Question

alguém me ajuda pelo amor de Deus

Answer 1

Resposta:

a)  F

log a + log b = log a*b

log 20 + log 6 = log 20*6 = log 120 ≠ log 26

b) V

log 5 + log 8 + log 2,5 = log 5*20*2,5 =  log 100 =log 10²=2 * log 10 = 2

c) V

log₂ 4¹⁸ = log₂ 2³⁶ =36 * log₂ 2 = 36 *1 =36

d) F

****√√√3=((3^(1/2))^(1/2))^(1/2) =3^(1/2*1/2*1/2)=3^(1/8)

log₃ 3^(1/8) =1/8 * log₃ 3 =1/8 =0,125 < 0,25

e) V

log₅ 35 - log₅ 7 = log₅ 5*7 - log₅ 7 = log₅ 5 + log₅ 7 - log₅ 7 =log₅ 5=1

f) V

log₃ (√2 +1) + log₃ (√2-1) =0

log₃ [(√2+1)*√2-1)] = 0

log₃ [√2²-1²)] = 0

log₃ [2 -1] = 0

log₃ 1 = 0   ==>1 =3⁰  ==> 1 = 1 ...OK