Question

log2(x)^2-7log2(x)+6=0

Answer 1

Resposta:

Se for  log₂(x)²-7log₂(x)+6=0

2*log₂(x)-7log₂(x)+6=0

-5*log₂(x) +6=0

log₂ x=6/5

x =2^(6/5)

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Se for  log²₂(x) -7log₂(x)+6=0

Fazendo y= log₂(x)

y²-7y+6=0

y=1= log₂(x) ==>x=2

y=6= log₂(x) ==>x=2^6 =64