What is the half-life of an isotope that decays to 6.25% of its original activity in 70.1 hours?

17.525 hours  
First, determine how many half lives the substance has undergone by taking the logarithm to base 2 of the amount left. So log(0.0625)/log(2)
 = -1.204119983/0.301029996
 = -4 
 So the substance has undergone exactly 4 half lives. This makes sense since 1/2^4 = 1/16 = 0.0625. 
 Since 4 half lives took 70.1 hours, the duration of 1 half-life will be 70.1/4 = 17.525 hours.