Bacteria in a certain culture increase at a rate proportional to the number present. If the number doubles in one hour, how long does it take for the number to triple?
To solve this problem, we need to understand that the rate of bacteria growth is proportional to the number present. Let's denote the number of bacteria at any given time as N(t), where t represents time in hours.
We are given that the number of bacteria doubles in one hour. Mathematically, this can be expressed as:
N(1) = 2N(0)
Where N(1) represents the number of bacteria after one hour, and N(0) represents the initial number of bacteria.
Now, we need to determine after how many hours the number of bacteria will triple. Let's call this time T.
To find T, we can set up the following equation based on the given information:
N(T) = 3N(0)
Since the rate of bacteria growth is proportional to the number present, we can write the equation as:
N(T) = N(0) * e^(kT)
Where e is the base of the natural logarithm, k is the proportionality constant, and T represents the time in hours.
Taking the natural logarithm of both sides of the equation, we get:
ln(N(T)) = ln(N(0)) + kT
Now, let's substitute the values we have:
ln(3N(0)) = ln(N(0)) + kT
Using the properties of logarithms, we can simplify further:
ln(3) + ln(N(0)) = ln(N(0)) + kT
ln(3) = kT
Finally, we can solve for T by dividing both sides of the equation by k:
T = ln(3) / k
So, to find how long it takes for the number of bacteria to triple, we need to determine the value of the proportionality constant k.
Let's denote the initial number of bacteria as N0.
Since the bacteria increase at a rate proportional to the number present, we can write the differential equation as:
dN/dt = kN
Where dN/dt represents the rate of change of the number of bacteria over time (t), and k is the proportionality constant.
To solve this differential equation, we can separate variables and integrate:
1/N dN = k dt
Integrating both sides:
∫(1/N) dN = ∫k dt
ln|N| = kt + C
Where C is the constant of integration.
Since we know that the number of bacteria doubles in one hour, we can use this information to find the value of k.
After one hour, the number of bacteria is 2N0, so:
ln|2N0| = k(1) + C
ln(2N0) = k + C
ln(2N0) - C = k
Now, let's find the time it takes for the number to triple. We know that when the number triples, it becomes 3N0.
Using the equation we derived earlier, we can plug in the values:
ln(3N0) - C = k
Since we want to find the time it takes for the number to triple, we need to find the time when N = 3N0.
At that time, the equation becomes:
ln(3N0) - C = ln(2N0) - C
C cancels out, and we're left with:
ln(3N0) = ln(2N0)
Taking the exponential of both sides:
e^(ln(3N0)) = e^(ln(2N0))
3N0 = 2N0
N0 cancels out, and we're left with:
3 = 2
This is a contradiction. Therefore, the equation for exponential growth is not satisfied in this case.
In summary, we cannot determine the exact time it takes for the number of bacteria to triple based on the information given.
clearly, after t hours, the population is 2^t.
So, when is 2^t = 3?