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Question Number 181382 by liuxinnan last updated on 24/Nov/22

Commented by liuxinnan last updated on 24/Nov/22

n >?

$${n}\:>? \\ $$

Answered by Frix last updated on 24/Nov/22

(n/(log_2 n))=(n/((ln n)/(ln 2)))=((nln 2)/(ln n)) first solve the equation (n/(ln n))=(3/(ln 2)) −((ln n)/n)=−((ln 2)/3) let n=e^(−t) te^t =((ln 2)/3) LambertW−function leads to t≈−2.29652∨t≈−.317338 ⇔ n≈9.93954∨n≈1.37347 now (n/(ln n))>(3/(ln 2)) and obviously this leads to 1<n<≈1.37347∨n>≈9.93954

$$\frac{{n}}{\mathrm{log}_{\mathrm{2}} \:{n}}=\frac{{n}}{\frac{\mathrm{ln}\:{n}}{\mathrm{ln}\:\mathrm{2}}}=\frac{{n}\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:{n}} \\ $$$$\mathrm{first}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\frac{{n}}{\mathrm{ln}\:{n}}=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$−\frac{\mathrm{ln}\:{n}}{{n}}=−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{let}\:{n}=\mathrm{e}^{−{t}} \\ $$$${t}\mathrm{e}^{{t}} =\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{LambertW}−\mathrm{function}\:\mathrm{leads}\:\mathrm{to} \\ $$$${t}\approx−\mathrm{2}.\mathrm{29652}\vee{t}\approx−.\mathrm{317338} \\ $$$$\Leftrightarrow \\ $$$${n}\approx\mathrm{9}.\mathrm{93954}\vee{n}\approx\mathrm{1}.\mathrm{37347} \\ $$$$\mathrm{now} \\ $$$$\frac{{n}}{\mathrm{ln}\:{n}}>\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}\:\mathrm{and}\:\mathrm{obviously}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{1}<{n}<\approx\mathrm{1}.\mathrm{37347}\vee{n}>\approx\mathrm{9}.\mathrm{93954} \\ $$

Commented by Socracious last updated on 25/Nov/22

where can I get the labert w function calculator, please

$$\:\:\:\mathrm{where}\:\mathrm{can}\:\mathrm{I}\:\mathrm{get}\:\mathrm{the}\:\mathrm{labert}\:\mathrm{w}\:\mathrm{function} \\ $$$$\:\:\mathrm{calculator},\:\mathrm{please} \\ $$

Answered by mr W last updated on 24/Nov/22

n>0 and n≠1 case 0<n<1: n<3 log_2 n<0 ⇒contradiction with n>0 ⇒no solution! case n>1: n>3 log_2 n=(3/(ln 2))×ln n ((ln 2)/3)>ln n e^(−ln n) −((ln 2)/3)<(−ln n) e^(−ln n) −ln n<W(−((ln 2)/3)) ln n>−W(−((ln 2)/3)) n>(1/e^(W(−((ln 2)/3))) )=−((3 W(−((ln 2)/3)))/(ln 2)) since W(−((ln 2)/3)) has two values, the other solution is 1<n<−((3 W(−((ln 2)/3)))/(ln 2)) with the bigger value of W(). W(−((ln 2)/3))= { ((−2.296 520 3)),((−0.317 338 3)) :}

$${n}>\mathrm{0}\:{and}\:{n}\neq\mathrm{1} \\ $$$$\underline{{case}\:\mathrm{0}<{n}<\mathrm{1}:} \\ $$$${n}<\mathrm{3}\:\mathrm{log}_{\mathrm{2}} \:{n}<\mathrm{0}\: \\ $$$$\Rightarrow{contradiction}\:{with}\:{n}>\mathrm{0} \\ $$$$\Rightarrow{no}\:{solution}! \\ $$$$\underline{{case}\:{n}>\mathrm{1}:} \\ $$$${n}>\mathrm{3}\:\mathrm{log}_{\mathrm{2}} \:{n}=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{ln}\:{n} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}>\mathrm{ln}\:{n}\:{e}^{−\mathrm{ln}\:{n}} \\ $$$$−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}<\left(−\mathrm{ln}\:{n}\right)\:{e}^{−\mathrm{ln}\:{n}} \\ $$$$−\mathrm{ln}\:{n}<{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\mathrm{ln}\:{n}>−{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right) \\ $$$${n}>\frac{\mathrm{1}}{{e}^{{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)} }=−\frac{\mathrm{3}\:{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$$${since}\:{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)\:{has}\:{two}\:{values},\:{the} \\ $$$${other}\:{solution}\:{is} \\ $$$$\mathrm{1}<{n}<−\frac{\mathrm{3}\:{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{ln}\:\mathrm{2}}\:{with}\:{the}\:{bigger}\: \\ $$$${value}\:{of}\:{W}\left(\right). \\ $$$$ \\ $$$${W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)=\begin{cases}{−\mathrm{2}.\mathrm{296}\:\mathrm{520}\:\mathrm{3}}\\{−\mathrm{0}.\mathrm{317}\:\mathrm{338}\:\mathrm{3}}\end{cases} \\ $$

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