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Question Number 172021 by Mikenice last updated on 23/Jun/22
solve:  2^x^2  =x^(2x)
$${solve}: \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } ={x}^{\mathrm{2}{x}} \\ $$
Answered by puissant last updated on 23/Jun/22
⇒  x^2 ln2 = 2xlnx  ⇒ xln2 = 2lnx  ⇒ ((lnx)/x)  =  ((ln2)/2)   ⇒   x=2
$$\Rightarrow\:\:{x}^{\mathrm{2}} {ln}\mathrm{2}\:=\:\mathrm{2}{xlnx} \\ $$$$\Rightarrow\:{xln}\mathrm{2}\:=\:\mathrm{2}{lnx} \\ $$$$\Rightarrow\:\frac{{lnx}}{{x}}\:\:=\:\:\frac{{ln}\mathrm{2}}{\mathrm{2}}\:\:\:\Rightarrow\:\:\:{x}=\mathrm{2} \\ $$
Commented by mr W last updated on 23/Jun/22
 ((lnx)/x)=((ln2)/2)=((2×ln 2)/(2×2))=((ln 4)/4)  ⇒x=2, x=4    from (a/b)=(c/d) we get not only  a=c, b=d. but generally  a=kc, b=kd.
$$\:\frac{{lnx}}{{x}}=\frac{{ln}\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{2}×\mathrm{ln}\:\mathrm{2}}{\mathrm{2}×\mathrm{2}}=\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{2},\:{x}=\mathrm{4} \\ $$$$ \\ $$$${from}\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\:{we}\:{get}\:{not}\:{only} \\ $$$${a}={c},\:{b}={d}.\:{but}\:{generally} \\ $$$${a}={kc},\:{b}={kd}. \\ $$
Commented by Tawa11 last updated on 25/Jun/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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