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Question Number 109336 by Rasikh last updated on 22/Aug/20

Commented by Ari last updated on 22/Aug/20

x=2

Commented by Ari last updated on 22/Aug/20

x=4

Commented by mr W last updated on 22/Aug/20

and x≈−0.7667

$${and}\:{x}\approx−\mathrm{0}.\mathrm{7667} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Aug/20

2^x =x^2 xlog2=2logx e^(logx) =((2logx)/(log2)) e^(−logx) =((log2)/(2logx)) −logxe^(−logx) =−((log2)/2) −logx=W_0 (−((log2)/2)) x=e^(−W_0 (((−log2)/2))) { ((x=2,4)),((x=e^(−W_0 (−((log2)/2))) )) :}

$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}} \\ $$$${xlog}\mathrm{2}=\mathrm{2}{logx} \\ $$$${e}^{{logx}} =\frac{\mathrm{2}{logx}}{{log}\mathrm{2}} \\ $$$${e}^{−{logx}} =\frac{{log}\mathrm{2}}{\mathrm{2}{logx}} \\ $$$$−{logxe}^{−{logx}} =−\frac{{log}\mathrm{2}}{\mathrm{2}} \\ $$$$−{logx}={W}_{\mathrm{0}} \left(−\frac{{log}\mathrm{2}}{\mathrm{2}}\right) \\ $$$${x}={e}^{−{W}_{\mathrm{0}} \left(\frac{−{log}\mathrm{2}}{\mathrm{2}}\right)} \\ $$$$\begin{cases}{{x}=\mathrm{2},\mathrm{4}}\\{{x}={e}^{−{W}_{\mathrm{0}} \left(−\frac{{log}\mathrm{2}}{\mathrm{2}}\right)} }\end{cases} \\ $$$$ \\ $$

Commented by mr W last updated on 22/Aug/20

in 2^x =x^2 the x can be negative, but in x log 2=2 log x the x must be positive. it means some roots may get lost in your solution x=e^(−W_0 (((−log 2)/2))) . in fact you get only two roots: x=2, 4. the “others”you mean are not included in your solution x=e^(−W_0 (((−log 2)/2))) . in think the complete solution should be x=∓e^(−W_0 (±((log 2)/2))) . please check!

$${in}\:\mathrm{2}^{{x}} ={x}^{\mathrm{2}} \:{the}\:{x}\:{can}\:{be}\:{negative},\:{but}\:{in} \\ $$$${x}\:\mathrm{log}\:\mathrm{2}=\mathrm{2}\:\mathrm{log}\:{x}\:{the}\:{x}\:{must}\:{be}\:{positive}. \\ $$$${it}\:{means}\:{some}\:{roots}\:{may}\:{get}\:{lost}\:{in}\: \\ $$$${your}\:{solution}\:{x}={e}^{−{W}_{\mathrm{0}} \left(\frac{−\mathrm{log}\:\mathrm{2}}{\mathrm{2}}\right)} .\:{in}\:{fact} \\ $$$${you}\:{get}\:{only}\:{two}\:{roots}:\:{x}=\mathrm{2},\:\mathrm{4}.\:{the} \\ $$$$``{others}''{you}\:{mean}\:{are}\:\:{not}\:{included}\:{in} \\ $$$${your}\:{solution}\:{x}={e}^{−{W}_{\mathrm{0}} \left(\frac{−\mathrm{log}\:\mathrm{2}}{\mathrm{2}}\right)} .\:{in}\:{think} \\ $$$${the}\:{complete}\:{solution}\:{should}\:{be} \\ $$$${x}=\mp{e}^{−{W}_{\mathrm{0}} \left(\pm\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{2}}\right)} .\:{please}\:{check}! \\ $$

Commented by Dwaipayan Shikari last updated on 22/Aug/20

2^x =x^2 2^(x/2) =±x

$$\mathrm{2}^{\mathrm{x}} =\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{\mathrm{x}}{\mathrm{2}}} =\pm\mathrm{x} \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 22/Aug/20

yes!

$${yes}! \\ $$

Commented by Rasikh last updated on 22/Aug/20

thanks

$$\mathrm{thanks} \\ $$

Commented by Rasikh last updated on 22/Aug/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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