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

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