Log4^16+log1/2^ (3x+1) = log1/4^ (3x+1)

Question

Математика

Агриппична

25 апреля, 21:38

Log4^16+log1/2^ (3x+1) = log1/4^ (3x+1)

+3

Ответы (2)

Знаете ответ?

Машута · Answer 1

ОДЗ 3 х+1>0⇒3x>-1⇒x>-1/3⇒x∈ (-1/3; ∞)

log (1/4) (3x+1) - log (1/2) (3x+1) = 2

log (2) (3x+1) / (-2) - log (2) (3x+1) / (-1) = 2

-1/2log (2) (3x+1) + log (2) (3x+1) = 2

1/2log (2) (3x+1) = 2

log (2) (3x+1) = 4

3x+1=16

3x=16-1=15

x=15:3=5

Ответ х=5

Ареф · Answer 2

Log[4] (16) + log[1/2] (3x+1) = log[1/4] (3x+1)

2-log[2] (3x+1) = - log[2] (3x+1) / 2

2=log[2] (3x+1) / 2

4=log[2] (3x+1) = log[2] (16)

(3x+1) = 16

x=5