Question

Найти решение и корень уравнения. Задания на картинках.

Answer 1

1)
${2}^{ log_{8}(3x - 4) } = 5 \\ {2}^{ log_{ {2}^{3} }(3x - 4) } = 5 \\ {2}^{ \frac{1}{3} log_{2}(3x - 4) } = 5 \\ {2}^{ log_{2} {(3x - 4)}^{ \frac{1}{3} } } = 5 \\ {2}^{ log_{2}( \sqrt[3]{3x - 4} ) } = 5 \\ \sqrt[3]{3x - 4} = 5 \\ 3x - 4 = {5}^{3} \\ 3x - 4 = 125 \\3 x = 129 \\ x = 43$
Ответ: 43.

2)
$log_{3}(8 - x) = log_{3}(4 - x) + 1 \\ log_{3}(8 - x) = log_{3}(4 - x) + log_{3}(3) \\ log_{3}(8 - x) = log_{3}(3(4 - x)) \\ 8 - x = 3(4 - x) \\ 8 - x = 12 - 3x \\ - x + 3x = 12 - 8 \\ 2x = 4 \\ x = 2$

Проверка:
log_3(8 - 2) = log_3(4 - 2) + 1
log_3(6) = log_3(2) + log_3(3)
log_3(6) = log_3(6) -- верно => х = 2 является корнем уравнения.
Ответ: 2.

Answer 2

$1) \: \: {2}^{ log_{8}(3x - 4) } = 5 \\ \\ \\ log_{8}(3x - 4) = log_{2}(5) \\ \\ \frac{1}{ 3} log_{2}(3x - 4 ) = log_{2}(5) \\ \\ log_{2}( \sqrt[3]{3x - 4} ) = log_{2}(5) \\ \\ \sqrt[3]{3x - 4} = 5 \\ \\ 3x - 4 = {5}^{3} \\ \\ 3x - 4 = 125 \\ \\ 3x = 129 \\ \\ x = 43$

ОТВЕТ: 43

$2) \: log_{3}(8 - x) = log_{3}(4 - x) + 1 \\ \\ log_{3}(8 - x) = log_{3}(4 - x) + log_{3}(3) \\ \\ log_{3}(8 - x) = log_{3}(3 \times (4 - x)) \\ \\ 8 - x = 3 \times (4 - x) \\ \\ 8 - x = 12 - 3x \\ \\ 2x = 4 \\ \\ x = 2$

Проверка:

log( 3 )_( 8 - 2 ) = log( 3 )_( 4 - 2 ) + 1

log( 3 )_( 6 ) = log( 3 )_( 2 ) + log( 3 )_( 3 )

log( 3 )_( 6 ) = log( 3 )_( 6 )

Верно


ОТВЕТ: 2