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Question Number 7178 by aftab ahmad last updated on 14/Aug/16

Commented by aftab ahmad last updated on 14/Aug/16

Very nice approach sir..Thanks a lot.

$$\boldsymbol{{Very}}\:\boldsymbol{{nice}}\:\boldsymbol{{approach}}\:\boldsymbol{{sir}}..\boldsymbol{{T}}{h}\boldsymbol{{anks}}\:\boldsymbol{{a}}\:\boldsymbol{{lot}}. \\ $$

Commented by Yozzia last updated on 14/Aug/16

x^x^(1/3) =(9^(1/3) /3)=(1/3^(1/3) ) Let u=x^(1/3) ⇒x=u^3 ∴ u^(3u) =3^(−1/3) ⇒u^u =3^(−1/9) ulnu=ln3^(−1/9) e^(lnu) lnu=((−1)/9)ln3 ⇒W(e^(lnu) lnu)=W(((−1)/9)ln3) ⇒lnu=W(((−ln3)/9)) ⇒u=e^(W(((−ln3)/9))) ∴ x^(1/3) =e^(W(((−ln3)/9))) x=e^(3W(((−ln3)/9))) ≈0.6561 W=product−log function

$${x}^{{x}^{\mathrm{1}/\mathrm{3}} } =\frac{\mathrm{9}^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$${Let}\:{u}={x}^{\mathrm{1}/\mathrm{3}} \Rightarrow{x}={u}^{\mathrm{3}} \\ $$$$\therefore\:{u}^{\mathrm{3}{u}} =\mathrm{3}^{−\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow{u}^{{u}} =\mathrm{3}^{−\mathrm{1}/\mathrm{9}} \\ $$$${ulnu}={ln}\mathrm{3}^{−\mathrm{1}/\mathrm{9}} \\ $$$${e}^{{lnu}} {lnu}=\frac{−\mathrm{1}}{\mathrm{9}}{ln}\mathrm{3} \\ $$$$\Rightarrow{W}\left({e}^{{lnu}} {lnu}\right)={W}\left(\frac{−\mathrm{1}}{\mathrm{9}}{ln}\mathrm{3}\right) \\ $$$$\Rightarrow{lnu}={W}\left(\frac{−{ln}\mathrm{3}}{\mathrm{9}}\right) \\ $$$$\Rightarrow{u}={e}^{{W}\left(\frac{−{ln}\mathrm{3}}{\mathrm{9}}\right)} \\ $$$$\therefore\:{x}^{\mathrm{1}/\mathrm{3}} ={e}^{{W}\left(\frac{−{ln}\mathrm{3}}{\mathrm{9}}\right)} \\ $$$${x}={e}^{\mathrm{3}{W}\left(\frac{−{ln}\mathrm{3}}{\mathrm{9}}\right)} \approx\mathrm{0}.\mathrm{6561} \\ $$$${W}={product}−{log}\:{function} \\ $$$$ \\ $$

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