln 3 ⇒((π(√3))/9)<ln 3<((π(√3))/3) ✓ =========== proof for (5/2)<e<3 e=1+1+(1/(2!))+(1/(3!))+...>1+1+(1/2)=(5/2) e=1+1+(1/(2!))+(1/(3!))+(1/(4!))... =1+1+(1/2)+(1/(2×3))+(1/(2×3×4))+.. <1+1+(1/2)+(1/(2×2))+(1/(2×2×2))+.. =1+(1/(1−(1/2)))=3
$$\mathrm{3}=\frac{\mathrm{12}}{\mathrm{4}}>\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}>\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{2}.\mathrm{5}<{e}<\mathrm{3}\:\left({proof}\:{see}\:{below}\right) \\ $$$$ \\ $$$${e}^{\mathrm{3}} >\mathrm{2}×\mathrm{2}×\mathrm{2}.\mathrm{5}=\mathrm{10}>\mathrm{9} \\ $$$$\Rightarrow{e}^{\frac{\mathrm{3}}{\mathrm{2}}} >\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{3} \\ $$$$ \\ $$$${e}<\mathrm{3}<{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{ln}\:{e}<\mathrm{ln}\:\mathrm{3}<\mathrm{ln}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{1}<\mathrm{ln}\:\mathrm{3}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}}<\frac{\mathrm{4}×\mathrm{2}}{\mathrm{9}}=\frac{\mathrm{8}}{\mathrm{9}}<\mathrm{1}<\mathrm{ln}\:\mathrm{3} \\ $$$$\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}>\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}}=\sqrt{\mathrm{3}}>\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{ln}\:\mathrm{3} \\ $$$$\Rightarrow\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}}<\mathrm{ln}\:\mathrm{3}<\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\:\checkmark \\ $$$$ \\ $$$$=========== \\ $$$${proof}\:{for}\:\frac{\mathrm{5}}{\mathrm{2}}<{e}<\mathrm{3} \\ $$$${e}=\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+…>\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${e}=\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}… \\ $$$$\:\:\:=\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}+.. \\ $$$$\:\:\:<\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}×\mathrm{2}}+.. \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3} \\ $$
Commented by Tawa11 last updated on 15/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Shrinava last updated on 15/Oct/22
cool dear professor thank you
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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