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Question Number 96321 by Mathudent last updated on 31/May/20

It is given that x^2 =2^x . Find x.

$${It}\:{is}\:{given}\:{that}\:{x}^{\mathrm{2}} =\mathrm{2}^{{x}} .\:{Find}\:{x}. \\ $$

Commented by bemath last updated on 31/May/20

look qn 96241

$$\mathrm{look}\:\mathrm{qn}\:\mathrm{96241} \\ $$

Answered by selea last updated on 31/May/20

2^x =x^2 ln(2^x )=ln(x^2 ) xln(2)=2ln∣x∣ ((lnx)/x)=(l/2)ln(2) ln(x)e^(ln((1/x))) =ln(√2) ln(x)e^(−lnx) =ln(√2) −ln(x)e^(−lnx) =−ln(√2) since w(xe^x )=x where w() is lambert function w(−lnxe^(−lnx) )=w(−ln(√2)) lnx=−w(−ln(√2)) x=e^(−w(−ln(√2))) for x>0

$$\mathrm{2}^{\mathrm{x}} =\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)=\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{xln}\left(\mathrm{2}\right)=\mathrm{2ln}\mid\mathrm{x}\mid \\ $$$$\frac{\mathrm{lnx}}{\mathrm{x}}=\frac{\mathrm{l}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)\mathrm{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} =\mathrm{ln}\sqrt{\mathrm{2}} \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)\mathrm{e}^{−\mathrm{lnx}} =\mathrm{ln}\sqrt{\mathrm{2}} \\ $$$$−\mathrm{ln}\left(\mathrm{x}\right)\mathrm{e}^{−\mathrm{lnx}} =−\mathrm{ln}\sqrt{\mathrm{2}} \\ $$$$\mathrm{since} \\ $$$$\mathrm{w}\left(\mathrm{xe}^{\mathrm{x}} \right)=\mathrm{x}\:\:\mathrm{where}\:\mathrm{w}\left(\right)\:\mathrm{is}\:\mathrm{lambert}\:\mathrm{function} \\ $$$$\mathrm{w}\left(−\mathrm{lnxe}^{−\mathrm{lnx}} \right)=\mathrm{w}\left(−\mathrm{ln}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{lnx}=−\mathrm{w}\left(−\mathrm{ln}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{x}=\mathrm{e}^{−\mathrm{w}\left(−\mathrm{ln}\sqrt{\mathrm{2}}\right)} \:\:\:\mathrm{for}\:\mathrm{x}>\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Mathudent last updated on 31/May/20

Can you solve without using De l′Ambert function?

$${Can}\:{you}\:{solve}\:{without}\:{using}\:{De}\:{l}'{Ambert}\:{function}? \\ $$

Commented by selea last updated on 31/May/20

U can use Newton Raphson Method

$$\mathrm{U}\:\mathrm{can}\:\mathrm{use}\:\mathrm{Newton}\:\mathrm{Raphson}\:\mathrm{Method} \\ $$

Commented by mr W last updated on 31/May/20

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