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oobleck lose the "urgency" plea. We help when we happen to see a problem. If you wait till the last minute, it's on you.
If we are using base 10 logs, then
1+(1/logx) < 0
1/logx < -1
1/logx = -1 if x = 1/10
since log(1) = 0, 1/logs has an asymptote at x=1
so, we want 1/10 < x < 1
for x > 1, logx > 0, so 1 + 1/logx will always be positive