1)log1/2(2x-1)-log1/2(16)=5 ОДЗ:2x-1> 0
log1/2 (2x-1)/16 =5 2x > 1
(2x-1)/16 = (1/2)^5 x > 0,5
(2x-1)/16 =1/32
2x-1=0,5
2x=1,5
x=7,5
2)lg(5x+2)=1/2lg36+lg2
Одз 5x+2>0
lg(5x+2)=lg6+lg2
lg(5x+2)=lg12
5x+2=`12
5x=10
x=2
или некак!?!?!