Question Number 27321 by 803jaideep@gmail.com last updated on 05/Jan/18
Commented by 803jaideep@gmail.com last updated on 05/Jan/18
$$\mathrm{plz}\:\mathrm{solve}\:\mathrm{it} \\ $$
Answered by mrW1 last updated on 05/Jan/18
$${x}>\mathrm{0} \\ $$$$\Rightarrow\mid{x}\mid={x} \\ $$$$\mathrm{log}_{\mathrm{0}.\mathrm{5}} \:{x}=\mid{x}\mid \\ $$$$\Rightarrow\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{0}.\mathrm{5}}={x} \\ $$$$\frac{\mathrm{ln}\:{x}}{−\mathrm{ln}\:\mathrm{2}}={x}={e}^{\mathrm{ln}\:{x}} \\ $$$$\left(−\mathrm{ln}\:{x}\right){e}^{−\mathrm{ln}\:{x}} =\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow−\mathrm{ln}\:{x}={W}\left(\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}={e}^{{W}\left(\mathrm{ln}\:\mathrm{2}\right)} =\frac{\mathrm{ln}\:\mathrm{2}}{{W}\left(\mathrm{ln}\:\mathrm{2}\right)} \\ $$$$\Rightarrow{x}=\frac{{W}\left(\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}=\frac{{W}\left(\mathrm{0}.\mathrm{639}\right)}{\mathrm{0}.\mathrm{639}}=\frac{\mathrm{0}.\mathrm{444436}}{\mathrm{0}.\mathrm{639}}=\mathrm{0}.\mathrm{641} \\ $$