A bacteria has a growth rate constant of 0.02. If initially, the number of bacteria is 1000, what is the time before it reaches a population of 100,000?
So, you have P(t) = 1000 e^(.02t)
so solve
1000 e^(.02t) = 100,000
e^(.02t) = 100
.02t = ln100
t = ln100/.02 = 230.26
Well, let's see how long it takes for these bacteria to throw a big party and invite all their friends!
Given that the growth rate constant is 0.02, it means that the number of bacteria will multiply by 1.02 every unit of time.
Now, we need to find out how many times the bacteria's population needs to multiply by 1.02 to reach 100,000 starting from 1000.
Mathematically, we can say:
1.02^x = 100,000/1000
Let's put on our thinking caps and solve this equation...
Applying a little mathematical mojo, the value of x comes out to be approximately 345.15.
So, it will take approximately 345.15 units of "bacterial party time" for our bacterial buddies to reach a population of 100,000.
Now, I must warn you, these bacteria can be quite unpredictable, so you might want to double-check their invitations before you plan any surprise parties!
To find the time it takes for the bacterial population to reach 100,000, we can use the exponential growth formula:
N(t) = N₀ * e^(kt)
where:
N(t) is the population size at time t,
N₀ is the initial population size,
e is the mathematical constant approximately equal to 2.71828, and
k is the growth rate constant.
In this case, N(t) needs to be 100,000, N₀ is 1000, and k is 0.02. We can rearrange the formula to solve for t:
100,000 = 1000 * e^(0.02t)
To solve for t, we can take the natural logarithm (ln) of both sides:
ln(100,000) = ln(1000 * e^(0.02t))
Using the logarithm properties, we can simplify further:
ln(100,000) = ln(1000) + ln(e^(0.02t))
Since ln(e^x) = x, the equation becomes:
ln(100,000) = ln(1000) + 0.02t
Now, we can isolate t by subtracting ln(1000) from both sides:
ln(100,000) - ln(1000) = 0.02t
Using a calculator, we can calculate the values of ln(100,000) and ln(1000).
ln(100,000) ≈ 11.5129
ln(1000) ≈ 6.9078
Substituting these values back into the equation:
11.5129 - 6.9078 = 0.02t
Simplifying further:
4.6051 ≈ 0.02t
To solve for t, we divide both sides by 0.02:
4.6051 / 0.02 ≈ t
The approximate value of t is:
t ≈ 230.25
Therefore, it will take approximately 230.25 units of time for the bacterial population to reach 100,000.
To find the time it takes for the population of bacteria to reach 100,000, we can use the exponential growth formula. This formula describes the relationship between the initial size of a population, the growth rate constant, and the time it takes for the population to reach a certain size.
The formula for exponential growth is:
N = N0 * e^(kt)
Where:
- N is the final population size
- N0 is the initial population size
- e is the base of the natural logarithm, approximately 2.71828
- k is the growth rate constant
- t is the time it takes for the population to reach N
In this case, we know that the initial population (N0) is 1000, the growth rate constant (k) is 0.02, and we want to find the time it takes to reach a population (N) of 100,000.
So, the equation becomes:
100000 = 1000 * e^(0.02t)
To solve for t, we can rearrange the equation as follows:
e^(0.02t) = 100000 / 1000
e^(0.02t) = 100
Now, take the natural logarithm (ln) of both sides to solve for t:
ln(e^(0.02t)) = ln(100)
0.02t = ln(100)
Finally, divide both sides of the equation by 0.02 to isolate t:
t = ln(100) / 0.02
Using a calculator, evaluate the right side of the equation to find the value of t. The approximate value is:
t ≈ 230.26
So, it will take approximately 230.26 units of time for the population of bacteria to reach 100,000.