Question

Alguem pode me ajudar pfvr urgente!

Answer 1

Resposta:

a) log₅ 14 == log₅ (2*7) =  log₅ 2 +  log₅ 7 =0,43+1,21 = 1,64

b)  log₅ 3,5 =  log₅ 5*7/10 =  log₅ 5 +  log₅ 7 -  log₅ 10

=  log₅ 5 +  log₅ 7 -  log₅ 2 - log₅ 5 = 1,21-0,43 = 0,78

c)

log₂ 7 = log 7/ log 2  = (log 7/log 5)/(log 2/log 5)=log₅ 7/ log₅ 2

=1,21/0,43 =2,814

d)

log₁₄ 7^(1/2) =(1/2)*log₁₄ 7=(1/2)* log 7/ log 14

=(1/2)* (log 7/log 5)/(log 14/log 5)

=(1/2)* log₅ 7 / (log₅ 14)

=(1/2)*log₅ 7/(log₅ 2 + log₅ 7)

=(1/2)*1,21/(0,43+1,21) = 0,3689

e)

log₇ 2 = log 2/log 7 =(log 2/ log 5) /(log 7/log 5)

= log₅ 2/ log₅ 7 =0,43/1,21 = 0,355372

Propriedades utilizadas :

log[a]  b  = log b/log a

log b^x =x*log b

log x*b=log x +log b

log x/b =log x - log b