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

1. Приведем к одному основанию:

log4 (16) + log (1/2) (3x + 1) = log (1/4) ^ (3x + 1); 2 + log (1/2) (3x + 1) = log (1/2) ^ (3x + 1) / log (1/2) (1/4); 2 + log (1/2) (3x + 1) = log (1/2) ^ (3x + 1) / 2; 4 + 2log (1/2) (3x + 1) = log (1/2) ^ (3x + 1); 2log (1/2) (3x + 1) - log (1/2) ^ (3x + 1) = - 4; log (1/2) ^ (3x + 1) = - 4.

2. По определению логарифма получим:

3x + 1 = (1/2) ^ (-4); 3x + 1 = 2^4; 3x + 1 = 16; 3x = 16 - 1; 3x = 15; x = 15 : 3; x = 5.

Ответ: 5.