Question Number 186996 by Mingma last updated on 12/Feb/23 | ||
Answered by aba last updated on 12/Feb/23 | ||
$$\mathrm{0}<\mathrm{ln}\left(\mathrm{3}/\mathrm{2}\right)<\mathrm{6}/\mathrm{13} \\ $$ | ||
Commented by MJS_new last updated on 12/Feb/23 | ||
$$\mathrm{you}\:\mathrm{must}\:\mathrm{show}\:\mathrm{it}\:\mathrm{without}\:\mathrm{a}\:\mathrm{calculator} \\ $$ | ||
Commented by aba last updated on 12/Feb/23 | ||
$$ \\ $$$$\frac{\mathrm{13ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{6}}=\mathrm{ln}\left(\sqrt[{\mathrm{6}}]{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{13}} }\right)<\mathrm{ln}\left(\mathrm{e}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)<\frac{\mathrm{6}}{\mathrm{13}} \\ $$ | ||
Commented by Frix last updated on 12/Feb/23 | ||
$$\mathrm{You}\:\mathrm{seriously}\:\mathrm{claim}\:\mathrm{to}\:\mathrm{see}\:\sqrt[{\mathrm{6}}]{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{13}} }<\mathrm{e} \\ $$$$\mathrm{without}\:\mathrm{calculating}\:\mathrm{the}\:\mathrm{value}?! \\ $$ | ||
Commented by aba last updated on 12/Feb/23 | ||
$$\forall\mathrm{n}\geqslant\mathrm{1}\::\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} \leqslant\mathrm{e} \\ $$$$\sqrt[{\mathrm{6}}]{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{13}} }=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{13}}{\mathrm{6}}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \mathrm{e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\left(\mathrm{1}.\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} .\mathrm{e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\mathrm{e} \\ $$ | ||
Commented by Mingma last updated on 12/Feb/23 | ||
Good! | ||
Commented by mr W last updated on 12/Feb/23 | ||
$${wrong}\:{sir}! \\ $$$${from} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\left(\mathrm{1}.\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} .\mathrm{e} \\ $$$${you}\:{can}\:{not}\:{get} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<\mathrm{e} \\ $$$${because}\:\left(\mathrm{1}.\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} >\mathrm{1}. \\ $$ | ||
Commented by aba last updated on 12/Feb/23 | ||
Commented by Frix last updated on 12/Feb/23 | ||
$$\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{2}}\:<\frac{\mathrm{6}}{\mathrm{13}} \\ $$$$\frac{\mathrm{13}}{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{2}}\:<\mathrm{1} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{13}}{\mathrm{6}}} <\mathrm{e} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} <\mathrm{e} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \frac{\mathrm{9}}{\mathrm{4}}<\frac{\mathrm{27}}{\mathrm{10}}<\mathrm{e} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} <\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}<\frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{5}^{\mathrm{6}} } \\ $$$$\frac{\mathrm{3}×\mathrm{5}^{\mathrm{6}} }{\mathrm{2}×\mathrm{6}^{\mathrm{6}} }<\mathrm{1} \\ $$$$\frac{\mathrm{5}^{\mathrm{6}} }{\mathrm{4}×\mathrm{6}^{\mathrm{5}} }<\mathrm{1} \\ $$$$\frac{\mathrm{15625}}{\mathrm{31104}}<\mathrm{1}\:\mathrm{true} \\ $$ | ||