Population of a pond

Question

An endangered species of fish has a population that is decreasing exponentially. The population 8 years ago was 3700. Today, only 1700 of the fish are alive. Once the population drops below 75, the situation will be irreversible. When will this happen, according to the model. (Round your answer to the nearest whole year.)

Answer 1

The trick with this one is to treat eight years ago as year zero, and now as year eight. That gives us two points:

(0,3700) & (8, 1700)

I have seen a variety of variables used for exponential equations, so I will simply use the ones I was taught years ago and you can substitute the ones your book uses:

A_t = A₀ * K^t       where A_t is the amount at any future time, A₀ is the original amount, K is the growth or decay factor and T is of course the time.

That gives you two equations:

3700 = A₀ * K⁰                 and          1700 = A₀ * K⁸

In this case, we have an easy approach, since a zero power makes nearly anything 1. The first equation collapses to:

  3700 = A₀      and we now have a part of your final equation. Substitute it into the other one:

1700 = 3700 * K⁸   then solve by first dividing both sides by 3700, then taking the 8^th route to isolate the K.

You will get .459 = K⁸     and then .907 = K

So now, you have:

A_t = 3700 * (.907)^t   . Substitute the 75 into the left side, then solve for T by first dividing by 3700. You get roughly:

.02 = .907^t                 Bring the "T" down by taking a log on both sides (Ln is fine too), then divide by the log .907 on both sides. This yields a time of 40.1 years, rounded to 40.