At noon, Professor Simon has a petri dish with 10,000 cells of Bacteria A. In another petri dish, he
has 33,000 cells of Bacteria B. Every hour, Bacteria A grows by 8%. Every hour Bacteria B cells die off,
decreasing the number of cells by 6%. After how many hours will there be more cells of Bacteria A than of
Bacteria B?
after t hours:
count of A = 10000(1.08)^t , where t is in hours
count of B = 33000(.94)^t
when is the count the same ?
10000(1.08)^t = 33000(.94)^t
1.08^t = 3.3 (.94)^t
take log of both sides
log 1.08^t = log 3.3 + log .94^t
t log1.08 = log 3.3 + tlog .94
t log 1.08 - t log.94 = log 3.3
t(log 1.08 - log .94) = log 3.3
t = log 3.3/(log 1.08 - log .94)
= appr 8.6 hours
state your conclusion
To solve this problem, we need to find when the number of cells of Bacteria A exceeds the number of cells of Bacteria B. We can do this by setting up a formula to calculate the number of bacteria for each hour and finding when the number of cells of A exceeds the number of cells of B.
Let's assume the number of hours is represented by 'h'.
Initially, we have 10,000 cells of Bacteria A and 33,000 cells of Bacteria B.
To calculate the number of cells of Bacteria A after 'h' hours, we can use the formula:
Number of cells of Bacteria A = Initial number of cells of A * (1 + growth rate of A)^h
Similarly, to calculate the number of cells of Bacteria B after 'h' hours, we can use the formula:
Number of cells of Bacteria B = Initial number of cells of B * (1 - death rate of B)^h
We want to find the value of 'h' when the number of cells of A exceeds the number of cells of B, so we set up an equation:
Number of cells of A = Number of cells of B
Plugging in the formulas above, we get:
10,000 * (1 + 0.08)^h = 33,000 * (1 - 0.06)^h
To solve this equation, we can take the logarithm of both sides to isolate 'h':
log(10,000) + h * log(1.08) = log(33,000) + h * log(0.94)
Using algebraic manipulation, we can rearrange this equation to solve for 'h':
h * (log(1.08) - log(0.94)) = log(33,000) - log(10,000)
h = (log(33,000) - log(10,000)) / (log(1.08) - log(0.94))
Using a calculator, we can find the value of 'h':
h ≈ 18.84
Therefore, it will take approximately 18.84 hours for the number of cells of Bacteria A to exceed the number of cells of Bacteria B.