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Question Number 41390 by mondodotto@gmail.com last updated on 06/Aug/18
solve  2^(x−2) =8x
$$\boldsymbol{\mathrm{solve}}\:\:\mathrm{2}^{\boldsymbol{{x}}−\mathrm{2}} =\mathrm{8}\boldsymbol{{x}} \\ $$
Answered by MrW3 last updated on 06/Aug/18
(2^x /2^2 )=8x  2^x =32x  e^(x ln 2) =32x  32x e^(−x ln 2) =1  (−x ln 2) e^(−x ln 2) =−((ln 2)/(32))  ⇒−x ln 2=W(−((ln 2)/(32)))  ⇒x=−((W(−((ln 2)/(32))))/(ln 2))= { ((−((−0.0221)/(ln 2))=0.0319)),((−((−5.5452)/(ln 2))=8)) :}
$$\frac{\mathrm{2}^{{x}} }{\mathrm{2}^{\mathrm{2}} }=\mathrm{8}{x} \\ $$$$\mathrm{2}^{{x}} =\mathrm{32}{x} \\ $$$${e}^{{x}\:\mathrm{ln}\:\mathrm{2}} =\mathrm{32}{x} \\ $$$$\mathrm{32}{x}\:{e}^{−{x}\:\mathrm{ln}\:\mathrm{2}} =\mathrm{1} \\ $$$$\left(−{x}\:\mathrm{ln}\:\mathrm{2}\right)\:{e}^{−{x}\:\mathrm{ln}\:\mathrm{2}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}} \\ $$$$\Rightarrow−{x}\:\mathrm{ln}\:\mathrm{2}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right)}{\mathrm{ln}\:\mathrm{2}}=\begin{cases}{−\frac{−\mathrm{0}.\mathrm{0221}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{0319}}\\{−\frac{−\mathrm{5}.\mathrm{5452}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{8}}\end{cases} \\ $$
Commented by mondodotto@gmail.com last updated on 06/Aug/18
thank you sir,but what does W stand for
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir},\mathrm{but}\:\mathrm{what}\:\mathrm{does}\:\mathrm{W}\:\mathrm{stand}\:\mathrm{for} \\ $$
Commented by $@ty@m last updated on 06/Aug/18
Lambert W function.  If f(x)=xe^x   then x=W{f(x)}  i.e. W≡f^(−1)
$${Lambert}\:{W}\:{function}. \\ $$$${If}\:{f}\left({x}\right)={xe}^{{x}} \\ $$$${then}\:{x}={W}\left\{{f}\left({x}\right)\right\} \\ $$$${i}.{e}.\:{W}\equiv{f}^{−\mathrm{1}} \\ $$
Commented by mondodotto@gmail.com last updated on 06/Aug/18
thanx
$$\mathrm{thanx} \\ $$

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