Question Number 183825 by Michaelfaraday last updated on 30/Dec/22
$${solve}\:{for}\:{x}\:{by}\:{using}\:{lambert}\:{function} \\ $$$$\mathrm{2}^{{x}} +\mathrm{3}{x}=\mathrm{8} \\ $$
Answered by MJS_new last updated on 30/Dec/22
$$\mathrm{generally} \\ $$$${b}^{{x}} ={ax}+{c} \\ $$$$\mathrm{inserting}\:{x}=−\frac{{t}}{\mathrm{ln}\:{b}}−\frac{{c}}{{a}}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{e}^{{t}} {t}=−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}} \\ $$$$\Rightarrow\:{x}=−\frac{{c}}{{a}}−\frac{\mathrm{1}}{\mathrm{ln}\:{b}}{W}\:\left(−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}}\right) \\ $$
Commented by Michaelfaraday last updated on 30/Dec/22
$${thanks}\:{sir} \\ $$
Commented by MJS_new last updated on 31/Dec/22
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$