Let N(t) be the number of bacteria after t days. Then N(t) = Pa^t for some constants P and a. Measurements indicate that N(4) = 5600 and N(8) = 362, 000.
a. Before working the problem, estimate the value of N(7), the number of bacteria at noon on the fourth day.
b. Write down two equations for P and a, one when t = 4 and the other when t = 8.
Thanks!
To solve this problem, we need to find the values of P and a in the equation N(t) = Pa^t using the given information.
a. To estimate the value of N(7), we can interpolate between the known values of N(4) and N(8). Since 7 is exactly halfway between 4 and 8, we can take the geometric mean of N(4) and N(8) to get an estimate.
The geometric mean is calculated by taking the square root of the product of two numbers. So, the estimate for N(7) is:
√(N(4) * N(8)) = √(5600 * 362000) ≈ √(2,027,200,000) ≈ 45,041 (rounded to the nearest whole number)
Therefore, the approximate value of N(7) is 45,041.
b. We can write down two equations for P and a using the given information.
From N(4) = 5600:
5600 = Pa^4
From N(8) = 362,000:
362000 = Pa^8
These equations represent a system of equations that can be solved to find the values of P and a.
By rearranging the first equation, we have:
P = 5600 / a^4
Substituting this value of P into the second equation, we have:
362000 = (5600 / a^4) * a^8
Simplifying, we get:
362000 = 5600 * a^4
Dividing both sides by 5600:
362000 / 5600 = a^4
64.642857 ≈ a^4
Taking the fourth root of both sides:
√(64.642857) ≈ a
a ≈ 2.522
Now, substituting this value of a back into the first equation to solve for P:
5600 = P * (2.522)^4
5600 = P * 40.091088
Dividing both sides by 40.091088:
P = 5600 / 40.091088
P ≈ 139.685
Therefore, the values of P and a are approximately P ≈ 139.685 and a ≈ 2.522.