Question

Resolva a equacao log2x+log4x+log8x+log16x= -6,25

Answer 1

log2x+log4x+log8x+log16x=-6,25

Mudança de bases, vamos colocar todos em base 2:

log2 x + (log2 x/log2 4) + (log2 x/log2 8) + (log2 x/log2 16) = -6,25

log2 x + (log2 x / 2) + (log2 x / 3) + (log2 x / 4) = -6,25

Resolvendo por MMC = 12, temos que:

(12 log2 x + 6 log2 x + 4 log2 x + 3 log2 x) / 12 = -6,25

25 log 2 x = -6,25*12

25 log2 x = -75

log2 x = -75 / 25

log2 x = -3

Agora, propriedade dos logs:

2^(-3) = x

x = 1/8