Решите уравнение [tex](\sqrt[3]{x})^{lg x} = 10^{6+lg x}[/tex]. Created by Frameinworker. algebra-ru

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ОДЗ : x > 0

$(\sqrt[3]{x} )^{lgx}=10^{6+lgx}\\\\x^{\frac{1}{3}lgx}=10^{6+lgx}\\\\lgx^{\frac{1}{3}lgx}=lg10^{6+lgx}\\\\\frac{1}{3}lgx*lgx=(6+lgx)*lg10\\\\\frac{1}{3}lg^{2}x=6+lgx\\\\lg^{2}x-3lgx-18=0\\\\lgx=m\\\\m^{2}-3m-18=0\\\\m_{1}=-3\\\\m_{2}=6$

$lgx=-3\\\\x_{1}=10^{-3}=0,001\\\\lgx=6\\\\x_{2}=10^{6}=1000000\\\\Otvet:\boxed{0,001;1000000}$

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Frameinworker спасибо большое

novinanovina Здравствуйте. Помогите с заданием https://znanija.com/task/33639717?utm_source=android&utm_medium=share&utm_campaign=question. Буду очень благодарна.

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$(\sqrt[3]{x})^{lgx}=10^{6+lgx}\; \; ,\; \; \; ODZ:\; x>0\\\\lg\Big (x^{\frac{lgx}{3}}\Big )=lg\Big (10^{6+lgx}\Big )\\\\\frac{lgx}{3}\cdot lgx=(6+lgx)\cdot \underbrace {lg10}_{1}\\\\\frac{1}{3}\, lg^2x=6+lgx\\\\lg^2x-3\, lgx-18=0\\\\lgx=-3\; \; ,\; \; lgx=6\; \; \; (teorema\; Vieta)\\\\x=10^{-3}=0,001\; \; ,\; \; \; x=10^6=1000000\\\\Otvet:\; \; x=0,001\; \; ,\; \; x=1000000\; .$

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mabilacom Здравствуйте. Помогите с заданием. https://znanija.com/task/33639782?utm_source=android&utm_medium=share&utm_campaign=question

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