Question

admitindo-se que log5 2=0,43 ; log5 3=0,68 e log5 7=0,76, determine:


a) log5 (4 raiz quadrada de 3 sobre 49)

b)log5 raiz (2 raiz de 5 sobre 21)

Answer 1

$log _{5}4. \frac{ \sqrt{3}}{49} = log _{5} 4 + (log _{5} \sqrt{3} - log _{5}49) \\ 2log_{5}2 + ( \frac{1}{3}log _{5}3 - 2log_{5}7) \\ 2.0,43 + ( \frac{1}{3} .0,68 - 2.0,76) \\ 0,86+(0,226-1,52) \\ 0,86 - 1,294 = -0,434$

Espero ter ajudado.

Answer 2

a)
log₅ 4.√(3/49) = log₅ 4 + log₅ √3 / 7 = log₅ 2.2 + log₅ 3¹/₂ - log₅ 7 = log₅ 2 + log₅ 2 + 1/2 . log₅ 3 - log₅ 7 = 0,43 + 0,43 + 1/2 . 0,68 - 0,76 = 0,44  

b) log₅ √(2.√(5/21)) = log₅ (2.√(5/21))¹/₂ = 1/2.log₅ 2.√(5/21) = 1/2.(log₅ 2 + log₅ √(5/21)) = 1/2.(log₅ 2 + log₅ (5/21)¹/₂) = 1/2.(log₅ 2 + 1/2.log₅ 5/21) = 1/2.(log₅ 2 + 1/2.(log₅ 5 - log₅ 21)) = 1/2.(log₅ 2 + 1/2.log₅ 5 - 1/2.log₅ 21) = 1/2 log₅ 2 + 1/4.log₅ 5 - 1/4.log₅ 3.7 = 1/2 log₅ 2 + 1/4.log₅ 5 - 1/4.(log₅ 3 - log₅ 7) =

1/2 log₅ 2 + 1/4 - 1/4.log₅ 3 - 1/4.log ₅ 7 = 0,43/2 + 0,25 - 0,68/4 - 0,76/4 = 0,215 + 0,25 - 0,17 - 0,19 = 0,105