Страница 99 номер 313, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\log_{2}^{2} x - 9\log_{8} x = 4\)
\(\log_{2}^{2} x - \dfrac{9}{3}\log_{2} x = 4,\)
\(\log_{2} x = z, z^{2} - 3z - 4 = 0,\)
\(z_{1{,}2} = \dfrac{3 \pm \sqrt{9 + 16}}{2},\)
\(z_{1} = 4, z_{2} = -1;\)
\(\log_{2} x = 4, x_{1} = 2^{4} = 16;\)
\(\log_{2} x = -1, x_{2} = 2^{-1} = \dfrac{1}{2};\)

2) \(16\log_{16}^{2} x + 3\log_{4} x - 1 = 0\)
\(\dfrac{16}{4}\log_{4}^{2} x + 3\log_{4} x - 1 = 0,\)
\(\log_{4} x = z, 4z^{2} + 3z - 1 = 0,\)
\(z_{1{,}2} = \dfrac{-3 \pm \sqrt{9 + 16}}{8},\)
\(z_{1} = \dfrac{1}{4}, z_{2} = -1;\)
\(\log_{4} x = \dfrac{1}{4}, x_{1} = 4^{\frac{1}{4}} = \sqrt{2};\)
\(\log_{4} x = -1, x_{2} = 4^{-1} = \dfrac{1}{4};\)

3) \(\log_{3}^{2} x + 5\log_{9} x - 1{,}5 = 0\)
\(\log_{3}^{2} x + \dfrac{5}{2}\log_{3} x - 1{,}5 = 0,\)
\(\log_{3} x = z, z^{2} + 2{,}5z - 1{,}5 = 0,\)
\(z_{1{,}2} = \dfrac{-2{,}5 \pm \sqrt{6{,}25 + 6}}{2},\)
\(z_{1} = \dfrac{1}{2}, z_{2} = -3;\)
\(\log_{3} x = \dfrac{1}{2}, x_{1} = 3^{\frac{1}{2}} = \sqrt{3};\)
\(\log_{3} x = -3, x_{2} = 3^{-3} = \dfrac{1}{27};\)

4) \(\log_{3}^{2} x - 15\log_{27} x + 6 = 0\)
\(\log_{3}^{2} x - \dfrac{15}{3}\log_{3} x + 6 = 0,\)
\(\log_{3} x = z, z^{2} - 5z + 6 = 0,\)
\(z_{1{,}2} = \dfrac{5 \pm \sqrt{25 - 24}}{2},\)
\(z_{1} = 3, z_{2} = 2;\)
\(\log_{3} x = 3, x_{1} = 3^{3} = 27;\)
\(\log_{3} x = 2, x_{2} = 3^{2} = 9.\)