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?
:
The half-life formula: A = Ao*2^(-t/h)
Where
A = Resulting amt after t hrs
Ao = initial amt
t = time (in hrs)
h = half-life of the substance
:
20*2^(-t/6.8) = 12.6
:
2^(-t/6.8) = {{{12.6/20}}}
2^(-t/6.8) = .63
use logs here
log(2^(-t/6.8)) = log(.63)
log equiv of exponents
{{{-t/6.8}}}*log(2) = log(.63)
Find the logs of 2 and .63
{{{-t/6.8}}}*.301 = -.201
{{{-.301t/6.8}}} = -.201
:
-.301t = -.201 * 6.8
-.301t = -1.3645
t = {{{(-1.3645)/(-.301)}}}
t = 4.5 hrs, for 12.6 kg to remain
:
:
Check solution on a calc; enter: 20*2^(4.5/6.8) results 12.6