# A marine biologist monitors the population of sunfish in a small lake. He records 800 sunfish in his first year, 600 sunfish in his fourth year, 450 sunfish in his sixth year, and 350 sunfish in his tenth year. Using the exponential regression equation (rounded to the hundredths), 1. Predict when the population of sunfish in the lake will be 200. 2. Predict the number of sunfish in the lake in the 25 thyear.

###### Question:

1. Predict when the population of sunfish in the lake will be 200.

2. Predict the number of sunfish in the lake in the 25 thyear.

## Answers

Answer:Let's use the variable y to represent the number of years passed since the first year.The population on the first year (y = 0) was 800The population in the second year (y = 1) was 600.The ratio in which the population decreased can be calculated as:R = 600/800 = 0.75This means that 600 is the 75% of 800, this also means that between the first year and the second year, the population decreased by the 25%Now let's look at the ratio between the second and third year (y = 2), the population the third year was 450Now the ratio is:R = 450/600 = 0.75Same as before, then we already can see that the population will decrease by 25% each year.The generic exponential decay equation is:f(y) = A*(1 - r)^ywhere:A = initial population, in this case, is 800y = our variable, in this case, represents the number of yearsr = the amount that decreases per each unit of our variable, this must be written in decimal form. In this case, we know that the population decreases by 25% each year, and 25% written in decimal form is 0.25Also, (1 - r) = R, where R is the ratio we found earlier.To do this, we just divide 25% by 100% to get (25%/100% = 0.25)Then our equation will be:f(y) = 800*(1 - 0.25)^yf(y) = 800*( 0.75)^y1) We want to predict when the population will be 200.to do this, we set:f(y) = 200 = 800*( 0.75)^yand solve it for y.(200/800) = 0.75^y(1/4) = 0.75^yNow we can use the relationship:Ln(a^x) = x*ln(a)Then let's apply Ln( ) in both sides to getln(1/4) = y*ln(0.75)ln(1/4)/ln(0.75) = y = 4.8 years.This means that 4.8 years after the first year, the population will be around 200.2) The population in the 25 th year (this is 24 years after the first one, so we take y = 24) is:f(24) = 800*(0.75)^(24) = 0.80Around this point, we will have no more sunfish in the lake.