Question

URGENTE ! ALGUÉM PODE AJUDAR ?

9bbe818a4d104b79c5197457ab77ed93.docx

Answer 1

Usando as propriedades dos logaritmos, exponenciais e radicais, podemos reescrever esta equação:

Lembre-se:

log a.b = log a + log b
log a/b = log a - log b
log a² = 2 log a

$\sqrt{a} \ = \ a^{\frac{1}{2}$

0,001 = $\frac{1}{1000}$



$S = log_{4} \ (\frac{2}{\frac{1}{2}}) \ + \ log_{4} \ (2)^{ \frac{1}{2}} \ + \ log_{100} \ ( \frac{1}{1000})^{\frac{1}{2}} \ + \ log_{ \frac{1}{9}} \ 3 \ + \ log_{\frac{1}{9}} \ (3)^{\frac{1}{2}}$


$S = log_{4} 2 - log_{4} \frac{1}{2} + \frac{1}{2} . log_{4} 2 + \frac{1}{2} . log_{100} 1 - log_{100} 1000 + log_{ \frac{1}{9}} 3 + \frac{1}{2} . log_{\frac{1}{9}} 3$


$S = log_{4} 2 - log_{4} 1 - log_{4} 2 + \frac{1}{2} . log_{4} 2 + \frac{1}{2} . (log_{100} 1 - log_{100} 1000) + log_{ \frac{1}{9}} 3 + \frac{1}{2} . log_{\frac{1}{9}} 3$

$S = \frac{1}{2} \ - \ (0 \ - \ \frac{1}{2}) \ + \ \frac{1}{2} \ . \ \frac{1}{2} \ + \ \frac{1}{2} \ . \ (0 \ - \ \frac{3}{2}) \ + \ (-\frac{1}{2}) \ + \ \frac{1}{2} \ . \ (-\frac{1}{2})$


$S = \frac{1}{2} \ + \ \frac{1}{2} \ + \ \frac{1}{2} \ . \ \frac{1}{2} \ + \ \frac{1}{2} \ . \ (- \ \frac{3}{2}) \ + \ (-\frac{1}{2}) \ + \ \frac{1}{2} \ . \ (-\frac{1}{2})$


$S = \frac{1}{2} \ + \ \frac{1}{2} \ + \ \frac{1}{4} \ - \ \frac{3}{4} \ + \ (-\frac{1}{2}) \ - \ \frac{1}{4}$

$S = \frac{1}{2} \ + \ \frac{1}{2} \ + \ \frac{1}{4} \ - \ \frac{3}{4} \ - \ \frac{1}{2} \ - \ \frac{1}{4}$


Cancelando (1/2) com (-1/2) e (1/4) com (-1/4):

$S = \frac{1}{2} \ - \ \frac{3}{4}$


Tirando o MMC:


$S = \frac{2 \ - \ 3}{4}$


$S = -\frac{1}{4}$


A resposta correta é a primeira alternativa.

Bons estudos!