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

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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 the original ?

The accurate number of years is calculated as follows:

m(t)=85e^(-t/29)

85/10 = 85 * 2 (-t/29), solve for t

t=96.34

Solve for t over the real numbers:
17/2 = 85 2^(-t/29)

17/2 = 85 2^(-t/29) is equivalent to 85 2^(-t/29) = 17/2:
85 2^(-t/29) = 17/2

Divide both sides by 85:
2^(-t/29) = 1/10

Take reciporicals of both sides:
2^(t/29) = 10

Take the logarithm base 2 of both sides:
t/29 = (log(10))/(log(2))

Multiply both sides by 29:
Answer: |t = (29 log(10))/(log(2))=96.34 years.

ElectricPavlov: See my explanation here:

http://web2.0calc.com/questions/m-t-85e-t-409-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

Also, please read this so that you can understand how radioactive materials decay with time:

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

No!. k is not (-1/29). k is calculated as ln(1/2) / 29=-0.02390162. k is ALWAYS a number < 0.

So, your equation becomes: 85/10 =85 * e^(=-0.02390162 x t). Now, if you solve for t, you should get:

t=96.34 years. Remember that 29 is half-life. So that: 1/2 ^(96.34/29)=~ 1/10 of the original amount.

NO!!. 29 IS HALF-LIFE!! Half-life of any radioactive substance is calculated as: ln(1/2) / k, when k is known. IF it is not known, then it is calculated as: ln(1/2) / half-life of the substance.

I think you should have Alan explain it to you if you don't understand the technical page I referred you to. http://math.usask.ca/emr/examples/expdeceg.html

P.S. The whole problem stems from the way the question is posed. When e is used, it automatically implies that you have to calculate k, so that it should read: e^(kt). When they give you the half-life, it automatically implies that you must calculate k. You can get the same result by using: 2^(-t/half-life), or: 1/2 ^ (t/half-life).

Hmm!

Radioactive decay equation often written as:  \(N=N_0e^{-kt}\) where k is the decay constant.

When \(N=\frac{N_0}{2}\) we have \(\frac{1}{2}=e^{-k\tau}\) where \(\tau\) is the half-life.

Taking logs we get \(\ln{\frac{1}{2}}=\ln{e^{-k\tau}}\)  or \(-\ln2=-k\tau\)  so the relation between half-life and decay constant is \(\tau=\frac{\ln2}{k}\)  or \(k=\frac{\ln2}{\tau}\)