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Question Number 127139 by MathSh last updated on 27/Dec/20

2^y = y^2 ⇒ y = ?

$$\mathrm{2}^{{y}} \:=\:{y}^{\mathrm{2}} \:\Rightarrow\:{y}\:=\:? \\ $$

Answered by Ar Brandon last updated on 27/Dec/20

2

$$\mathrm{2} \\ $$

Commented by MathSh last updated on 27/Dec/20

Thanks, solution sir

$${Thanks},\:{solution}\:{sir} \\ $$

Answered by Dwaipayan Shikari last updated on 27/Dec/20

2^y =y^2 ⇒y=±2^(y/2) ⇒y=e^(±(y/2)log(2)) ⇒±(y/2)log(2)=(1/2)e^(±(y/2)log(2)) log(2) ⇒±(y/2)log(2)e^(±(y/2)log(2)) =±(1/2)log(2) ⇒±(y/2)log(2)=W_0 (±((log2)/2))⇒y=±((2W_0 (±((log2)/2)))/(log(2)))

$$\mathrm{2}^{{y}} ={y}^{\mathrm{2}} \Rightarrow{y}=\pm\mathrm{2}^{\frac{{y}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}={e}^{\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right)} \\ $$$$\Rightarrow\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right)} {log}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right){e}^{\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right)} =\pm\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\pm\frac{{y}}{\mathrm{2}}{log}\left(\mathrm{2}\right)={W}_{\mathrm{0}} \left(\pm\frac{{log}\mathrm{2}}{\mathrm{2}}\right)\Rightarrow{y}=\pm\frac{\mathrm{2}{W}_{\mathrm{0}} \left(\pm\frac{{log}\mathrm{2}}{\mathrm{2}}\right)}{{log}\left(\mathrm{2}\right)} \\ $$

Commented by MathSh last updated on 27/Dec/20

Thanks sir, answer:?

$${Thanks}\:{sir},\:{answer}:? \\ $$

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