Question

1. Calcule o valor dos logaritmos a seguir.
a) log3 81
b) log 1/5 625
c) log 8 √32
d) log 3√2 5√16
e) log 0.01
f) log 2/3 (32/243)
g) log 5 (1/7√125)
h) log 4/25 4√125/8)
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Answer 1

Resposta:

a) 4

b) -4

c) 5/6

d) log(400) / log(18)

e) -2

f) 5

g) 3/2 - log(7)/log(5)

h) (3/2 log(5) - log(2)) / (2log(2) - 2log(5))

Explicação passo-a-passo:

a) log3 81  = x

3^x = 81

3^x = 3^4

x = 4

b) log 1/5 625  = x

(1/5)^x = 625

5^-x = 625

5^-x = 5^4

x = -4

c) log 8 √32  = x

8^x = 32^(1/2)

(2^3)^x = (2^5)^(1/2)

2^(3x) = 2^(5/2)

3x = 5/2

x = 5/6

d) log 3√2 5√16  = x

(3*2^(1/2))^x = 5*4

18^(1/2)^x = 20

18^(x/2) = 20

18^x = 400

log (18^x) = log (400)

x * log(18) = log(400)

x = log(400) / log(18)

e) log 0.01  = x

10^x = 0,01

10^x = 10^-2

x = -2

f) log 2/3 (32/243)  = x

(2/3)^x = 32/243

(2/3)^x = 2^5/3^5

(2/3)^x = (2/3)^5

x = 5

g) log 5 (1/7√125)  = x

5^x = 5/7*5^(1/2)

log (5^x) = log (5/7*5^(1/2))

x * log(5) = log(5) - log(7) + 1/2 * log(5)

x = (3/2 log (5) - log(7)) / log(5)

x = 3/2 - log(7)/log(5)

h) log 4/25 4√125/8)  = x

(4/25)^x = (125^(1/2))/2

(2/5)^(2x) = 5/2 * 5^(1/2)

log ((2/5)^(2x)) = log (5/2 * 5^(1/2))

2x * (log(2) - log(5)) = log(5) - log(2) + 1/2 * log(5)

x = (3/2 log(5) - log(2)) / (2log(2) - 2log(5))

​

Answer 2

Calculando o valor dos logaritmos, obtemos:

a) log₃ 81 = 4

b) log(1/5) 625 = -4

c) log₈ √32 = 5/6

d) log(3√2) 5√16 = log 400/log 18

e) log₁₀ 0,01 = -2

f) log(2/3) 32/243 = 5

g) log₅ 1/7·√125 = (-3/2) - log 7/log 5

h) log 4/25 4√125/8) = log 2/log (5/2) - 3/4

Logaritmos

Pela definição de logaritmo, sabemos que a base do logaritmo elevado ao resultado do mesmo é igual ao logaritmando, ou seja:

logₐ x = b

aᵇ = x

Para responder essa questão, devemos aplicar a definição acima e calcular os logaritmos.

a)

81 = 3ᵇ

3⁴ = 3ᵇ

b = 4

log₃ 81 = 4

b)

(1/5)ᵇ = 625

5⁻ᵇ = 5⁴

b = -4

log(1/5) 625 = -4

c)

log₈ √32

8ᵇ = √32

(2³)ᵇ = √2⁵

Temos que √2⁵ = 2^(5/2), então:

3b = 5/2

b = 5/6

log₈ √32 = 5/6

d)

(3√2)ᵇ = 5√16

Elevando ao quadrado:

18ᵇ = 400

Aplicando o logaritmo:

log 18ᵇ = log 400

b·log 18 = log 400

b = log 400/log 18

log(3√2) 5√16 = log 400/log 18

e)

10ᵇ = 0,01

10ᵇ = 10⁻²

b = -2

log₁₀ 0,01 = -2

f)

(2/3)ᵇ = (32/243)

(2/3)ᵇ = (2/3)⁵

b = 5

log(2/3) 32/243 = 5

g)

5ᵇ = 1/7·√125

5ᵇ = 1/7·5^(3/2)

log 5ᵇ = log 5^(-3/2)/7

b·log 5 = (-3/2)·log 5 - log 7

b = (-3/2) - log 7/log 5

log₅ 1/7·√125 = (-3/2) - log 7/log 5

h)

(4/25)ᵇ = 4√125/8

[(2/5)²]ᵇ = 2²·(5/2)^(3/2)

log (2/5)²ᵇ = log 2² + log (5/2)^(3/2)

log (5/2)-²ᵇ = log 2² + log (5/2)^(3/2)

-2b·log (5/2) = 2 log 2 + (3/2)·log (5/2)

b = [2 log 2 + (3/2)·log (5/2)]/-2·log(5/2)

b = log 2/log (5/2) - (3/4)

log 4/25 4√125/8) = log 2/log (5/2) - 3/4

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