Exponential growth and decay word problem

Question

A population of 80 cougars decreases at a rate of 5% per year.

A. How many cougars will there be after 6 years?

B. After how many years will the population first drop below 40 cougars?

Answer 1

So the formula to know for these kind of problems is :

A = P(1 ± n/r)^rt

A: Final amount

P: Starting amount

n: Rate of growth/decay (the ± is + for growth and - for decay)

r: Rate of compounding (if it's 2 times per year this number is 2, if its every month per year its 12)

t: Time elapsed

So you could plug this problem into that formula to get:

A = 80(1 - 0.05)⁶ (Because n is just 1, it cancels out)

Solving for A you get 58.81 cougars, which rounds up to 59 (Can't have .8 cougars after all!)

To solve part b, you once again substitute into the formula, but this time the unknown will be t:

40 = 80(1 - 0.05)^t

First, divide both sides by 80 to get:

0.5 = (0.95)^t

Then, take the log of both sides to get:

log(0.5) = t * log(0.95)

log(0.5) / log(0.95) = t

t = 13.51

We can check this by plugging back into the equation and making sure A ends up as 40:

A = 80(1 - 0.05)^13.51 = 40

Which works, so it will be 13.51 years.

Hope this helped!