Although this is a differential equation topic, many students come across this topic while studying basic integrals. You do not need to know anything other than integrals to understand where the equations come from. If you are given the equation and not expected to derive it, you need only logarithms and algebra to work many problems.
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quick notes |
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rate of change is proportional to quantity \( y' = ky ~~~ \to ~~~ y = Ae^{kt} \) |
You will definitely need to be sharp with your logarithms for this topic. The precalculus logarithms page will help you get up to speed. |
The population dynamics page expands on this discussion of exponential growth and decay applied specifically to population change. |
What Does Exponential Growth/Decay Mean?
What does it mean for something to grow or decay exponentially? The idea is that the independent variable is found in the exponent rather than the base. For example, comparing \(f(t)=t^2\) and \(g(t)=2^t\), notice that \(t\) is in the exponent of the \(g(t)\), so \(g(t)\) is considered an example of exponential growth but \(f(t)\) is not (since \(t\) is not in the exponent).
The general form of an exponential growth equation is \(y = a(b^t)\) or \(y=a(1+r)^t\). These equations are the same when \(b=1+r\), so our discussion will center around \(y = a(b^t)\) and you can easily extend your understanding to the second equation if you need to.
When \(b > 1\), we call the equation an exponential growth equation.
When \(b < 1\), it is called exponential decay.
When \(b = 1\), we have \(y=a(1^t)=a\), which is just linear equation and it is not considered an exponential equation.
So it is easy to see that we have the same equation in both cases and the value of \(b\) tells us if the exponential is growing or decaying. Let's watch a quick video before we go on.
video by PatrickJMT |
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A Special Type of Exponential Growth/Decay
A specific type of exponential growth is when \(b=e^{rt}\) and \(r\) is called the growth/decay rate. The equation comes from the idea that the rate of change is proportional to the quantity that currently exists. This produces the autonomous differential equation
autonomous equation |
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A special type of differential equation of the form \(y' = f(y)\) where the independent variable does not explicitly appear in the equation. |
\( y' = ky \) |
where \(k\) is a constant called the growth/decay constant/rate.
When \(k < 0\), we use the term exponential decay.
When \(k > 0\), we use the term exponential growth.
To solve this differential equation, there are several techniques available to us. We will use separation of variables. Although not explicitly written in this equation, the independent variable we usually use in these types of equations is t to represent time.
\[\begin{array}{rcl}
y' & = & ky \\
\displaystyle{\frac{dy}{dt} } & = & ky \\
\displaystyle{\frac{dy}{y} } & = & k~dt \\
\displaystyle{\int{ \frac{dy}{y} }} & = & \displaystyle{\int{ k~dt }} \\
\ln|y| & = & kt+C \\
e^{\ln|y|} & = & e^{kt+C} \\
y & = & Ae^{kt}
\end{array}\]
The constant A is usually called the initial amount since at time \(t=0\), we have \(y = Ae^0 = A\).
The key to solving these types of problems usually involves determining \(k\). Depending on what your instructor wants, you can usually just start with the equation \(y = Ae^{kt}\), if you know that you have an exponential decay or growth problem.
Here is a quick video explaining this and showing a graph to give you a feel for these equations.
video by Brightstorm |
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Half-Life
Half-life is the time it takes for half the substance to decay. The idea is to take the equation \(y = Ae^{kt}\), set the left side to \(A/2\) and solve for t. Notice that you don't have to know the initial amount A since in the equation \(A/2 = Ae^{kt}\), the A cancels leaving \(1/2 = e^{kt}\). You can then use basic logarithms to solve for t.
Uniqueness of Solution
So it may not have occurred to you but have you thought that maybe it is possible there is another solution to the differential equation \(x'=ax\)? This question is part of a very deep discussion in differential equations involving existence and uniqueness. Here is a really good video that shows that the solution discussed on this page is unique.
video by Dr Chris Tisdell |
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Practice
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?
Problem Statement |
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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?
Final Answer |
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The amount of the substance after 75 days is approximately \(0.743\) grams.
Problem Statement
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?
Solution
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.
Final Answer
The amount of the substance after 75 days is approximately \(0.743\) grams.
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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?
Problem Statement |
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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?
Final Answer |
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approximately 16.98 years
Problem Statement
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?
Solution
video by PatrickJMT |
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Final Answer
approximately 16.98 years
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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?
Problem Statement |
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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?
Final Answer |
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a. 1280 bacteria; b. 16.6 hours
Problem Statement
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?
Solution
video by PatrickJMT |
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Final Answer
a. 1280 bacteria; b. 16.6 hours
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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?
Problem Statement |
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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?
Final Answer |
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135 grams
Problem Statement
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?
Solution
video by PatrickJMT |
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Final Answer
135 grams
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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?
Problem Statement |
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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?
Final Answer |
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The half-life is approximately 169.8 years.
Problem Statement
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?
Solution
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 \(t\).
\(\begin{array}{rcl} 1250 & = & 2500 e^{0.1t\ln(24/25)} \\ 0.5 & = & e^{0.1t\ln(24/25)} \\ \ln(0.5) & = & 0.1t\ln(24/25) \\ t & = & \displaystyle{\frac{\ln(0.5)}{0.1\ln(24/25)}} \\ & \approx & 169.8 ~ \text{years} \end{array} \)
Final Answer
The half-life is approximately 169.8 years.
