Question

A population of geese doubles every 18 months. If there are 5 geese initially, how many years will it take for there to be 111 geese?

Answer 1

Answer: 7 years

Step-by-step explanation:

Let t be the number of periods in which geese doubles .

The exponential growth function:$y=Ab^t$ , where A = initial value , b= growth factor.

As per given , A = 5 , b=2

i.e. Number of geese after t periods = $y=5(2)^t$

Put y=111, we get

$111=5(2)^t\\\\\Rightarrow\ \dfrac{111}{5}=2^t\\\\\Rightarrow\ 22.2=2^t$

Taking log on both sides , we get

$\log 22.2 = t\log 2\\\\\Rightarrow\ 1.346353=t(0.30103)\\\\\Rightarrow\ t=\dfrac{1.346353}{0.30102}\\\\\Rightarrow\ t=4.47263636\approx4.47$

Now , total months = 4.47 x 18 =80.46≈ 80 months

since 1 year =12 months

Number of years it will take $=\dfrac{80}{12}=\dfrac{20}{3}=6.66666666667\approx7$

Hence, it will take 7 years (approx.)