Question

log16 (x) +log4 (x) +log2 (x) =7

Answer 1

log₁₆X + log₄X + log₂X = 7
ОДЗ: x>0

$frac{ log_{2}X }{4} + frac{ log_{2}X }{2} + log_{2}X = 7$

log₂X + 2log₂X + 4log₂X = 28
log₂X  + log₂X² + log₂X⁴ = 28
log₂X⁷ = 28
x⁷ = 2²⁸
x = 2⁴
x= 16
Ответ: 16