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.