Question Number 182644 by mr W last updated on 12/Dec/22
$${solve}\:{for}\:{x}>\mathrm{0} \\ $$ $$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} >\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented byFrix last updated on 12/Dec/22
$$\mathrm{Obviously}\:{x}>\mathrm{1} \\ $$
Commented bymr W last updated on 12/Dec/22
$${it}\:{should}\:{not}\:{be}\:{a}\:{too}\:{special}\:{case}, \\ $$ $${therefore}\:{i}\:{have}\:{modified}\:{the}\: \\ $$ $${question}. \\ $$
Commented byFrix last updated on 12/Dec/22
$${x}=\frac{\mathrm{1}}{{t}} \\ $$ $${t}+\mathrm{1}>\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{t}} \\ $$ $$\mathrm{This}\:\mathrm{should}\:\mathrm{be}\:\mathrm{possible}\:\mathrm{with}\:\mathrm{LambertW} \\ $$
Commented byMJS_new last updated on 12/Dec/22
$$\mathrm{I}\:\mathrm{get} \\ $$ $${t}+\mathrm{1}={q}^{{t}} \:\Rightarrow\:{t}=−\left(\mathrm{1}+\frac{{W}\left(−\frac{\mathrm{ln}\:{q}}{{q}}\right)}{\mathrm{ln}\:{q}}\right) \\ $$ $${q}=\frac{\mathrm{5}}{\mathrm{2}}\:\Leftrightarrow\:\begin{cases}{{t}=\mathrm{0}}\\{{t}\approx.\mathrm{188114820}}\end{cases}\:\Rightarrow\:{x}\approx\mathrm{5}.\mathrm{31590237} \\ $$
Commented bymr W last updated on 13/Dec/22
$${thanks}\:{sirs}! \\ $$