SOLUTION: A scientist has started his experiment with 5,000,000 cells. He noticed that 45% of the cells die every minute. How long does it take for the number of cells to be less than 1,000?

Question 1188339: A scientist has started his experiment with 5,000,000 cells. He noticed that 45% of the cells die every minute. How long does it take for the number of cells to be less than 1,000?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
he starts with 5,000,000 cells.
45% of the cells die every minute.
you want to know how long it takes for the number of cells to be less than 1000.

the formula to use here is f = p * (1 + r) ^ n.

f is the future value
p is the present value
n is the number of time periods.
r is the rate of increase or decrease per time period.

your time periods are in minutes.

in your problem.

f = 1000
p = 5000000
r = -.45

equation becomes 1000 = 5000000 * (1 - .45) ^ n

divide both sides of the equation by 5000000 and simplify to get:

1/5000 = .55 ^ n

take the log of both sides of the equation to get:

log(1/5000) = log(.55 ^ n)

since log(.55 ^ n) is equal to n * log(.55), the equation becomes:

log(1/5000) = n * log(.55)

divide both sides of the equation by log(.55) to get:

log(1/5000) / log(.55) = n

solve for n to get:

n = log(1/5000) / log(.55) = 14.24668125.

confirm this is correct by replacing n in the original equation and solving for f to get:

f = 5000000 * (1 - .45) ^ 14.24668125 = 1000.

the solution is confirmed to be good.
the solution is that 5 million cells will be reduced to 1000 cells in 14.24668125 minutes.
anything more than 14.24668125 minutes will reduce the number of cells left to less than 1000.

the graph of the equation looks like this.

he coordinates on the graph are in (x,y) format.
y is the number of cells left.
x is the number of minutes from when the count started.

when x = 0, the number of cells is 5,000,000
when x = 14.24668125, the number of cells is 1,000.

on the graph, numbers are rounded to 3 decimal places.
14.24668125 becomes 14.247.

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