Algebra Questions

Question Number 10248 by j.masanja06@gmail.com last updated on 31/Jan/17

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\$$$$\:\:\:\mathrm{logx}^{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{25}} \\$$

Answered by mrW1 last updated on 01/Feb/17

$${I}.\:{if}\:{x}>\mathrm{0}: \\$$$$\mathrm{log}\:{x}^{\mathrm{2}} =\frac{{x}}{\mathrm{25}} \\$$$$\Rightarrow\mathrm{2log}\:{x}=\frac{{x}}{\mathrm{25}} \\$$$$\mathrm{log}\:{x}=\frac{{x}}{\mathrm{50}} \\$$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{10}}=\frac{{x}}{\mathrm{50}} \\$$$$\mathrm{ln}\:{x}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}{x}=−{ax} \\$$$${with}\:{a}=−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}} \\$$$$\mathrm{ln}\:{x}=−{ax} \\$$$${x}={e}^{−{ax}} \\$$$${xe}^{{ax}} =\mathrm{1} \\$$$$\left({ax}\right){e}^{{ax}} ={a} \\$$$$\Rightarrow{ax}={W}\left({a}\right)\:\:\:\:{Lambert}\:{W}\:{function} \\$$$$\Rightarrow{x}=\frac{{W}\left({a}\right)}{{a}}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}}=−\frac{{W}\left(−\mathrm{0}.\mathrm{046052}\right)}{\mathrm{0}.\mathrm{046052}} \\$$$$\begin{cases}{=\frac{−\mathrm{0}.\mathrm{048332}}{−\mathrm{0}.\mathrm{046052}}=\mathrm{1}.\mathrm{049516}}\\{=\frac{−\mathrm{4}.\mathrm{605162}}{−\mathrm{0}.\mathrm{046052}}=\mathrm{100}}\end{cases} \\$$$$\\$$$${II}.\:{if}\:{x}<\mathrm{0}: \\$$$$\mathrm{log}\:{x}^{\mathrm{2}} =\frac{{x}}{\mathrm{25}} \\$$$${let}\:{t}=−{x},\:{t}>\mathrm{0} \\$$$$\mathrm{log}\:\left(−{t}\right)^{\mathrm{2}} =−\frac{{t}}{\mathrm{25}} \\$$$$\mathrm{log}\:\left({t}\right)^{\mathrm{2}} =−\frac{{t}}{\mathrm{25}} \\$$$$\mathrm{2log}\:{t}=−\frac{{t}}{\mathrm{25}} \\$$$${see}\:{above} \\$$$${t}=\frac{{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}} \\$$$${x}=−{t}=−\frac{{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}}=−\frac{{W}\left(\mathrm{0}.\mathrm{046052}\right)}{\mathrm{0}.\mathrm{046052}} \\$$$$=−\frac{\mathrm{0}.\mathrm{044067}}{\mathrm{0}.\mathrm{046052}}=−\mathrm{0}.\mathrm{956903} \\$$$$\\$$$$\Rightarrow\:{x}=−\frac{{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}}\approx\begin{cases}{−\mathrm{0}.\mathrm{956903}}\\{\mathrm{1}.\mathrm{049516}}\\{\mathrm{100}}\end{cases} \\$$

Answered by arge last updated on 04/Feb/17

$$\mathrm{2}{log}\:{x}=\frac{{x}}{\mathrm{25}} \\$$$$\\$$$${log}\:{x}={y} \\$$$$\\$$$${x}=\mathrm{50}{y} \\$$