Math, asked by juliapereiraalves597, 1 month ago

Considerando log 2 = 0,3; log 5 = 0,7 e log 7 = 0,8 , qual o valor do log 700. *
5 pontos
0,035
2,8
2
2,4
Me ajudem, por favor ​

Answers

Answered by narayanisahu82
0

Answer:

Vamos utilizar as propriedades de logaritmo e a fatoração para reescrever os logaritmos de modo que possamos utilizar as informações dadas no enunciado para calcular o valor dos logaritmos.

a)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_3}2~=~\dfrac{\log2}{\log3}\\\\\\\log_{_3}2~=~\dfrac{0,30}{0,48}\\\\\\\log_{_3}2~=~\dfrac{30}{48}\\\\\\\boxed{\log_{_3}2~=~\dfrac{5}{8}~~ou~~0,625}\end{gathered}

Utilizando a propriedade da

troca de base

log

3

2 =

log3

log2

log

3

2 =

0,48

0,30

log

3

2 =

48

30

log

3

2 =

8

5

ou 0,625

b)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_2}5~=~\dfrac{\log5}{\log2}\\\\\\\log_{_2}5~=~\dfrac{0,70}{0,30}\\\\\\\log_{_2}5~=~\dfrac{70}{30}\\\\\\\boxed{\log_{_2}5~=~\dfrac{7}{3}~~ou~~2,333...}\end{gathered}

Utilizando a propriedade da

troca de base

log

2

5 =

log2

log5

log

2

5 =

0,30

0,70

log

2

5 =

30

70

log

2

5 =

3

7

ou 2,333...

c)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_5}3~=~\dfrac{\log3}{\log5}\\\\\\\log_{_5}3~=~\dfrac{0,48}{0,70}\\\\\\\log_{_5}3~=~\dfrac{48}{70}\\\\\\\boxed{\log_{_5}3~=~\dfrac{24}{35}}\end{gathered}

Utilizando a propriedade da

troca de base

log

5

3 =

log5

log3

log

5

3 =

0,70

0,48

log

5

3 =

70

48

log

5

3 =

35

24

d)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_3}100~=~\dfrac{\log100}{\log3}\\\\\\Fatorando~o~~ logaritmando~~''100''\\\\\\\log_{_3}100~=~\dfrac{\log\,(2\cdot2\cdot5\cdot5)}{\log3}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_3}100~=~\dfrac{\log2~+~\log2~+~\log5~+~\log5}{\log3}\\\\\\\log_{_3}100~=~\dfrac{0,30~+~0,30~+~0,70~+~0,70}{0,48}\\\\\\\log_{_3}100~=~\dfrac{2,00}{0,48}\end{gathered}

Utilizando a propriedade da

troca de base

log

3

100 =

log3

log100

Fatorando o logaritmando

′′

100

′′

log

3

100 =

log3

log(2⋅2⋅5⋅5)

Aplicando a propriedade do

logaritmo do produto

log

3

100 =

log3

log2 + log2 + log5 + log5

log

3

100 =

0,48

0,30 + 0,30 + 0,70 + 0,70

log

3

100 =

0,48

2,00

\begin{gathered}\log_{_3}100~=~\dfrac{200}{48}\\\\\\\boxed{\log_{_3}100~=~\dfrac{25}{6}~~ou~~4,166...}\end{gathered}

log

3

100 =

48

200

log

3

100 =

6

25

ou 4,166...

e)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_4}18~=~\dfrac{\log18}{\log4}\\\\\\Fatorando~os~~ logaritmandos~~''18''~e~''4''\\\\\\\log_{_4}18~=~\dfrac{\log\,(2\cdot3\cdot3)}{\log\,(2\cdot2)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_4}18~=~\dfrac{\log2~+~\log3~+~\log3}{\log2~+~\log2}\\\\\\\log_{_4}18~=~\dfrac{0,30~+~0,48~+~0,48}{0,30~+~0,30}\\\\\\\log_{_4}18~=~\dfrac{1,26}{0,60}\end{gathered}

Utilizando a propriedade da

troca de base

log

4

18 =

log4

log18

Fatorando os logaritmandos

′′

18

′′

e

′′

4

′′

log

4

18 =

log(2⋅2)

log(2⋅3⋅3)

Aplicando a propriedade do

logaritmo do produto

log

4

18 =

log2 + log2

log2 + log3 + log3

log

4

18 =

0,30 + 0,30

0,30 + 0,48 + 0,48

log

4

18 =

0,60

1,26

\begin{gathered}\log_{_4}18~=~\dfrac{126}{60}\\\\\\\boxed{\log_{_4}18~=~\dfrac{21}{10}~~ou~~2,1}\end{gathered}

log

4

18 =

60

126

log

4

18 =

10

21

ou 2,1

f)

\begin{gathered}Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log0,5}{\log36}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log\frac{1}{2}}{\log36}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log2^{-1}}{\log36}\\\\\\Fatorando~o~~ logaritmando~~''36''\\\\\\\log_{_{36}}0,5~=~\dfrac{\log\,2^{-1}}{\log\,(2\cdot2\cdot3\cdot3)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~da~potencia}\end{gathered}

Utilizando a propriedade da

troca de base

log

36

0,5 =

log36

log0,5

log

36

0,5 =

log36

log

2

1

log

36

0,5 =

log36

log2

−1

Fatorando o logaritmando

′′

36

′′

log

36

0,5 =

log(2⋅2⋅3⋅3)

log2

−1

Aplicando a propriedade do

logaritmo da potencia

\begin{gathered}\log_{_{36}}0,5~=~\dfrac{-1\cdot\log\,2}{\log\,(2\cdot2\cdot3\cdot3)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_{36}}0,5~=~\dfrac{-1\cdot\log2}{\log2~+~\log2~+~\log3~+~\log3}\\\\\\\log_{_{36}}0,5~=~\dfrac{-1\cdot0,30}{0,30~+~0,30~+~0,48~+~0,48}\\\\\\\log_{_{36}}0,5~=~\dfrac{-0,30}{1,56}\\\\\\\log_{_{36}}0,5~=\,-\dfrac{30}{156}\\\\\\\boxed{\log_{_{36}}0,5~=\,-\dfrac{5}{26}}\end{gathered}

log

36

0,5 =

log(2⋅2⋅3⋅3)

−1⋅log2

Aplicando a propriedade do

logaritmo do produto

log

36

0,5 =

log2 + log2 + log3 + log3

−1⋅log2

log

36

0,5 =

0,30 + 0,30 + 0,48 + 0,48

−1⋅0,30

log

36

0,5 =

1,56

−0,30

log

36

0,5 =−

156

30

log

36

0,5 =−

26

5

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