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Question Number 27321 by 803jaideep@gmail.com last updated on 05/Jan/18

Commented by 803jaideep@gmail.com last updated on 05/Jan/18

plz solve it

$$\mathrm{plz}\:\mathrm{solve}\:\mathrm{it} \\ $$

Answered by mrW1 last updated on 05/Jan/18

x>0 ⇒∣x∣=x log_(0.5) x=∣x∣ ⇒((ln x)/(ln 0.5))=x ((ln x)/(−ln 2))=x=e^(ln x) (−ln x)e^(−ln x) =ln 2 ⇒−ln x=W(ln 2) ⇒(1/x)=e^(W(ln 2)) =((ln 2)/(W(ln 2))) ⇒x=((W(ln 2))/(ln 2))=((W(0.639))/(0.639))=((0.444436)/(0.639))=0.641

$${x}>\mathrm{0} \\ $$$$\Rightarrow\mid{x}\mid={x} \\ $$$$\mathrm{log}_{\mathrm{0}.\mathrm{5}} \:{x}=\mid{x}\mid \\ $$$$\Rightarrow\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{0}.\mathrm{5}}={x} \\ $$$$\frac{\mathrm{ln}\:{x}}{−\mathrm{ln}\:\mathrm{2}}={x}={e}^{\mathrm{ln}\:{x}} \\ $$$$\left(−\mathrm{ln}\:{x}\right){e}^{−\mathrm{ln}\:{x}} =\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow−\mathrm{ln}\:{x}={W}\left(\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}={e}^{{W}\left(\mathrm{ln}\:\mathrm{2}\right)} =\frac{\mathrm{ln}\:\mathrm{2}}{{W}\left(\mathrm{ln}\:\mathrm{2}\right)} \\ $$$$\Rightarrow{x}=\frac{{W}\left(\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}=\frac{{W}\left(\mathrm{0}.\mathrm{639}\right)}{\mathrm{0}.\mathrm{639}}=\frac{\mathrm{0}.\mathrm{444436}}{\mathrm{0}.\mathrm{639}}=\mathrm{0}.\mathrm{641} \\ $$

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