→ hwanger : 0 > log(x)-log_x(10) = log(x)-1/log(x) 得 11/21 01:15
→ hwanger : 1/log(x) > log(x) 進一步考慮兩個cases 11/21 01:15
→ hwanger : Case1. 若log(x)>0 則得 0 > (log(x)-1)*(log(x)+1) 11/21 01:18
→ hwanger : 故 0 < log(x) < 1 推得1<x<10 11/21 01:19
→ hwanger : Case2. 若log(x)<0 則得 0 < (log(x)-1)*(log(x)+1) 11/21 01:20
→ hwanger : 故 log(x) < -1 推得0 < x < 1/10 11/21 01:21
→ susiseptem : 感恩 11/21 01:32