Question

me ajudem a responder essas questões! 20 PONTOS

Answer 1

1)

$log_{3}( {x}^{2} - 3x - 6 ) = log_{3}(x + 6) \\ {x}^{2} - 3x - 6 = x + 6 \\{x}^{2} - 3x - x - 6 - 6 = 0 \\ {x}^{2} - 4x - 12 = 0$

$s = \frac{ - ( - 4)}{1} = 4 \\ p = \frac{ - 12}{1} = - 12$

x=-2 e x=6,→ alternativa d

2)

$log_{ \frac{1}{3} }(81) = log_{ {3}^{ - 1} }( {3}^{4} ) = \frac{4}{ - 1} \\ = - 4$

$log( \sqrt[3]{10} ) = log( {10}^{ \frac{1}{3} } ) = \frac{1}{3}$

$log(10000) = log( {10}^{4} ) \\ = 4 log(10) = 4$

$log1 = 0$

$log_{ \frac{1}{3} }(81) + log(10000) \\ + log( \sqrt[3]{10}) + log1$

$- 4 + 4 + \frac{1}{3} + 0 = \frac{1}{3}$

Alternativa a

4)

$log(90) = log(2. {3}^{2}.5 ) \\ = log(2) +2 log(3) + log(5) \\ = 0,30 + 2.0,48 + 0,70 \\ = 0,30 + 0,96 + 0,70$

$= 1,96$

Alternativa a