Question

1) log (x+2) по основанию 3 =5 2)log (6-x) по основанию 7 =0 3) log (15-2x) по основанию 1\4 =-3 4) log (x^2-4x) по основанию 2 =log (6x-16) по основанию 2 5)log3 (21-x) = log3 (x-7) +2 6)log3 (4x-7)=3log3 6

Answer 1

$1) \:\:\: log_2 (x+2) =5 \\ \\ ( x+2\ \textgreater \ 0; \:\:\:\:\: x \ \textgreater \ -2 ) \\ \\ log_2 (x+2) =5log_3 3 = log_3 3^5 \\ \\ x+2 = 3^5 = 243 \\ \\ x = 241 \\ \\$

$2) \:\:\: log_7 (6-x) = 0 \\ \\ (6-x\ \textgreater \ 0; \:\:\:\:\: x \ \textless \ 6) \\ \\ log_7 (6-x) = log_7 1 \\ \\ 6-x = 1 \\ \\ x = 5$

$3) \:\:\: log_{ \frac{1}{4} } (15-2x) = -3 \\ \\ (15 - 2x \ \textgreater \ 0; \:\:\:\:\: x \ \textless \ 7,5 ) \\ \\ log_{ \frac{1}{4} } (15-2x) = -3 log_{ \frac{1}{4} } \frac{1}{4} = log_{ \frac{1}{4} } ( \frac{1}{4})^{-3} \\ \\ 15 - 2x = 4^3 = 64 \\ \\ 2x = -49 \\ \\ x = -24,5$

$4) \:\:\: log_2 (x^2-4x) = log_2 (6x-16) \\ \\ (x^2-4x\ \textgreater \ 0; \:\:\:\:\: x(x-4) \ \textgreater \ 0; \:\:\:\:\: x\ \textless \ 0; \:\:\:\:\: x\ \textgreater \ 4 \\ \\ 6x -16 \ \textgreater \ 0; \:\:\:\:\: x \ \textgreater \ \frac{8}{3} \\ \\=\ \textgreater \ x \ \textgreater \ 4 ) \\ \\ x^2 - 4x = 6x -16 \\ \\ x^2 - 10x + 16 = 0 \\ \\ x_{1,2} = 5 \pm \sqrt{5^2 -1*16} = 5 \pm 3 \\ \\ x_1 =2; \:\:\:\:\: x_2 = 8$

Первый корень не входит в ОДЗ, поэтому единственное решение x = 8.

$5)\:\:\: log_3 (21-x) = log_3 (x-7) +2 \\ \\ (21-x \ \textgreater \ 0; \:\:\:\:\: x \ \textless \ 21 \\ x-7 \ \textgreater \ 0; \:\:\:\:\: x \ \textgreater \ 7 \\ =\ \textgreater \ \:\:\: 7 \ \textless \ x \ \textless \ 21) \\ \\ log_3 (21-x) = log_3 (x-7) +2log_3 3 \\ \\ log_3 (21-x) = log_3 (x-7) +log_3 3^2 \\ \\ log_3 (21-x) = log_3 9(x-7) \\ \\ 21 - x = 9x -63 \\ \\ 10x = 84 \\ \\ x = 8,4$

$6) \:\:\: log_3 (4x-7)=3log_3 6 \\ \\ (4x - 7 \ \textgreater \ 0; \:\:\:\:\: x \ \textgreater \ \frac{7}{4} ) \\ \\ log_3 (4x-7)=log_3 6^3 \\ \\ 4x - 7 = 216 \\ \\ 4x = 223 \\ \\ x = 55.75$