Question

1) Log (x+2) по основанию 3 =5

 

2)Log (6-x) по основанию 7 =0

 

3) Log (15-2x) по основанию 14 =-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.

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

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