The rate of increase of bacteria in a culture is proportional to the number of bacteria present .if the original number of bacteria double in the two hours, in how many hours will it be five times?
this is exponential growth. Since it doubles in 2 hours, the population is after t hours is
p(t) = A*2^(t/2)
So, it will be 5 times as big as at first when
2^(t/2) = 5
or, you can start from scratch, using
dp/dt = kp
dp/p = k dt
lnp = kt+lnA
p = A*e^(kt)
and work it from there
To solve this problem, we can use the formula for exponential growth:
N(t) = N₀ * e^(kt)
where:
- N(t) is the number of bacteria at time t
- N₀ is the initial number of bacteria
- k is the growth rate constant
- e is the base of the natural logarithm
We know that the rate of increase of bacteria is proportional to the number of bacteria present. This means we can write the following proportionality:
dN/dt = kN
To solve for the growth rate constant (k), we can integrate both sides of the equation:
∫(1/N)dN = ∫kdt
ln(N) = kt + C
Now, let's use the given information that the original number of bacteria doubles in two hours. This means that after two hours, N becomes 2N₀. We can substitute these values into the equation:
ln(2N₀) = k * 2 + C
Next, let's solve for C:
C = ln(2N₀) - 2k
Now, we can substitute C back into the equation:
ln(N) = kt + ln(2N₀) - 2k
To find the time at which the number of bacteria will be five times the original number, we can substitute N = 5N₀ into the equation:
ln(5N₀) = kt + ln(2N₀) - 2k
Now, rearrange the equation to solve for t:
kt + ln(2N₀) - 2k = ln(5N₀)
kt = ln(5N₀) - ln(2N₀) + 2k
kt = ln(5/2)N₀ + 2k
t = (ln(5/2)N₀ + 2k) / k
Now, let's substitute the given information that the original number of bacteria doubles in two hours:
t = (ln(5/2) * N₀ + 2 * ln(2) / 2) / ln(2)
After evaluating this expression, we can find the value of t, which represents the number of hours it will take for the bacteria to increase by a factor of five.
To solve this problem, we can use the concept of exponential growth.
Let's assume that the original number of bacteria is N0, and the rate of increase is k.
According to the given information, the original number of bacteria doubles in two hours. This means that after two hours, the number of bacteria will be 2N0.
We can write this relationship as: 2N0 = N0 * e^(2k), where e is the base of the natural logarithm.
Simplifying the equation, we get: 2 = e^(2k).
To find the value of k, we can take the natural logarithm of both sides of the equation: ln(2) = 2k.
Now, we can find the value of k: k = ln(2) / 2.
Now, let's find how many hours it will take for the number of bacteria to be five times the original value.
Let's assume that after t hours, the number of bacteria will be 5N0.
We can write this relationship as: 5N0 = N0 * e^(kt).
Substituting the value of k, we get: 5 = e^((ln(2)/2) * t).
Taking the natural logarithm of both sides of the equation, we get: ln(5) = (ln(2)/2) * t.
Now, we can find the value of t: t = (2 * ln(5)) / ln(2).
Calculating this expression, we find that t is approximately equal to 3.47 hours.
Therefore, it will take approximately 3.47 hours for the number of bacteria to be five times the original value.