+78 M(t)=85e^(-t/719) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/6 of the original ? The response can be written as t=alnb where a and b are integers and a=1
M(t)=85e^(-t/719)
85/6 =85 * 2^(-t/719), solve for t
t=1,858.59 years.
Solve for t over the real numbers:
85/6 = 85/2^(t/719)
85/6 = 85/2^(t/719) is equivalent to 85/2^(t/719) = 85/6:
85/2^(t/719) = 85/6
Divide both sides by 85:
2^(-t/719) = 1/6
Take reciporicals of both sides:
2^(t/719) = 6
Take the logarithm base 2 of both sides:
t/719 = (log(6))/(log(2))
Multiply both sides by 719:
Answer: |t = (719 log(6))/(log(2))=1,858.59 years.
