Логарифмы 10 класс

1) lg (3-x) + 2lg V(-x) = 1
ОДЗ:
(-x)>0 ------> x<0
(3-x)>0 -----> x<3
Общее ОДЗ: x < 0
lg (3-x) + 2*(1/2)*lg (-x) = 1
lg (3-x) + lg (-x) = lg 10
lg (3-x)*(-x) = lg 10
(3-x)*(-x) = 10
x^2 - 3x - 10 = 0
x(1,2) = [3 + - V(3^2 - 4*(-10)]/2= (3 + - 7)/2
x1 = (3+7)/2 = 5 - не удовл.ОДЗ
x2 = (3-7)/2 = -2 - ответ

2) log4 ((log3 x)^2 - 2*log3 x - 1) = 1/2
ОДЗ:
((log3 x)^2 - 2*log3 x - 1) > 0
x > 0
=>
log4 (....) = 1/2
(...) = 4^(1/2) = 2
или
((log3 x)^2 - 2*log3 x - 1) = 2 ----> log3 x = t
t^2 - 2t - 3 = 0 -----> t1= -1; t2 =3
t1 = - 1 ----> log3 x = - 1 -----> x1 = 3^(-1) = 1/3
t2 = 3 ------> log3 x = 3 -------> x2 = 3^3 = 27

3) log4 (9x^2) + {log(1/2) (3x)}^2 = 2
ОДЗ: x > 0
__ log4 (9x^2) = log(2^2) ((3x)^2) = 2 * 1/2 * log2 (3x) = log2 (3x)
__ log(1/2) (3x) = log(2^(-1)) (3x) = 1/(-1) * log2 (3x) = - log2 (3x)
=>
log2 (3x) + {- log2 (3x)}^2 = 2 -----> log2 (3x) = t
t^2 + t - 2 = 0 -----> t1 = -2; t2 = 1
t1 = - 2 -----> log2 (3x) = - 2 ----> 3x = 2^(-2) = 1/4 ---> x1 = 1/12
t2 = 1 -------> log2 (3x) = 1 ------> 3x = 2^1 --------------> x2 = 2/3

4) log(3x) x^3 = log(9x) x^5
3 * log(3x) x = 5 * log(9x) x
3 / log(x) (3x) = 5 / log(x) (9x)
3 * log(x) (9x) = 5 * log(x) (3x)
3 * [log(x) 9 + log(x) x] = 5 * [log(x) 3 + log(x) x]
3 * [log(x) 9 + 1] = 5 * [log(x) 3 + 1]
3*log(x) 9 + 3 = 5*(log(x) 3 + 5
2*log(x) 3 = 2
log(x) 3 = 1
x^1 = 3^1 ------> x = 3

5) x^(log2 x) = 8x^2
log2 x = a ------> x = 2^(a)
=>
(2^(a))^(a) = 8*(2^(a))^2
2^(a^2) = 2^3 * 2^(2a)
2^(a^2) = 2^(3+2a)
a^2 = 3+2a
a^2 - 2x - 3 = 0 -----> a1 = -1; a2 = 3
=>
a1 = -1 -----> x1 = 2^(-1) = 1/2
a2 = 3 ------> x2 = 2^(3) = 8

6) log(x) (x-6)^2 + log(6-x) x^4 = 6
log(x) (6-x)^2 + log(6-x) x^4 = 6
2 * log(x) (6-x) + 4 * log(6-x) x = 6
log(x) (6-x) + 2 * log(6-x) x = 3
log(x) (6-x) + 2 / log(x) (6-x) = 3 -----> log(x) (6-x) = t
t + 2/t = 3
t^2 - 3t + 2 = 0 -----> t1 = 1; t2 = 2
t1 = 1 ----> log(x) (6-x) = 1 ----> x^1 = 6 - x ---> 2x = 6 ---> x1 = 3
t2 = 2 ----> log(x) (6-x) = 2 ----> x^2=6-x----> x^2+x-6=0 --> x2=-2; x3=3

4) А что тут мудрить. ОДЗ х > 0, ≠ 1/3, ≠1/9 (плохо видно). Перейдем к основанию 3 и положим у = log(3)x => 3y/(y+1) = 5y/(y + 2) => или у = 0, или 3у + 6 = 5у + 5 =>
x = 1, или х = √3.
5) у = log(2)x => x = 2^y => 2^(y^2) = 2^3*2^y => y^2 -2y -3 = 0 => y=3 или y=-1 =>
x = 8 или х = 1/2.