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Question Number 103047 by bemath last updated on 12/Jul/20

$$findtheproductofroots$$$$\sqrt[3]{2021}{x}^{{\log }_{2021}\left(x\right)}=x^3$$

Commented by mr W last updated on 12/Jul/20

$$\mathrm{\Pi }x={2021}^3?$$

Answered by floor(10²Eta[1]) last updated on 12/Jul/20

$${\log }_{2021}\left(\mathrm{x}\right)=\mathrm{y}\Rightarrow \mathrm{x}={2021}^{\mathrm{y}}$$$$\Rightarrow \sqrt[3]{2021}.{\left({2021}^{\mathrm{y}}\right)}^{\mathrm{y}}={\left({2021}^{\mathrm{y}}\right)}^3$$$${2021}^{1/3}.{2021}^{{\mathrm{y}}^2}={2021}^{3\mathrm{y}}$$$$\Rightarrow {\mathrm{y}}^2+\frac{1}{3}=3\mathrm{y}\Rightarrow {\mathrm{y}}^2-3\mathrm{y}+\frac{1}{3}=0$$$$\mathrm{product}={\mathrm{x}}_1.{\mathrm{x}}_2={2021}^{{\mathrm{y}}_1}.{2021}^{{\mathrm{y}}_2}$$$$={2021}^{{\mathrm{y}}_1+{\mathrm{y}}_2}={2021}^3$$

Answered by bemath last updated on 14/Jul/20

$$setx={2021}^y\iff y=\log {}_{2021}\left(x\right)$$$$\Rightarrow {2021}^{\frac{1}{3}}{\left({2021}^y\right)}^y={2021}^{3y}$$$${2021}^{y^2-3y+\frac{1}{3}}=1$$$$\Rightarrow y^2-3y+\frac{1}{3}=0;\mathrm{\Delta }=9-\frac{4}{3}>0$$$$\Rightarrow (y-a)(y-b)=0\Rightarrow a+b=3$$$$A={2021}^a;B={2021}^b$$$$\iff A.B={2021}^{a+b}={2021}^3$$$$={21}^3\left(mod100\right)=261(mod$$$$1000)$$

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