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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?
Problem Statement |
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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?
Final Answer |
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5000 bacteria
Problem Statement
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?
Solution
video by PatrickJMT |
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Final Answer
5000 bacteria
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A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?
Problem Statement |
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A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?
Final Answer |
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3.87 hours
Problem Statement
A bacteria population increases sixfold in 10 hours. Assuming normal growth, how long did it take for their population to double?
Solution
video by Krista King Math |
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Final Answer
3.87 hours
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What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Problem Statement |
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What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Final Answer |
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t = 1590 yrs
Problem Statement
What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Solution
video by Krista King Math |
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Final Answer
t = 1590 yrs
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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?
Problem Statement |
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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?
Final Answer |
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a) 236g
b) 1104 years
Problem Statement
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?
Solution
Carbon-14 has a half-life of 5730 years. |
This first line of the problem statement gives us enough information to determine the decay rate, \(k\), in the equation \( A(t) = A_o e^{kt} \) |
Half-life means the time it takes for half of the initial amount to decay. The initial amount is \(A_o\), so the amount \(A(t) = A_o/2 \) occurs when \(t=5730\). |
Plugging these into the equation gives us \( A_o/2 = A_o e^{5730k} \) |
The initial amount \(A_o\) cancels out, so we don't need to know the initial amount. |
Now we solve for the decay rate \(k\). |
\( 1/2 = e^{5730k} \) |
To get \(k\) out of the exponent, we take the natural logarithm of both sides. We chose the natural logarithm because we have \(e\) as the base of the term \(e^{5730k}\). |
\( \ln(1/2) = \ln( e^{5730k} ) \) |
\( -\ln 2 = 5730k \) |
\( k = \dfrac{-\ln 2}{5730} \) |
We will leave \(k\) in this form for now to make our calculations more precise in the rest of the problem. |
a) If the initial amount is 300g, how much is left after 2000 years? |
Now for part a) we are given the initial amount of 300g. So \( A_o = 300 \). We know this because the word initial tells us that \(t=0\) and so \( A(0) = A_o e^0 \to A(0) = A_o = 300 \) |
We are asked to find \(A(2000)\). |
Plugging in what we know gives us \( A(2000) = 300e^{2000(-\ln2)/5730} \) |
So our answer is \( A(2000) \approx 236 \)g |
b) If the initial amount is 400g, when will there be 350g left? |
Since the initial amount is given as 400g, we know that \(A_o = 400\) |
Now we are asked to find the time \(t\) when \(A(t) = 350\) |
Our partial equation is \( 350 = 400 e^{kt} \) |
We know \(k\) but we will solve for \(t\) first and then plug in our value for \(k\). |
\( 350/400 = e^{kt} \to \ln (350/400) = \ln(e^{kt}) \) |
\( \ln(350/400) = kt \to t = \ln(350/400)/k \) |
Now we can plug in our exact value of \(k\). |
\( t = \dfrac{\ln(350/400)}{(-\ln 2)/5730} \) |
We can plug these values into our calculator at this point but it might help to simplify this a bit. |
\( t = \dfrac{-5730 \ln(350/400)}{\ln 2} \) |
\( t = \dfrac{-5730 (\ln 350 - \ln 400)}{\ln 2} \) |
\( t \approx 1104 \) years |
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.
Final Answer
a) 236g
b) 1104 years
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Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Problem Statement |
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Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Final Answer |
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\(t = 13.86\) years
Problem Statement
Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Solution
The decay rate is the magnitude of \(k\) in the equation \( A(t) = A_o e^{kt}\). Since the problem states it the compound decays, the value of \(k\) is negative. So \( k = -0.05 \). |
Since we are talking about half-life, we know that \( 1/2 = e^{-0.05t} \). |
Now we need to solve for \(t\). |
\( \ln( 1/2 ) = \ln( e^{-0.05t} ) \) |
\( -\ln( 2 ) = -0.05t \) |
\( \ln( 2 )/0.05 = t \) |
Plugging these into a calculator, we get \(t=13.86\) years. |
Final Answer
\(t = 13.86\) years
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After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?
Problem Statement |
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After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?
Final Answer |
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1355 grams
Problem Statement
After 1000 years, we have 500 grams of a substance with a decay rate of 0.001. What was the initial amount?
Solution
Final Answer
1355 grams
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Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?
Problem Statement |
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Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?
Final Answer |
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80.75%
Problem Statement
Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?
Solution
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.
video by Khan Academy |
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Final Answer
80.75%
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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?
Problem Statement |
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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?
Hint |
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Part c requires calculus.
Problem Statement |
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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?
Final Answer |
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a. \(\displaystyle{ A(t) = 100e^{t \ln(4.20)} }\) bacteria
b. \(7409\) bacteria
c. \(10 632\) bacteria/hr
d. 3.2 hours
Problem Statement
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?
Hint
Part c requires calculus.
Solution
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.
Final Answer
a. \(\displaystyle{ A(t) = 100e^{t \ln(4.20)} }\) bacteria
b. \(7409\) bacteria
c. \(10 632\) bacteria/hr
d. 3.2 hours
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Really UNDERSTAND Differential Equations
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these topics are not required for this page but will help you understand where the equations come from |
external links you may find helpful |
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Math Insight - Exponential Growth and Decay: A Differential Equation |
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