Лагорифмы. помогите пожалуйста

1.1)-4
2)15
3) log3(243)=5
2.>

И правильно писать не забудь слово "логарифм"

4)
(log3 x)^2 - 2*log3 x =< 3 ----------> ОДЗ: x > 0
log3 x = t
t^2 - 2t - 3 =< 0
(t +1)(t -3) =< 0 ----------> -1 =< t =< 3 =>
-1 =< log3 x =< 3
log3 (3^(-1)) =< log3 x =< log3 (3^3)
1\3 =< x =< 27

log(1\3) (x-5) > 1 --------> ОДЗ: (x-5)>0 ----> x > 5
log(1\3) (x-5) > log(1\3) (1\3)
(x-5) < 1\3
3x - 15 < 1
x < 16\3 => x < 5 1\3
С учётом ОДЗ: 5 < x < 5 1\3

3) log5 (2x-1) = 2 -----> ОДЗ: (2x-1) > 0 -------> x > 1\2
(2x-1) = 5^2
2x = 26 ----------> x = 13

log2 (x-2) + log2 x = 3 -----> ОДЗ: (x-2)>0 и x>0 ----> x>2 и x>0 -------> x > 2
log2 (x-2)*x = 3
x*(x-2) = 2^3
x^2 - 2x - 8 = 0 -----> x1 = -4 (не удовл. ОДЗ); x2 = 6 - ответ

log8 x + log(V2) x = 14 --------> ОДЗ: x > 0
log(2^3) x + log(2^(1\2)) x = 14
(1\3) * log2 x + (1\(1\2)) * log2 x = 14
1\3 * log2 x + 2*log2 x = 14
log2 x + 6*log2 x = 14*3
7 * log2 x = 14*3
log2 x = 6 ----------------> x = 2^6 = (2^3)^2 = 8^2 = 64

2) log(1\2) (3\4) <------------> log(1\2) (2\5)
3\4 - 5\2 = 3\4 - 10\4 = - 13\4 < 0 => log(1\2) 2\5 > log(1\2) 3\4
так как
log(1\2) (3\4) = log(2^(-1)) (3\4) = - log2 (3\4)
log(1\2) (5\4) = log(2^(-1) (2\5) = - log2 (2\5)
(- log2 (3\4)) - (- log2 (2\5)) =
= - log2 (3\4) + log2 (2\5) =
= log2 (2\5) - log2 (3\4) =
= log2 (2\5 : 3\4) = log2 (2\5) * 4\3) = log2 16\15, где 16\15 > 1

1) log(1\2) 16 = 1\2 * log2 16 = 1\2 * 4 = 2
5^(1 + log5 3) = 5 * 5^(log5 3) = 5 * 3 = 15
log3 135 - log3 20 + 2*log3 6 =
= log3 (135\20) + log3 6^2 =
= log3 (135*36\20) = log3 243 = log3 (3^5) = 5 * log3 3 = 5