Question

10)no caderno escreva na forma de um único logaritmo.
$a) log_{5}(6) + log_{5}(11)$
$b) log_{7}(28) - log_{7}(4)$
$c)4 \times log(3)$
$d) \frac{ log_{2}(3) }{ log_{8}(7) }$
$e) \frac{1}{3} \times log_{3}(7) - log_{3}(2)$
$f)1 + log_{5}(4)$
​

Answer 1

Resposta:

Explicação passo a passo:

a)

log₅(6)+log₅(11)=log₅(6.11)=log₅(66)

b)

log₇(28)-log₇(4)=log₇(28/4)=log₇7

c)

4.log(3)=log(3)⁴

d)

log₂(3)/log₈(7)

mudança de base:

log₈(7)=log₂(7)/log₂(8)=log₂(7)/x

log₂(8)=x => 2ˣ=8 => 2ˣ=2³ => x=3

log₈(7)=log₂(7)/3

log₂(3)/log₈(7)=log₂(3)/[log₂(7)/3]=3.log₂(3)/log₂(7)=3.log₇(3)=log₇(3)³

e)

log₃(7)/3-log₃(2)=[log₃(7)-3.log₃(2)]/3=[log₃(7)-log₃(2)³]/3=log₃(7/2³)/3=log₃(7/8)/3

f)

1+log₅(4)=log₅(5)+log₅(4)=log₅(5.4)=log₅(20)

Propriedades:

log(a)ⁿ=n.log(a)

log(a.b)=log(a)+log(b)

log(a/b)=log(a)-log(b)

logₓ(a)=logₙ(a)/logₙ(x), mudança de base

logₓ(x)=1

Answer 2

Resposta:

Bons estudos! ☺

Explicação passo-a-passo:

$a) \: log_{5}(6) + log_{5}(11) \\ log_{5}(6 \times 11) \\ log_{5}(66) \\ \\ b) \: log_{7}(28) - log_{7}(4) = log_{7}( \frac{28}{4} ) \\ log_{7}(7) = 1 \\ \\ c) \: 4 \times log_{10}(3) \\ 4 log_{10}(3) = log_{10}(81) \\ \\ d) \: \frac{ log_{2}(3) }{ log_{8}(7) } \\ \frac{ log_{2}(3) }{ log_{ {2}^{3} }(7) } = \frac{ log_{2}(3) }{ log_{2}( \sqrt[3]{7} ) } \\ \frac{3}{ \sqrt[3]{7} } \\ \\ e) \: \frac{1}{2} \times log_{3}(7) - log_{3}(2) \\ \frac{1}{2} \times log_{3}( \frac{7}{2} ) \\ log_{3}( \sqrt{ \frac{7}{2} } ) \\ \\ f) \: 1 + log_{5}(4) \\ log_{5}(5) + log_{5}(4) = log_{5}(5 \times 4) \\ log_{5}(20)$

Obs: As alíneas c e d ainda são questionáveis.