Question-20132

Question Number 20132 by mondodotto@gmail.com last updated on 22/Aug/17

Answered by mrW1 last updated on 22/Aug/17

log x^2 =(x/(25)) x^2 =10^(x/(25)) x=±10^(x/(50)) 50×(x/(50))=±10^(x/(50)) 50t=±10^t with t=(x/(50)) 50t=±e^(tln 10) 50te^(−tln 10) =±1 (−tln 10)e^(−tln 10) =±((ln 10)/(50)) ⇒−tln 10=W(±((ln 10)/(50))) ⇒t=−(1/(ln 10))W(±((ln 10)/(50))) ⇒x=−((50)/(ln 10))W(±((ln 10)/(50))) = { ((−((50)/(ln 10))W(((ln 10)/(50)))=−0.95698)),((−((50)/(ln 10))W(−((ln 10)/(50)))= { ((1.05952)),((100)) :})) :}

$$\mathrm{log}\:\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{25}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{25}}} \\ $$$$\mathrm{x}=\pm\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{50}}} \\ $$$$\mathrm{50}×\frac{\mathrm{x}}{\mathrm{50}}=\pm\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{50}}} \\ $$$$\mathrm{50t}=\pm\mathrm{10}^{\mathrm{t}} \:\mathrm{with}\:\mathrm{t}=\frac{\mathrm{x}}{\mathrm{50}} \\ $$$$\mathrm{50t}=\pm\mathrm{e}^{\mathrm{tln}\:\mathrm{10}} \\ $$$$\mathrm{50te}^{−\mathrm{tln}\:\mathrm{10}} =\pm\mathrm{1} \\ $$$$\left(−\mathrm{tln}\:\mathrm{10}\right)\mathrm{e}^{−\mathrm{tln}\:\mathrm{10}} =\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}} \\ $$$$\Rightarrow−\mathrm{tln}\:\mathrm{10}=\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right) \\ $$$$\Rightarrow\mathrm{t}=−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{10}}\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{50}}{\mathrm{ln}\:\mathrm{10}}\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right) \\ $$$$=\begin{cases}{−\frac{\mathrm{50}}{\mathrm{ln}\:\mathrm{10}}\mathbb{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)=−\mathrm{0}.\mathrm{95698}}\\{−\frac{\mathrm{50}}{\mathrm{ln}\:\mathrm{10}}\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)=\begin{cases}{\mathrm{1}.\mathrm{05952}}\\{\mathrm{100}}\end{cases}}\end{cases} \\ $$

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