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Question Number 25787 by ibraheem160 last updated on 14/Dec/17

2^x =x^(2 ) ,hence x=2 . prove

$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}\:} ,{hence}\:{x}=\mathrm{2}\:.\:{prove} \\ $$

Commented by mrW1 last updated on 14/Dec/17

It′s not true, Sir.  If 2^x =x^2 ,then x=2 or x=4 or x=−0.77.

$${It}'{s}\:{not}\:{true},\:{Sir}. \\ $$$${If}\:\mathrm{2}^{{x}} ={x}^{\mathrm{2}} ,{then}\:{x}=\mathrm{2}\:{or}\:{x}=\mathrm{4}\:{or}\:{x}=−\mathrm{0}.\mathrm{77}. \\ $$

Commented by mrW1 last updated on 15/Dec/17

solution of 2^x =x^2 :    if x>0:  2^(x/2) =x  e^((xln 2)/2) =x  1=xe^(−((xln 2)/2))   −((ln 2)/2)=(−((xln 2)/2))×e^(−((xln 2)/2))   ⇒−((xln 2)/2)=W(−((ln 2)/2))  ⇒x=−((W(−((ln 2)/2)))/((ln 2)/2))=−((W(−ln (√2)))/(ln (√2)))  if x<0:  2^(x/2) =−x  e^((xln 2)/2) =−x  1=−xe^(−((xln 2)/2))   ((ln 2)/2)=(−((xln 2)/2))×e^(−((xln 2)/2))   ⇒−((xln 2)/2)=W(((ln 2)/2))  ⇒x=−((W(−((ln 2)/2)))/((ln 2)/2))=−((W(ln (√2)))/(ln (√2)))  all together:  ⇒x=−((W(±ln (√2)))/(ln (√2)))  = { ((−((0.2657)/(0.3466))=−0.7666)),((−((−0.6932)/(0.3466))=2)),((−((1.3864)/(0.3466))=4)) :}    in general the solution of  a^x =x^n   is x=−((W(−((ln a)/n)))/((ln a)/n))  or x=−((W(±((ln a)/n)))/((ln a)/n)) if n is even

$${solution}\:{of}\:\mathrm{2}^{{x}} ={x}^{\mathrm{2}} : \\ $$$$ \\ $$$${if}\:{x}>\mathrm{0}: \\ $$$$\mathrm{2}^{\frac{{x}}{\mathrm{2}}} ={x} \\ $$$${e}^{\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} ={x} \\ $$$$\mathrm{1}={xe}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}=\left(−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)×{e}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\Rightarrow−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}=\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)}{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}}=−\frac{\mathbb{W}\left(−\mathrm{ln}\:\sqrt{\mathrm{2}}\right)}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$${if}\:{x}<\mathrm{0}: \\ $$$$\mathrm{2}^{\frac{{x}}{\mathrm{2}}} =−{x} \\ $$$${e}^{\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =−{x} \\ $$$$\mathrm{1}=−{xe}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}=\left(−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)×{e}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\Rightarrow−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}=\mathbb{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)}{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}}=−\frac{\mathbb{W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\right)}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$${all}\:{together}: \\ $$$$\Rightarrow{x}=−\frac{\mathbb{W}\left(\pm\mathrm{ln}\:\sqrt{\mathrm{2}}\right)}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$$=\begin{cases}{−\frac{\mathrm{0}.\mathrm{2657}}{\mathrm{0}.\mathrm{3466}}=−\mathrm{0}.\mathrm{7666}}\\{−\frac{−\mathrm{0}.\mathrm{6932}}{\mathrm{0}.\mathrm{3466}}=\mathrm{2}}\\{−\frac{\mathrm{1}.\mathrm{3864}}{\mathrm{0}.\mathrm{3466}}=\mathrm{4}}\end{cases} \\ $$$$ \\ $$$${in}\:{general}\:{the}\:{solution}\:{of} \\ $$$${a}^{{x}} ={x}^{{n}} \\ $$$${is}\:{x}=−\frac{\mathbb{W}\left(−\frac{{ln}\:{a}}{{n}}\right)}{\frac{\mathrm{ln}\:{a}}{{n}}} \\ $$$${or}\:{x}=−\frac{\mathbb{W}\left(\pm\frac{{ln}\:{a}}{{n}}\right)}{\frac{\mathrm{ln}\:{a}}{{n}}}\:{if}\:{n}\:{is}\:{even} \\ $$

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