A-x-x-Bx-x-C-R-Make-x-the-subject-of-the-formular-

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

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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