A^x^x +Bx^x +C=R, Make x the subject of the formular


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Question Number 170948 by Mastermind last updated on 04/Jun/22

$A^{x^{x}} + {Bx}^{x} + C = R, Make x the subject$ $of the formular$
Commented by mr W last updated on 04/Jun/22
$let t=x^x A^t +Bt=R−C A^t =R−C−Bt e^(tln A) =B(((R−C)/B)−t) e^(tln A) e^(−((R−C)/B)ln A) =Be^(−((R−C)/B)ln A) (((R−C)/B)−t) e^((t−((R−C)/B))ln A) =Be^(−((R−C)/B)ln A) (((R−C)/B)−t) (((R−C)/B)−t)ln Ae^((((R−C)/B)−t)ln A) =((A^((R−C)/B) ln A)/B) t=((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B)) x^x =t x=t^(1/x) =e^((ln t)/x) −((ln t)/x)e^(−((ln t)/x)) =−ln t −((ln t)/x)=W(−ln t) x=−((ln t)/(W(−ln t))) ⇒x=−((ln [((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B))])/(W{−ln [((R−C)/B)−(1/(ln A))W(((A^((R−C)/B) ln A)/B))]}))$
$l e t t = x^{x}$ $A^{t} + B t = R - C$ $A^{t} = R - C - B t$ $e^{t \ln A} = B (\frac{R - C}{B} - t)$ $e^{t \ln A} e^{- \frac{R - C}{B} \ln A} = {B e}^{- \frac{R - C}{B} \ln A} (\frac{R - C}{B} - t)$ $e^{(t - \frac{R - C}{B}) \ln A} = {B e}^{- \frac{R - C}{B} \ln A} (\frac{R - C}{B} - t)$ $(\frac{R - C}{B} - t) \ln {A e}^{(\frac{R - C}{B} - t) \ln A} = \frac{A^{\frac{R - C}{B}} \ln A}{B}$ $t = \frac{R - C}{B} - \frac{1}{\ln A} W (\frac{A^{\frac{R - C}{B}} \ln A}{B})$ $x^{x} = t$ $x = t^{\frac{1}{x}} = e^{\frac{\ln t}{x}}$ $- \frac{\ln t}{x} e^{- \frac{\ln t}{x}} = - \ln t$ $- \frac{\ln t}{x} = W (- \ln t)$ $x = - \frac{\ln t}{W (- \ln t)}$ $\Rightarrow x = - \frac{\ln [\frac{R - C}{B} - \frac{1}{\ln A} W (\frac{A^{\frac{R - C}{B}} \ln A}{B})]}{W {- \ln [\frac{R - C}{B} - \frac{1}{\ln A} W (\frac{A^{\frac{R - C}{B}} \ln A}{B})]}}$
Commented by Mastermind last updated on 04/Jun/22

$T h a n k s s o m u c h P r o f . m r W$
Commented by Tawa11 last updated on 04/Jun/22

$Great sir .$
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