Question

A lake has a small population of a rare endangered fish. The lake currently has a population of 10 fish. The number of fish is increasing at a rate of 4% per year. When will the population double? How long will it take the population to be 80 fish?

Answer 1

Answer: 1) After 17.673 years ( Approx) the population will be doubled.

2) After 53.019 years the population will be 80.

Step-by-step explanation:

1) Let x represents the number of years after which the population will doubled,

Since rate of interest = 4%

And, initial number of fish = 10,

Hence the fish after x years = $10(1+\frac{4}{100})^x =10(1+0.04)^x =10(1.04)^x$

⇒ $10(1.04)^x = 20$

⇒ $(1.04)^x = 2$

⇒ $x log (1.04) = log(2)$

⇒ $x = \frac{log(2)}{log(1.04)}=17.6729877\approx 17.673$

Hence, After 17.673 years ( Approx) the population will be doubled.

2) Let after y years the population will be 80.

⇒ $10(1.04)^x = 80$

⇒ $(1.04)^x = 8$

⇒ $x log (1.04) = log(8)$

⇒ $x = \frac{log(8)}{log(1.04)}=53.0189631\approx 53.019$

Hence, After 53.019 years the population will be 80.