Question Number 170948 by Mastermind last updated on 04/Jun/22
$$\mathrm{A}^{\mathrm{x}^{\mathrm{x}} } +\mathrm{Bx}^{\mathrm{x}} +\mathrm{C}=\mathrm{R},\:\mathrm{Make}\:\mathrm{x}\:\mathrm{the}\:\mathrm{subject} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{formular} \\ $$
Commented by mr W last updated on 04/Jun/22
![let t=x^x A^t +Bt=R−C A^t =R−C−Bt e^(tln A) =B(((R−C)/B)−t) e^(tln A) e^(−((R−C)/B)ln A) =Be^(−((R−C)/B)ln A) (((R−C)/B)−t) e^((t−((R−C)/B))ln A) =Be^(−((R−C)/B)ln A) (((R−C)/B)−t) (((R−C)/B)−t)ln Ae^((((R−C)/B)−t)ln A) =((A^((R−C)/B) ln A)/B) t=((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B)) x^x =t x=t^(1/x) =e^((ln t)/x) −((ln t)/x)e^(−((ln t)/x)) =−ln t −((ln t)/x)=W(−ln t) x=−((ln t)/(W(−ln t))) ⇒x=−((ln [((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B))])/(W{−ln [((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B))]}))](https://www.tinkutara.com/question/Q170949.png)
$${let}\:{t}={x}^{{x}} \\ $$$${A}^{{t}} +{Bt}={R}−{C} \\ $$$${A}^{{t}} ={R}−{C}−{Bt} \\ $$$${e}^{{t}\mathrm{ln}\:{A}} ={B}\left(\frac{{R}−{C}}{{B}}−{t}\right) \\ $$$${e}^{{t}\mathrm{ln}\:{A}} {e}^{−\frac{{R}−{C}}{{B}}\mathrm{ln}\:{A}} ={Be}^{−\frac{{R}−{C}}{{B}}\mathrm{ln}\:{A}} \left(\frac{{R}−{C}}{{B}}−{t}\right) \\ $$$${e}^{\left({t}−\frac{{R}−{C}}{{B}}\right)\mathrm{ln}\:{A}} ={Be}^{−\frac{{R}−{C}}{{B}}\mathrm{ln}\:{A}} \left(\frac{{R}−{C}}{{B}}−{t}\right) \\ $$$$\left(\frac{{R}−{C}}{{B}}−{t}\right)\mathrm{ln}\:{Ae}^{\left(\frac{{R}−{C}}{{B}}−{t}\right)\mathrm{ln}\:{A}} =\frac{{A}^{\frac{{R}−{C}}{{B}}} \mathrm{ln}\:{A}}{{B}} \\ $$$${t}=\frac{{R}−{C}}{{B}}−\frac{\mathrm{1}}{\mathrm{ln}\:{A}}{W}\left(\frac{{A}^{\frac{{R}−{C}}{{B}}} \mathrm{ln}\:{A}}{{B}}\right) \\ $$$${x}^{{x}} ={t} \\ $$$${x}={t}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:{t}}{{x}}} \\ $$$$−\frac{\mathrm{ln}\:{t}}{{x}}{e}^{−\frac{\mathrm{ln}\:{t}}{{x}}} =−\mathrm{ln}\:{t} \\ $$$$−\frac{\mathrm{ln}\:{t}}{{x}}={W}\left(−\mathrm{ln}\:{t}\right) \\ $$$${x}=−\frac{\mathrm{ln}\:{t}}{{W}\left(−\mathrm{ln}\:{t}\right)} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{ln}\:\left[\frac{{R}−{C}}{{B}}−\frac{\mathrm{1}}{\mathrm{ln}\:{A}}{W}\left(\frac{{A}^{\frac{{R}−{C}}{{B}}} \mathrm{ln}\:{A}}{{B}}\right)\right]}{{W}\left\{−\mathrm{ln}\:\left[\frac{{R}−{C}}{{B}}−\frac{\mathrm{1}}{\mathrm{ln}\:{A}}{W}\left(\frac{{A}^{\frac{{R}−{C}}{{B}}} \mathrm{ln}\:{A}}{{B}}\right)\right]\right\}} \\ $$
Commented by Mastermind last updated on 04/Jun/22
$${Thanks}\:{so}\:{much}\:{Prof}.\:{mrW} \\ $$
Commented by Tawa11 last updated on 04/Jun/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$