A certain culture of bacteria intially contains 10,000 bacteria and increases by 20% every hour. Identify the explicit nth term of the sequence. how many bacteria are in the culture at the end of 10 hours? <-- thats basically exactly what the problem says. I could do it like hour per hour but there is probably a formula to solve these kinds of problems.. *it could come in handy too.
Hello, arturju!
A certain culture of bacteria intially contains 10,000 bacteria
and increases by 20% every hour.
Identify the explicit nth term of the sequence.
How many bacteria are in the culture at the end of 10 hours?
Let's baby-talk through this . . .
There are X bacteria.
At the end of an hour, there is 20% more bacteria, 0.20X more.
. . So there is: .X + 0.20X .= .1.2X bacteria
So the original amount in multiplied by 1.2 to give us the new amount.
Since this happens every hour,
. . at the end of two hours, there are: .(1.2)²X bacteria
. . and at the end of three hours, there are: .(1.2)³X bacteria.
The problem says that: X = 10,000
Hence, at the end of n hours, there are: .10,000·(1.2)^n bacteria
At the end of 10 hours: .10,000·(1.2)^10 .≈ .61,917 bacteria.
There is a very handy formula for expotential growth and decay: y=ce^(kt), where y is the new amount, c is the original, t is the time and k is a constant given by ln(f)/p where f is the factor of growth per time period p.
y=(10000)e^(ln(1.2)*10) (ln(1.2) is divided by 1 here) I don't have a good calc around but I bet that gives you the right answer.
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