Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 345

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\[1)\ 2^{3\lg x} \bullet 5^{\lg x} = 1600\]

\[8^{\lg x} \bullet 5^{\lg x} = 1600\]

\[(8 \bullet 5)^{\lg x} = 1600\]

\[40^{\lg x} = 40^{2}\]

\[\lg x = 2\]

\[\lg x = \lg 10^{2}\]

\[x = 10^{2} = 100\]

\[Ответ:\ \ x = 100.\]

\[2)\ 2^{\log_{3}x^{2}} \bullet 5^{\log_{3}x} = 400\]

\[2^{2\log_{3}x} \bullet 5^{\log_{3}x} = 400\]

\[4^{\log_{3}x} \bullet 5^{\log_{3}x} = 400\]

\[(4 \bullet 5)^{\log_{3}x} = 400\]

\[20^{\log_{3}x} = 20^{2}\]

\[\log_{3}x = 2\]

\[\log_{3}x = \log_{3}3^{2}\]

\[x = 3^{2} = 9\]

\[Ответ:\ \ x = 9.\]

\[3)\ \frac{1}{4 + \lg x} + \frac{2}{2 - \lg x} = 1\]

\[\frac{2 - \lg x + 2\left( 4 + \lg x \right)}{\left( 4 + \lg x \right)\left( 2 - \lg x \right)} = 1\]

\[2 - \lg x + 8 + 2\lg x =\]

\[= \left( 4 + \lg x \right)\left( 2 - \lg x \right)\]

\[10 + \lg x =\]

\[= 8 - 4\lg x + 2\lg x - \lg^{2}x\]

\[\lg^{2}x + 3\lg x + 2 = 0\]

\[Пусть\ y = \lg x:\]

\[y^{2} + 3y + 2 = 0\]

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[y_{1} = \frac{- 3 - 1}{2} = - 2;\text{\ \ }\]

\[y_{2} = \frac{- 3 + 1}{2} = - 1.\]

\[1)\ \lg x = - 2\]

\[\lg x = \lg 10^{- 2}\]

\[x = 10^{- 2}\]

\[x = 0,01.\]

\[2)\ \lg x = - 1\]

\[\lg x = \lg 10^{- 1}\]

\[x = 10^{- 1}\]

\[x = 0,1.\]

\[Ответ:\ \ x_{1} = 0,01;\ \ x_{2} = 0,1.\]

\[4)\ \frac{1}{5 - \lg x} + \frac{2}{1 + \lg x} = 1\]

\[\frac{1 + \lg x + 2\left( 5 - \lg x \right)}{\left( 5 - \lg x \right)\left( 1 + \lg x \right)} = 1\]

\[1 + \lg x + 10 - 2\lg x =\]

\[= \left( 5 - \lg x \right)\left( 1 + \lg x \right)\]

\[11 - \lg x =\]

\[= 5 + 5\lg x - \lg x - \lg^{2}x\]

\[\lg^{2}x - 5\lg x + 6 = 0\]

\[Пусть\ y = \lg x:\]

\[y^{2} - 5y + 6 = 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }y_{2} = \frac{5 + 1}{2} = 3\]

\[1)\ \lg x = 2\]

\[\lg x = \lg 10^{2}\]

\[x = 10^{2}\]

\[x = 100.\]

\[2)\ \lg x = 3\]

\[\lg x = \lg 10^{3}\]

\[x = 10^{3}\]

\[x = 1000.\]

\[Ответ:\ \ x_{1} = 100;\ \ x_{2} = 1000.\]