Question

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Answer 1

Ответ на фото............

B)

Answer 2

Ответ:  27 .

Пользуемся свойством логарифмов :  $log_{a}x^{2n}=2n\cdot log_{a}\, |x|$  .

$log^2_3(x^4)=12\, log_3(x^2)+16\ \ ,\ \ \ ODZ:\ x\ne 0\ \ ,\\\\\star \ \ log_3(x^4)=4\, log_3\, |x|\ \ ,\ \ log_3(x^2)=2\, log_3\, |x|\ \ \star \\\\(4\, log_3\, |x|)^2-12\cdot 2\, log_3\, |x|-16=0\ \ ,\\\\16\, log_3^2\, |x|-24\, log_3|x|-16=0\ \Big|:8\\\\2log_3^2\, |x|-3log_3|x|-2=0\\\\log_3\, |x|=t\ \ \to \ \ \ 2t^2-3t-2=0\ \ ,\ \ D=b^2-4ac=9+16=25\ ,\\\\t_1=\dfrac{3-5}{4} =-\dfrac{1}{2}\ \ ,\ \ t_2=\dfrac{3+5}{4}=2$

$a)\ \ log_3\, |x|=-\dfrac{1}{2}\ \ \to \ \ \ |x|=3^{-1/2}\ \ ,\ \ |x|=\dfrac{1}{\sqrt3}\ \ ,\ \ x_{1,2}=\pm \dfrac{1}{\sqrt3}\\\\b)\ \ log_3\, |x|=2\ \ \to \ \ \ |x|=3^2\ \ ,\ \ |x|=9\ \ ,\ \ x_{3,4}=\pm 9\\\\c)\ \ x_1\cdot x_2\cdot x_3\cdot x_4=\dfrac{1}{\sqrt3}\cdot \dfrac{-1}{\sqrt3}\cdot 9\cdot (-9)=\dfrac{81}{3}=27$