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:

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