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Question Number 172029 by Mikenice last updated on 23/Jun/22

solve: x^2 =2^x

$${solve}: \\ $$$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Jun/22

x=2,4

$${x}=\mathrm{2},\mathrm{4} \\ $$

Commented by Mikenice last updated on 23/Jun/22

please show the solution

$${please}\:{show}\:{the}\:{solution} \\ $$

Commented by mr W last updated on 23/Jun/22

x=−0.76666 as well

$${x}=−\mathrm{0}.\mathrm{76666}\:{as}\:{well} \\ $$

Commented by puissant last updated on 23/Jun/22

x^2 =2^(x ) ⇒ 2lnx = xln2 ⇒ ((lnx)/x) = ((ln2)/2) ((lnx)/x) = ((2ln2)/4) = ((ln4)/4). S={2 ; 4}

$${x}^{\mathrm{2}} =\mathrm{2}^{{x}\:} \:\Rightarrow\:\:\mathrm{2}{lnx}\:=\:{xln}\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{lnx}}{{x}}\:=\:\frac{{ln}\mathrm{2}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\frac{{lnx}}{{x}}\:=\:\frac{\mathrm{2}{ln}\mathrm{2}}{\mathrm{4}}\:=\:\frac{{ln}\mathrm{4}}{\mathrm{4}}. \\ $$$$\:\:\:\:\:{S}=\left\{\mathrm{2}\:;\:\mathrm{4}\right\} \\ $$

Commented by mr W last updated on 23/Jun/22

from (a/b)=(c/d) we get not only a=c, b=d. generally we get a=kc, b=kd.

$${from}\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\:{we}\:{get}\:{not}\:{only} \\ $$$${a}={c},\:{b}={d}.\: \\ $$$${generally}\:{we}\:{get} \\ $$$${a}={kc},\:{b}={kd}. \\ $$

Answered by mr W last updated on 25/Jun/22

x^2 =2^x x=±2^(x/2) =±e^((xln 2)/2) (−((xln 2)/2))e^(−((xln 2)/2)) =±((ln 2)/2) −((xln 2)/2)=W(±((ln 2)/2)) ⇒x=−(2/(ln 2))W(±((ln 2)/2)) = { ((−(2/(ln 2))W(−((ln 2)/2))= { (2),(4) :})),((−(2/(ln 2))W(((ln 2)/2))=−0.766665)) :}

$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$${x}=\pm\mathrm{2}^{\frac{{x}}{\mathrm{2}}} =\pm{e}^{\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\left(−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right){e}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}={W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\begin{cases}{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{\mathrm{2}}\\{\mathrm{4}}\end{cases}}\\{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=−\mathrm{0}.\mathrm{766665}}\end{cases} \\ $$

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