Для решения уравнения \(\lg(x - 6) - 0,5 ext{lg}2 = ext{lg}3 + ext{lg}√{x} - 10\) найдем область допустимых значений (ОДЗ) и преобразуем уравнение, используя свойства логарифмов.
\[ ext{lg}(x - 6) - ext{lg}(√{x}) - 0,5 ext{lg}2 = ext{lg}3 - 10 \]
\[ ext{lg}rac{x - 6}{√{x}} - ext{lg}(2^{0,5}) = ext{lg}3 - 10 \]
\[ ext{lg}rac{x - 6}{√{x}} - ext{lg}(√{2}) = ext{lg}3 - 10 \]
\[ ext{lg}rac{x - 6}{√{x} √{2}} = ext{lg}3 - 10 \]
\[ ext{lg}rac{x - 6}{√{2x}} = ext{lg}3 - 10 \]
\[ ext{lg}rac{x - 6}{√{2x}} = ext{lg}3 - ext{lg}10^{10} \]
\[ ext{lg}rac{x - 6}{√{2x}} = ext{lg}rac{3}{10^{10}} \]
\[ rac{x - 6}{√{2x}} = rac{3}{10^{10}} \]
\[ rac{(x - 6)^2}{2x} = rac{9}{10^{20}} \]
\[ (x - 6)^2 = rac{18x}{10^{20}} \]
\[ x^2 - 12x + 36 = rac{18x}{10^{20}} \]
\[ x^2 - (12 + rac{18}{10^{20}})x + 36 = 0 \]
\[ ext{lg}(x - 6) - ext{lg}√{2} = ext{lg}3 + ext{lg}√{x} - 10 \]
\[ ext{lg}rac{x - 6}{√{2}} = ext{lg}(3√{x}) - 10 \]
\[ ext{lg}rac{x - 6}{√{2}} + 10 = ext{lg}(3√{x}) \]
\[ ext{lg}rac{x - 6}{√{2}} + ext{lg}10^{10} = ext{lg}(3√{x}) \]
\[ ext{lg}rac{(x - 6)10^{10}}{√{2}} = ext{lg}(3√{x}) \]
\[ rac{(x - 6)10^{10}}{√{2}} = 3√{x} \]
\[ (x - 6)10^{10} = 3√{2x} \]
\[ ext{lg}(x - 6) - ext{lg}3 - ext{lg}√{x} = 0,5 ext{lg}2 - 10 \]
\[ ext{lg}rac{x - 6}{3√{x}} = ext{lg}√{2} - 10 \]