Question Number 165226 by Neils last updated on 27/Jan/22
$${make}\:{x}\:{the}\:{subject}\:{of}\:{the}\:{formula}; \\ $$$${a}^{{x}} +{bx}+{c}=\mathrm{0} \\ $$
Answered by mr W last updated on 28/Jan/22
$${a}^{{x}} =−\left({bx}+{c}\right) \\ $$$$\left({a}^{\frac{\mathrm{1}}{{b}}} \right)^{{bx}} =−\left({bx}+{c}\right) \\ $$$$\left({a}^{\frac{\mathrm{1}}{{b}}} \right)^{{bx}+{c}} =−\left({bx}+{c}\right){a}^{\frac{{c}}{{b}}} \\ $$$${e}^{\left({bx}+{c}\right)\frac{\mathrm{ln}\:{a}}{{b}}} =−\left({bx}+{c}\right){a}^{\frac{{c}}{{b}}} \\ $$$$−\left({bx}+{c}\right)\frac{\mathrm{ln}\:{a}}{{b}}{e}^{−\left({bx}+{c}\right)\frac{\mathrm{ln}\:{a}}{{b}}} =\frac{\mathrm{ln}\:{a}}{{ba}^{\frac{{c}}{{b}}} } \\ $$$$−\left({bx}+{c}\right)\frac{\mathrm{ln}\:{a}}{{b}}={W}\left(\frac{\mathrm{ln}\:{a}}{{ba}^{\frac{{c}}{{b}}} }\right) \\ $$$$\Rightarrow{x}=−\frac{{c}}{{b}}−\frac{\mathrm{1}}{\mathrm{ln}\:{a}}{W}\left(\frac{\mathrm{ln}\:{a}}{{ba}^{\frac{{c}}{{b}}} }\right) \\ $$
Commented by Neils last updated on 27/Jan/22
$${A}\:{very}\:{nice}\:{solution}\:{but}\:{there}'{s}\:{a}\:{slight}\:{mistake}\:{from}\:{the}\:{fifth}\:{line} \\ $$$${You}\:{omitted}\:{a}\:{minus}\:{sign} \\ $$
Commented by mr W last updated on 28/Jan/22
$${thanks}\:{sir}! \\ $$