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

\(\ln(6^{719})\\ =719 \ln 6\\ \approx 1288.2750583749715456\)

At time equal to 0 years(t=0),the original isotope has 85

when dropped to 1/6,the isotope has 85/6

substitute it into M(t) and we have

\(85/6=85*e^\left (-t/719 \right )\)  

\(1/6=e^\left (-t/719 \right )\)

ln(1/6)=-t/719

-ln(6)=-t/719

ln6=t/719

t=719*ln6

t=ln(6^719)=1*ln(6^719)

6^719 about equal to infinity

NOTE: Anybody interested in understanding how radioactive substances decay with time, should read this page which very clearly explains it:

http://math.usask.ca/emr/examples/expdeceg.html

1/6 (85) = 85 e^(-t/719)         this is of the form   A = Ao e^kt    where k= -.0013908

1/6 = e^(-t/719)

ln(1/6) = (-t/719)

-719 ln(1/6) = t     = 1288.28 years

NO!. When you calculate k, you no longer use the half-life given, which is 719 years in this case. So that your equation should read: 85/6 =85 * e^(-0.000964 * t), which is equal to: 85/6 =85 * 2^(-t/719). If you now solve for t in both cases, you will get EXACTLY the same answer, which comes to:1,858.59 years.

ALAN WILL EXPLAIN IT TO YOU TOMORROW, HOPEFULLY. THE EQUATION, AS GIVEN, IS THE PROBLEM. THAT IS WHY I MODIFIED IT. IF YOU SUBSTITUTE 1/2 FOR e, THEN YOU WILL GET THE RIGHT ANSWER: SOLVE FOR t: 85/6=85 * 1/2^(t/719), WHERE t=ELAPSED NUMBER OF YEARS, 719=HALF-LIFE OF THE RADIOACTIVE SUBSTANCE. THAT IS THE END OF THIS MATTER FOR ME!.

See:  http://web2.0calc.com/questions/m-t-85e-t-29-describes-the-amount-of-a-radioactive-isotope-as-a-function-of-time-t-years-after-how-long-the-amount-has-dropped-to-1-10-of#r9

Nowhere does the original question call 719 the half-life.  

The structure of the equation implies that 1/719 is the decay constant, so the half-life would be ln(2)/(1/719) or just 719*ln(2).

Alan: Is the constant of decay ALWAYS a rational number in these examples given by this questioner?: 1/29, 1/719, 1/409.........etc.???!!!. I  thought when you obtain the constant of decay by taking the natural log of 2 or 1/2 and multiplying it or dividing it by half-life, you automatically get an irrational constant of decay. No?.