Домашние задания: Алгебра

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1) log3 1/27 = log3 (3^(-3)) = (-3) * log3 3 = (-3) * 1 = -3
(1/3)^(2 * log(1/3) 7) = (1/3)^(log(1/3) 7^2) = 7^2 = 49
log2 56 + 2*log2 12 - 2*log2 63 =
= log2 56 + log2 (12^2) - log2 (63^2) =
= log2 (56 * 12^2 / 63^2) =
= log2 (14 * 4 * 4^2 * 3^2 / 7 * 3^2) =
= log2 (4^4) = log2 (2^8) = 8

2) log(0,9) 1 1/2 = log(0,9) 3/2 = log(0,9) 9/6
log(0,9) 1 1/3 = log(0,9) 4/3 = log(0,9) 8/6
9/6 > 8/6 =>
log(0,9) 1 1/2 < log(0,9) 1 1/3

3) log4 (2x+3) = 3 ----> ОДЗ: (2x+3)>0 ----> x > - 3/2
(2x+3) = 4^3
2x = 61
x = 30,5

4) log(1/2) (x-3) > 2 ----> ОДЗ: (x-3)>0 ----> x > 3
log(1/2) (x-3) > log(1/2) ((1/2)^2)
(x-3) < (1/2)^2
x < 3 - 1/4
x < 2 3/4

5) log(V3) x + log(9) x = 10 ----> ОДЗ: x > 0
log(3^(1/2)) x + log(3^2) x = 10
1/(1/2) * log(3) x + 1/2 * log3 x = 10
log(3) x * (2 + 1/2) = 10
log(3) x = 10*2/5 = 4
x = 3^4 = 81

6) log(1/2) (x-3) + log(1/2) (9-x) >= - 3
ОДЗ: (x-3)>0 ---> x>3 и (9-x)>0 ----> x < 9 --------> 3 < x < 9
log(1/2) [(x-3)(9-x)] >= log(1/2) [(1/2)^(-3)]
(x-3)(9-x) =< (1/2)^(-3)
9x - 27 - x^2 + 3x =< 8
x^2 - 12x + 35 >= 0
(x - 5)(x - 7) >= 0
x =< 5 или x >= 7
С учетом ОДЗ:
x < 3; 7 =< x =< 9

(log2 x)^2 - 3*log2 x =< 4 ----> ОДЗ: x > 0
log2 x = t
t^2 - 3t - 4 =< 0
(t + 1)(t - 4) =< 0
1 =< t =< 4
1 =< log2 x =< 4
log2 (2^1) =< log2 x =< log2 (2^4)
2 =< x =< 16
Вера Яковлева
Вера Яковлева
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