Кто все сделает скину 100 р(правда)

1.1)log3 (1/27) = log3 (3^(-3)) = (-3) * log3 3 = (-3) * 1 = - 3

1.2) (1/3)^(2*log(1/3) 7) = (1/3)^(log(1/3) 7^2 = 7^2 = 49

1.3) log2 56 + 2*log2 12 - log2 63 = log2 (56 * 12^2 / 63) = 56*12^2*63 = считай

2) log(0,8) 1 1/2 - log(0,8) 1 1/3 = log(0,8) (3/2 : 4/3) = log(0,8) (3/2 * 3/4) =
= log(0,8) (9/8)
0,8 < 1 --->
если 0,8^n при n>0 ----> 0,8^n < 1
если 0,8^n при n< 0 ----> 0,8^n >1
и так как 0,8^n = 9/8 > 1 =>
log(0,8) 1 1/2 > log(0,8) 1 1/3

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

4) log(0,9) (x-2) > 1 ------> ОДЗ: (x-2)>0 ----> x > 2
log(0,9) (x-2) > log(0,9) 0,9
(x-2) < 0,9
x < 2,9
С учётом ОДЗ: 2 < x < 2,9

5) log(V3) x + log9 x = 10 ----> ОДЗ: x > 0
___ log(V3) x = log(3^(1/2)) x = 1/(1/2) * log3 x = 2 * log3 x
___ log9 x = log(3^2) x = 1/2 * log3 x = log3 x / 2
=>
2*log3 x + (log3 x)/2 = 10
log3 x * (2 + 1/2) = 10
log3 x * 5/2 = 10
log3 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
(9-x)>0 ------------> x < 9
Общее ОДЗ: 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^2 - 12x + 35 = 0 -----> x1= 5; x2 = 7
(x - 5)(x - 7) =< 0
5 =< x =< 7

(log2 x)^2 - 3*log2 x =< 4 ----> ОДЗ: x > 0
__ log2 x = t
t^2 - 3*t - 4 =< 0
__ t^2 - 3t - 4 = 0 ----> t1 = -1; t2 = 4
(t - t1)(t - t2) =< 0
(t + 1)(t - 4) =< 0
1 =< t =< 4 =>
1 =< log2 x =< 4
log2 2^1 =< log2 x =< log2 (2^4)
2 =< x =< 16

На предложенные тобой деньги купи маме цветы.

2 <
3. x=30,5

а) log3 1/27 = log3 1/(3^3) = log3 3^(-3) = - 3 * log3 3 = - 3;

б) (1/3)^2log1/3 7 = (1/3)^(log1/3 7^2) = (1/3)^(log1/3 49) = 49;

в) log2 56 + 2log2 12 - log2 63 = log2 56 + log2 12^2 - log2 63 = log2 (56 * 144/63) = log2 128 = log2 2^7 = 7 * log2 2 = 7.