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Question Number 160317 by quvonnn last updated on 27/Nov/21

Commented by quvonnn last updated on 27/Nov/21

x=?

$$\boldsymbol{{x}}=? \\ $$

Answered by MathsFan last updated on 27/Nov/21

x^x^(20) =(2)^(1/(√2)) x^(20) lnx=ln(2)^(1/(√2)) lnx•e^(20lnx) =ln(2)^(1/(√2)) 20lnx•e^(20lnx) =20ln(2)^(1/(√2)) 20lnx=w(20ln(2)^(1/(√2)) ) lnx=((w(20ln(2)^(1/(√2)) ))/(20)) x=e^((w(20ln(2)^(1/(√2)) ))/(20))

$${x}^{{x}^{\mathrm{20}} } =\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{20}} {lnx}={ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}} \\ $$$$\:{lnx}\bullet{e}^{\mathrm{20}{lnx}} ={ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}} \\ $$$$\:\mathrm{20}{lnx}\bullet{e}^{\mathrm{20}{lnx}} =\mathrm{20}{ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}} \\ $$$$\:\mathrm{20}{lnx}={w}\left(\mathrm{20}{ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}}\right) \\ $$$$\:{lnx}=\frac{{w}\left(\mathrm{20}{ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}}\right)}{\mathrm{20}} \\ $$$${x}={e}^{\frac{{w}\left(\mathrm{20}{ln}\sqrt[{\sqrt{\mathrm{2}}}]{\mathrm{2}}\right)}{\mathrm{20}}} \\ $$$$ \\ $$

Commented by quvonnn last updated on 27/Nov/21

w=?

$$\boldsymbol{{w}}=? \\ $$

Commented by quvonnn last updated on 28/Nov/21

You understand

$${You}\:{understand} \\ $$

Commented by quvonnn last updated on 28/Nov/21

w=???? what

$${w}=????\:{what} \\ $$

Commented by quvonnn last updated on 28/Nov/21

what is this concept

$${what}\:{is}\:{this}\:{concept} \\ $$

Commented by quvonnn last updated on 28/Nov/21

ok. w(x). f(x)=xe^x

$${ok}.\:{w}\left({x}\right).\:{f}\left({x}\right)={xe}^{{x}} \\ $$

Commented by mr W last updated on 28/Nov/21

W(x) stands for Lambert W function, which is defined as the inverse function of f(x)=xe^x . i.e. W(x)e^(W(x)) =x.

$${W}\left({x}\right)\:{stands}\:{for}\:{Lambert}\:{W}\:{function}, \\ $$$${which}\:{is}\:{defined}\:{as}\:{the}\:{inverse} \\ $$$${function}\:{of}\:{f}\left({x}\right)={xe}^{{x}} .\:{i}.{e}. \\ $$$${W}\left({x}\right){e}^{{W}\left({x}\right)} ={x}. \\ $$

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