Question

Salut, ma puteti ajuta la problema 112, am incercat sa o fac insa nu iasa.....

Answer 1

log_(2x) (4x) + log_(4x) 16x = 4

log_(2x) (2x)² - log_(2x) x + log_(4x) (4x)² - log_(4x) x = 4

=> 2+2 - log_(2x) x - log_(4x) x = 4

=> log_(2x) x + log_(4x) x = 0

(1) log_(2x) x = 0 si log_(4x) x = 0

=> x = 1

(2) x ≠ 1

=> log_(2x) x + log_(4x) x = 0 |:log_(2x) x

=> 1+ log_(4x) x / log_(2x) x = 0

=> 1+ log_(4x) x • log_x (2x) = 0

=> 1+ log_(4x) (2x) = 0

=> log_(4x) (2x) = -1

=> 2x = (4x)^-1

=> 2x = 1/(4x)

=> 8x² = 1

=> x² = 1/8

=> x = 1/(2√2) (x>0)

=> x ∈ {1/(2√2), 1}

Răspuns corect C.