Thread: exponential growth

The population of fish in an isolated lake is decreasing by 10% per year due to acid rain. Thus, 10% of the fish present at the beginning of any year are lost by the end of the year. At this rate, how many years will it take for the fish population to drop to 25% of its current level?

Hello, Harry!

The population of fish in an isolated lake is decreasing by 10% per year due to acid rain.
Thus, 10% of the fish present at the beginning of any year are lost by the end of the year.
At this rate, how many years will it take for the fish population to drop to 25% of its current level?

First, work out the general formula . . .

Let Po = original number of fish (at t = 0).

After the first year, 90% of the fish remain.
. . There are: .(0.90)·Po fish left.

After the second year, there are: .(0.90)²·Po fish left.

After the third year, there are: .(0.90)³·Po fish left.

After the nth year, there are: .(0.90)^n·Po fish left.


When will this equal 0.25·Po ?

We have: . (0.90)^n·Po .= .0.25·Po

Divide by Po: . (0.90)^n .= .0.25

Take logs: . ln(0.90)^n .= .ln(0.25)

. . and we have: . n·ln(0.90) .= .ln(0.25)

. . . . . . . . . . .ln(0.25)
Hence: . n .= .---------- .= .13.1576...
. . . . . . . . . . .ln(0.90)


Therefore, it will take a little more than 13 years.