Log_(√3) (x-2)*log₅x=2log₃(x-2)
ОДЗ:
{x-2>0 {x>2
x>0 x>0
x∈(2;∞)
log_(3¹/²) (x-2)*log₅x=2log₃(x-2)
log_(a^n) b=(1/n)*log_a b
(1:1/2)*log₃(x-2)*log₅x-2log₃(x-2)=0
2log₃(x-2)*(log₅x-1)=0
2log₃(x-2)=0 или log₅x-1=0
1. 2log₃(x-2)=0. x-2=3⁰. x-2=1. x=3
2. log₅x=1. x=5¹. x=5
ответ: x₁=3, x₂=5