Question

помогите пожалуйста..

Answer 1

Ответ:

$log_{3}^{2} (x) + 2 log_{3}( \sqrt{x} ) = 2 \\ log_{3}^{2} (x) + 2 log_{3}( {x}^{ \frac{1}{2} } ) = 2 \\ log_{3}^{2} (x) + 2 \times \frac{1}{2} \times log_{3}( x ) = 2 \\ log_{3}^{2} (x) + log_{3}( x ) = 2$

$Допустим \: log_{3}(x) = t$

${t}^{2} + t = 2$

${t}^{2} + t - 2 = 0 \\ t_{1} = - 2 \: \: \: \: \: \: \: t_{2} = 1$

$log_{3}(x) = - 2 \\ x = {3}^{ - 2} \\ x = \frac{1}{9}$

$log_{3}(x) = 1 \\ x = {3}^{1} \\ x = 3$

$log_{3}^{2} ( \frac{1}{9} ) + 2 log_{3}( \sqrt{ \frac{1}{9}} ) = 2 \\ 4 + 2 \times ( - 1) = 2 \\ 4 - 2 = 2 \\ 2 = 2$

$log_{3}^{2}(3) + 2 log_{3}( \sqrt{3} ) = 2 \\ 1 +2 \times \frac{1}{2} = 2 \\ 1 + 1 = 2 \\ 2 = 2$

$x_{1} = \frac{1}{9} \: \: \: \: \: \: \: \: x_{2} = 3$