Question

помогите решить:
log81 x + log9 x + log3 x = 7;
log2 x + 2 log4 x + 3 log8 x + 4 log16 x = 4;
log3 x + 2 log9 x + 3 log2Y x + 4 log81 x = 8;

Answer 1

log₈₁x+log₉x+log₃x=7  (1/4)log₃x+(1/2)log₃x+log₃x=7
(7/4)log₃x=7 log₃x=7/(7/4) log₃x=4  x=3⁴=81
log2 x + 2 log4 x + 3 log8 x + 4 log16 x = 4
log₂x+ (2/2)log₂x+(3/3)log₂x+(4/4)log₂x=4
4log₂x=4  log₂x=1  x=2
log3 x + 2 log9 x + 3 log2Y x + 4 log81 x = 8
log₃x+(2/2)log₃x+(3/3)log₃3x+(4/4)log₃x=8
4log₃x=8  log₃x=2  x=3²=9