Модуль x-13 модуль*log в основании 2 аргумент (x-3) = 3*(13-x)

Решите задачу:

$|x - 13|log_2(x - 3) = 3(13 - x) \\ \\ 1) \ x \ \textgreater \ 13 \\ \\ (x - 13)log_2(x - 3) = 3(13 - x) \\ \\ (x - 13)log_2(x - 3) + 3(x - 13) = 0 \\ \\ (x - 13)(log_2(x - 3) + 3) = 0 \\ \\ \boxed{x = 13} \ \ \ \ \ \ \ and \ \ \ log_2(x - 3) = -3 \\ \\ 2^{-3} = x - 3 \\ \\ x - 3 = \dfrac{1}{8} \\ \\ x = \dfrac{25}{8} - \ \ ne \ \ ud. \ \ uslobiju \ \ x \ \textgreater \ 13$

$2) \ x \ \textless \ 13 \\ \\ -(x - 13)log_2(x - 3) = 3(13 - x) \\ \\ (x - 13)log_2(x - 3) = 3(x - 13) \\ \\ (x - 13)log_2(x - 3) - 3(x - 13) = 0 \\ \\ (x - 13)(log_2(x - 3) - 3) = 0 \\ \\ x = 13 \ \ \ \ \ \ \ \ \ and \ \ \ \ \ log_2(x - 3) = 3 \\ \\ 2^3 = x - 3 \\ \\ 8 = x - 3 \\ \\ x = 11 \\ \\ OTBET: \ x = 11; \ \ 13.$