Suppose that the half-life of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?

The equation we use is \( A(t) = A_0 e^{kt} \) where
\(A(t)\) is the amount of the substance at time t in grams
\(A_0\) is the initial amount in grams
k is the decay rate constant
time t is in days

From the problem statement, we know that \(A_0 = 10g\). In order to answer the question about how much remains after 75 days, we use the half-life information to determine the constant k.
The statement that the half-life of the substance is 20 days tells us that in 20 days, half of the initial amount remains. That would be \(10/2=5\) grams at time \(t=20\) days. We can substitute both of those values into the original \(A(t)\) equation and see if it helps us.
\( 5 = 10 e^{20k} \)
Notice that we are left with just one unknown here, k, so we can solve for it.
\(\displaystyle{\begin{array}{rcl} 5 & = & 10 e^{20k} \\ 1/2 & = & e^{20k} \\ \ln(1/2) & = & \ln(e^{20k}) \\ \ln(1) - \ln(2) & = & (20k) \ln(e) \\ -\ln(2) & = & 20k \\ -\ln(2)/20 & = & k \end{array} }\)

Now that we know the value of k, our equation is \(\displaystyle{ A(t) = 10 e^{-t\ln(2)/20} }\)
So, we have an equation that tells us the amount of the substance at every time t. To determine the amount of the substance after 75 days, we just let \(t=75\) in this last equation. This gives us
\(\displaystyle{ A(75) = 10 e^{-75\ln(2)/20} \approx 0.743 }\) grams.

See the logarithms page for a review of the logarithm rules we used above.

The amount of the substance after 75 days is approximately \(0.743\) grams.

Suppose that radioactive ion-X decays at a constant annual rate of 4%. What is the half-life of the substance when the initial amount is 100g?

approximately 16.98 years

Suppose a bacteria population starts with 10 bacteria and that they divide every hour.
a. What is the population 7 hours later?
b. When will there be 1,000,000 bacteria?

a. 1280 bacteria; b. 16.6 hours

A radioactive material is know to decay at a yearly rate of 0.2 times the amount at each moment. Suppose there are 1000 grams of the material now. What is the amount after 10 years?

135 grams

Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. If there are 2400 grams now, what is the half-life?

Since the material decays proportional to the quantity of the material, the equation we need is \(A(t) = A_0e^{kt}\). With the given information we need to determine the decay rate, k. Then use that to help us determine the time \(t\) when the quantity is \((1/2)A_0\) (since we need to know the HALF life, i.e. the time when half the material remains).

Given - - at \(t=0\), \(A=2500g\) so \(A_0 = 2500\)
Also given - - \(t=10\), \(A(10) = 2400g\)
Use this to determine k.

\(\begin{array}{rcl} 2400 & = & 2500 e^{k(10)} \\ \displaystyle{\frac{24}{25}} & = & e^{10k} \\ \ln(24/25) & = & 10k \\ 0.1\ln(24/25) & = & k \end{array} \)

Half of the initial amount is \(2500/2 = 1250\), so we have \(\displaystyle{ 1250 = 2500 e^{0.1t\ln(24/25)} }\)
and we need to solve for \(t\).

\(\begin{array}{rcl} 1250 & = & 2500 e^{0.1t\ln(24/25)} \\ 0.5 & = & e^{0.1t\ln(24/25)} \\ \ln(0.5) & = & 0.1t\ln(24/25) \\ t & = & \displaystyle{\frac{\ln(0.5)}{0.1\ln(24/25)}} \\ & \approx & 169.8 ~ \text{years} \end{array} \)

The half-life is approximately 169.8 years.

Suppose a bacteria population grows at a rate proportional to the population. There were 200 bacteria 3 days ago and 1000 bacteria 1 day ago. How many bacteria will there be tomorrow?

5000 bacteria

A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?

3.87 hours

What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.

t = 1590 yrs

Carbon-14 has a half-life of 5730 years.
a) If the initial amount is 300g, how much is left after 2000 years?
b) If the initial amount is 400g, when will there be 350g left?

The reason we kept \(k\) in exact terms until the very end is that, if we found a decimal approximation too early, our final answer would have been off. This is a problem especially if you are expected to enter your answer into an online learning system.

a) 236g
b) 1104 years

Find the half-life of a compound where the decay rate is 0.05. (time is in years)

\(t = 13.86\) years

After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?

1355 grams

Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?

He works this problem a little differently in the video than you may have seen before.
The way to work this problem using the standard equation \(A(t) = A(0)e^{kt}\) is to determine that \(k = \ln(0.965)\) and then set \(A(6) = aA(0)\) and solve for a.

80.75%

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.
a. Find an expression for the number of bacteria after t hours.
b. Find the number of bacteria after 3 hours.
c. Find the rate of growth after 3 hours.
d. When will the population reach 10,000?

Part c requires calculus.

First, we use the information to find the equation. Then we use that equation to solve each part of the problem. Initially we have 100 cells, so \(A_0 = 100\). Since the cells grow at a rate proportional to its size, we know the equation is of the form \(A(t) = A_0 e^{kt}\) Since they tell us after an hour the population is at 420, we use this information to find \(k\). When \(t=1\), \(A = 420\). Plugging those values into the equation, we have \(A(1) = 100 e^{k(1)} = 420\). Notice that we used the initial value in this equation. Now we need to solve for \(k\).
\( \begin{array}{rcl} 420 & = & 100e^k \\ 4.2 & = & e^k \\ \ln(4.2) & = & \ln(e^k) \\ \ln(4.2) & = & k \end{array} \)
So now our equation is \( A(t) = 100 e^{t\ln(4.2)} \). It is best to leave the exponent in this exact form so that rounding problems don't occur later on. Okay, so let's answer each part.
a. Well, we've already solved this part. \( A(t) = 100 e^{t\ln(4.2)} \)
b. At 3 hours, \(t=3\), so \(A(3) = 100 e^{3\ln(4.2)} = 100 e^{\ln(4.2^3)} = 100(4.2^3) = 7408.8\)
If you are asked for an exact answer, 7408 is probably the better answer since we cannot have a partial number of cells. However, some instructors may want you to round, in which case your answer would be 7409.
c. To find the rate of growth after 3 hours, we need to use calculus to find \(dA/dt\) at \(t=3\).
\(A(t) = 100 e^{t\ln(4.2)}\)
\( \dfrac{dA}{dt} = 100 e^{t\ln(4.2)} \ln(4.2) \)
At \(t=3\), \( \dfrac{dA}{dt} = 100 e^{3\ln(4.2)} \ln(4.2) \approx 10632 \) bacteria/hr
d.To find when the population will reach 10000, we need to calculate \(t\) when \(A(t) = 10000\). Plugging this into the equation, we get
\( \begin{array}{rcl} 10000 & = & 100e^{t\ln(4.2)} \\ 100 & = & e^{t\ln(4.2)} \\ \ln(100) & = & t\ln(4.2) \\ t & = & \ln(100)/\ln(4.2) \approx 3.20899 \end{array} \)
Since we have only one decimal place in the value 4.2, we will also give our answer to one decimal place as \(t=3.2\) hours.

a. \(\displaystyle{ A(t) = 100e^{t \ln(4.20)} }\) bacteria
b. \(7409\) bacteria
c. \(10 632\) bacteria/hr
d. 3.2 hours