помогите решить логорифмы!!! http://cs418921.userapi.com/v418921349/3290/iIk-7KzW0mU.jpg

Ответ
log2 16 = log2 2^4 = 4
log3 1/9 = log3 3^(-2) = -2
log32 V32 = log32 32^(1/2) = 1/2
=> 4+(-2) + 1/2 = 2 1/2

log3 (x^2+6) = log3 5x
ОДЗ: x^2+6>0 и 5x>0 или x^2>-6 (естественно) =>
x>0
x^2+6 = 5x
x^2 - 5x + 6 = 0
решай с учетом ОДЗ.

log(1/4) (2x-1) = -1
ОДЗ: 2x-1>0 => x>1/2
log(1/4) (2x-1) = log(1/4) (1/4)^(-1)
(2x-1) = (1/4)^(-1)
2x-1 = 4
x = 5/2

log(1/4) (4x+3) >= -1
ОДЗ: 4x+3>0 => x>-3/4
log(1/4) (4x+3) >= log(1/4) (1/4)^(-1)
(4x+3) =< (1/4)^(-1)
4x+3 =< 4
x =< 1/4
Ответ: -3/4 < x < 1/4

log5 x > log5 (3x-4)
ОДЗ: x>0 и 3x-4>0 или x>0 и x>4/3 => x>4/3
x > 3x-4
2x < 4
x < 2
Ответ: 4/3 < x < 2