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Question Number 156709 by aliyn last updated on 14/Oct/21

solve (x^3 )^(1/(1/x)) = 27

$$\boldsymbol{{solve}}\:\sqrt[{\frac{\mathrm{1}}{\boldsymbol{{x}}}}]{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:\mathrm{27} \\ $$

Answered by mr W last updated on 14/Oct/21

(x^3 )^(1/(1/x)) =(x^3 )^x =(x^x )^3 =27=3^3 ⇒x^x =3 ⇒x=3^(1/x) =e^((ln 3)/x) ⇒ln 3=((ln 3)/x)e^((ln 3)/x) ⇒((ln 3)/x)=W(ln 3) ⇒x=((ln 3)/(W(ln 3)))≈((ln 3)/(0.60182928))=1.825455

$$\sqrt[{\frac{\mathrm{1}}{{x}}}]{{x}^{\mathrm{3}} }=\left({x}^{\mathrm{3}} \right)^{{x}} =\left({x}^{{x}} \right)^{\mathrm{3}} =\mathrm{27}=\mathrm{3}^{\mathrm{3}} \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{3} \\ $$$$\Rightarrow{x}=\mathrm{3}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:\mathrm{3}}{{x}}} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{3}=\frac{\mathrm{ln}\:\mathrm{3}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{3}}{{x}}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\mathrm{3}}{{x}}=\mathbb{W}\left(\mathrm{ln}\:\mathrm{3}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{3}}{\mathbb{W}\left(\mathrm{ln}\:\mathrm{3}\right)}\approx\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{0}.\mathrm{60182928}}=\mathrm{1}.\mathrm{825455} \\ $$

Commented by tabata last updated on 14/Oct/21

thank you sir. but can you give me formulla how can calculate ((any veriable))^(1/(1/(any veriable)))

$${thank}\:{you}\:{sir}.\:{but}\:{can}\:{you}\:{give}\:{me}\:{formulla} \\ $$$${how}\:{can}\:{calculate}\:\sqrt[{\frac{\mathrm{1}}{{any}\:{veriable}}}]{{any}\:{veriable}} \\ $$

Commented by mr W last updated on 14/Oct/21

i think you know (2)^(1/3) =2^(1/3) . generally (a)^(1/b) =a^(1/b) when you replace b with (1/c), you have (a)^(1/(1/c)) =a^(1/(((1/c)))) =a^c

$${i}\:{think}\:{you}\:{know}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} .\:{generally} \\ $$$$\sqrt[{{b}}]{{a}}={a}^{\frac{\mathrm{1}}{{b}}} \\ $$$${when}\:{you}\:{replace}\:{b}\:{with}\:\frac{\mathrm{1}}{{c}},\:{you}\:{have} \\ $$$$\sqrt[{\frac{\mathrm{1}}{{c}}}]{{a}}={a}^{\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{{c}}\right)}} ={a}^{{c}} \\ $$

Commented by tabata last updated on 14/Oct/21

thank you sir

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

Commented by mr W last updated on 15/Oct/21

oui!

$${oui}! \\ $$

Commented by Sky_b last updated on 15/Oct/21

la fontion W de lambert ??

$$ \\ $$$${la}\:{fontion}\:{W}\:{de}\:{lambert}\:?? \\ $$$$ \\ $$$$ \\ $$

Commented by Ruuudiy last updated on 16/Oct/21

Ans I think you know (y)^(1/x) =y^(1/x) But if you replace x with (1/z)... (y)^(1/(1/z)) =y^(1/(1/z)) =y^z

$$\mathrm{Ans} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{know}\:\sqrt[{{x}}]{{y}}={y}^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{you}\:\mathrm{replace}\:{x}\:\mathrm{with}\:\frac{\mathrm{1}}{{z}}... \\ $$$$\sqrt[{\frac{\mathrm{1}}{{z}}}]{{y}}={y}^{\frac{\mathrm{1}}{\frac{\mathrm{1}}{{z}}}} ={y}^{{z}} \\ $$

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