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Question Number 65115 by hovea cw last updated on 25/Jul/19

x^x =64 find x

$$\mathrm{x}^{\mathrm{x}} =\mathrm{64} \\ $$$$\mathrm{find}\:\mathrm{x} \\ $$

Answered by mr W last updated on 25/Jul/19

x^x =64 x=64^(1/x) x=e^((ln 64)/x) 1=(1/x)e^((ln 64)/x) ln 64=(((ln 64)/x))e^((ln 64)/x) ⇒((ln 64)/x)=W(ln 64) ⇒x=((ln 64)/(W(ln 64)))=((ln 64)/(1.223517))=3.399122

$${x}^{{x}} =\mathrm{64} \\ $$$${x}=\mathrm{64}^{\frac{\mathrm{1}}{{x}}} \\ $$$${x}={e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\mathrm{ln}\:\mathrm{64}=\left(\frac{\mathrm{ln}\:\mathrm{64}}{{x}}\right){e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\mathrm{64}}{{x}}=\mathbb{W}\left(\mathrm{ln}\:\mathrm{64}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{64}}{\mathbb{W}\left(\mathrm{ln}\:\mathrm{64}\right)}=\frac{\mathrm{ln}\:\mathrm{64}}{\mathrm{1}.\mathrm{223517}}=\mathrm{3}.\mathrm{399122} \\ $$

Commented by hovea cw last updated on 25/Jul/19

sir I don't know a lot of the Lambert omega function so am kinda stuck at the last few lines plz explain how the function comes about sir ����

Commented by mr W last updated on 25/Jul/19

Lambert W function is the inverse function from y=xe^x . i.e. if y=xe^x then x=W(y). in our example i have formed the original equation into (((ln 64)/x))e^((((ln 64)/x))) =ln 64 that means ((ln 64)/x)=W(ln 64) ⇒x=((ln 64)/(W(ln 64)))

$${Lambert}\:{W}\:{function}\:{is}\:{the}\:{inverse} \\ $$$${function}\:{from}\:{y}={xe}^{{x}} .\:{i}.{e}. \\ $$$${if}\:{y}={xe}^{{x}} \:{then}\:{x}={W}\left({y}\right). \\ $$$${in}\:{our}\:{example}\:{i}\:{have}\:{formed}\:{the} \\ $$$${original}\:{equation}\:{into} \\ $$$$\left(\frac{\mathrm{ln}\:\mathrm{64}}{{x}}\right){e}^{\left(\frac{\mathrm{ln}\:\mathrm{64}}{{x}}\right)} =\mathrm{ln}\:\mathrm{64} \\ $$$${that}\:{means} \\ $$$$\frac{\mathrm{ln}\:\mathrm{64}}{{x}}={W}\left(\mathrm{ln}\:\mathrm{64}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{64}}{{W}\left(\mathrm{ln}\:\mathrm{64}\right)} \\ $$

Commented by mr W last updated on 25/Jul/19

if you can reform an equation into the form ♠e^♠ =♣, then you can apply Lambert W function as ♠=W(♣) where ♠ is an expression containng x and ♣ is a constant.

$${if}\:{you}\:{can}\:{reform}\:{an}\:{equation}\:{into} \\ $$$${the}\:{form}\:\spadesuit{e}^{\spadesuit} =\clubsuit,\:{then}\:{you}\:{can}\:{apply} \\ $$$${Lambert}\:{W}\:{function}\:{as} \\ $$$$\spadesuit={W}\left(\clubsuit\right) \\ $$$${where}\:\spadesuit\:{is}\:{an}\:{expression}\:{containng}\:{x} \\ $$$${and}\:\clubsuit\:{is}\:{a}\:{constant}. \\ $$

Commented by hovea cw last updated on 25/Jul/19

Thank u sir and how do we approximate the value of w or is it a constant number (*-*)

Commented by mr W last updated on 25/Jul/19

in internet you may find online calculator for W function values. but since W(ln 64) is the root of xe^x =ln 64 you may get the root via graphic method. i use e.g. an app named Grapher.

$${in}\:{internet}\:{you}\:{may}\:{find}\:{online} \\ $$$${calculator}\:{for}\:{W}\:{function}\:{values}. \\ $$$${but}\:{since}\:{W}\left(\mathrm{ln}\:\mathrm{64}\right)\:{is}\:{the}\:{root}\:{of} \\ $$$${xe}^{{x}} =\mathrm{ln}\:\mathrm{64} \\ $$$${you}\:{may}\:{get}\:{the}\:{root}\:{via}\:{graphic} \\ $$$${method}.\:{i}\:{use}\:{e}.{g}.\:{an}\:{app}\:{named} \\ $$$${Grapher}. \\ $$

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