Question

4 и 6 решите 92 бала дам

Answer 1

Ответ:

#4.

x = 0

#6.

$- \frac{1}{6}$

Пошаговое объяснение:

#4.

${3}^{2x + 3} - 12 \times {3}^{2x} = 15 \\ {3}^{2x} ( {3}^{3} - 12) = 15 \\ {3}^{2x} \times 15 = 15 \\ {3}^{2x} = 1\\ {3}^{2x} = {3}^{0} \\ 2x = 0 \\ x = 0$

#6.

$log_{8}( log_{14}(196) )- log_{7}( \sqrt{7} ) = \\ = log_{ {2}^{3} } (log_{14}( {14}^{2} )) - log_{7}( {7}^{ \frac{1}{2} } ) = \\ = log_{ {2}^{3} } ( 2) - \frac{1}{2} = \\ = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = - \frac{1}{6} \:$

Используемые формулы:

$log_{a}( {a}^{b} ) = b \\ \sqrt{a} = {a}^{ \frac{1}{2} } \\ log_{ {a}^{b} }(a) = \frac{1}{b}$