Question

Решите пожалуйста очень надо срочно

Answer 1

$9. \ 27^{x} = 3 \\ \\ 3^{3x} = 3^{1} \\ \\ 3x = 1 \\ \\ \boxed{x = \dfrac{1}{3} }$ 


$10. \ 5^{x} = 0 \\ \\ x \in \varnothing$


$11. \ \ 4^{x + 1} + 4^{x + 2} = 20 \\ \\ 4^{x}(4 + 16) = 20 \\ \\ 4^{x} \cdot 20 = 20 \\ \\ 4^{x} = 1 \\ \\ 4^{x} = 4^{0} \\ \\ x = 0$


$12. \ 27 \cdot 4^x - 8 \cdot 9^{x} = 0 \\ \\ 3^{3} \cdot 2^{2x} = 2^{3} \cdot 3^{2x} \\ \\ \dfrac{3^3}{2^3} = \dfrac{3^{2x}}{2^{2x}} \\ \\ \bigg( \dfrac{3}{2} \bigg)^3 = \bigg ( \dfrac{3}{2} \bigg )^{2x} \\ \\ 2x = 3 \\ \\ \boxed{ x = 1,5}$


$13. \ log_{ \dfrac{1}{4}} (2x^2 - 7x - 6) =-2 \\ \\ \\ \bigg ( \dfrac{1}{4} \bigg )^{-2} = 2x^2 - 7x - 6 \\ \\ ODZ: \\ 2x^2 - 7x - 6 \ \textgreater \ 0 \\ \\ 2x^2 - 7x - 6 = 0 \\ \\ D = 49 + 4 \cdot 2 \cdot 6 = 49 + 48 = 97 \\ \\ x_1 = \dfrac{7 + \sqrt{97} }{4} \\ \\ x_2 = \dfrac{7 - \sqrt{97} }{4} \\ \\ (x - \dfrac{7 + \sqrt{97} }{4} )( x - \dfrac{7 - \sqrt{97} }{4}) \ \textgreater \ 0 \\ \\ x \in ( -\infty; \dfrac{7 - \sqrt{97} }{4} ) \ U \ (\dfrac{7 + \sqrt{97} }{4} ; \ + \infty)$

$2x^2 - 7x - 6 = 16 \\ \\ 2x^2 - 7x - 22 = 0 \\ \\ D = 49 + 4 \cdot 2 \cdot 22 = 225 = 15^2 \\ \\ x_1 = \dfrac{7 + 15}{4} = 6,5 - \ \ ne \ \ ud. \\ \\ x_2 = \dfrac{7 - 15}{4} = -2 \\ \\ \boxed{OTBET: \ x = -2. }$



$14. \ log_3(2x - 11) = 2 \\ \\ ODZ: \\ 2x \ \textgreater \ 11 \\ x \ \textgreater \ 5,5 \\ \\ 3^2 = 2x - 11 \\ \\ 2x - 11 = 9 \\ \\ 2x = 20 \\ \\ \boxed{x = 10}$



$15. log_2(7x - 5) = -2 \\ \\ ODZ: \\ 7x -5 \ \textgreater \ 0 \\ 7x \ \textgreater \ 5 \\ x \ \textgreater \ \dfrac{5}{7} \\ \\ 7x - 5 = 2^{-2} \\ \\ 7x - 5 = \dfrac{1}{4} \\ \\ 7x = 5+ \dfrac{1}{4} \\ \\ 7x = \dfrac{21}{4} \\ \\ \boxed{ x = \dfrac{3}{4}} \ \ \ \ \ \ ( \dfrac{3}{4} \ \textgreater \ \dfrac{5}{7})$