Question 324058
A radioactive substance has a half-life of 6.8 hours.  If 12.6 kg. remain from the original amount of 20 kg. How many hours have passed?

Nt = No * (1/2)^(t/hl)
Nt = N sub t = remaining quantity after time t = 12.6 kg
No = N sub o = initial (original) quantity of a decaying substance = 20kg
t = time, in this case time in hours
hl = t sub 1/2 = half-life of the decaying quantity = 6.8 hours
(details about this equation at http://en.wikipedia.org/wiki/Half-life)

12.6 = 20 * (1/2)^(t/6.8)
12.6/20 = (1/2)^(t/6.8)
0.63 = (1/2)^(t/6.8)
logarithmic rule: if b^x = y, then logb (y) = x
log1/2 (0.63) = t/6.8
6.8 * log1/2 (0.63) = t
logarithmic rule: to change bases --> logb (x) = (logk (x))/(logk (b)),
where b is the old base and k is the new base
log1/2 (0.63) = (log10 (0.63))/(log10 (1/2)) = 0.666576266274808225391191653851317 = approx. 0.666576 to 6 places
t = 6.8 * log1/2 (0.63) = 4.53271861066869593266010324618895
t = approximately 4.5 hours has passed