Question

calcule o valor de s = log4 (log3 9) + log2 (log81 3) + log0 8 (log16 32)

Answer 1

Queremos calcular:

S= log₄(log₃9) + log₂(log₈₁3) + log₀,₈(log₁₆32)

Lembrando que:

$log_ba = x$ ↔ $b^x = a$

Considere que:

log₃9 = x

$3^x = 9$

$3^x = 3^2$

x = 2

Assim:

log₄2 = y

$4^y = 2$

$2^{2y}=2$

2y = 1

$y = \frac{1}{2}$

Ou seja,

$log_{4}(log_39) = \frac{1}{2}$

Considere que:

log₈₁3 = a

$81^a = 3$

$3^{4a} = 3$

4a = 1

$a = \frac{1}{4}$

Assim,

$log_2 \frac{1}{4}=b$

$2^b = \frac{1}{4}$

$2^b = 2^{-2}$

b = -2

Ou seja,

log₂(log₈₁3) = -2

Por fim, considere que:

log₁₆32 = c

$16^c = 32$

$2^{4c} = 2^5$

4c = 5

$c = \frac{5}{4}$

Assim,

$log_{0,8}\frac{5}{4} = d$

$0,8^d = \frac{5}{4}$

$(\frac{4}{5})^d = (\frac{4}{5})^{-1}$

d = -1

Ou seja,

log₀,₈(log₁₆32) = -1

Portanto:

$S = \frac{1}{2}-2-1$

$S = -\frac{5}{2}$