The population of a parish is growing with an annual percentage rate compounded continuously. The population reaches 1.1 times its previous size in 2 years. Find the annual percentage rate according to the exponential growth function.Write your answer in exact form, using ln (do not round).

Answer :

Answer:

[tex]4.8\%[/tex]

Step-by-step explanation:

The population that grows with an annual percentage rate compounded continuously is given by:

[tex]P=P_0e^{rt}[/tex]

The population reaches 1.1 times its previous size in 2 years.

This means that, when t=2,

[tex] P=1.1P_0[/tex]

We substitute to obtain:

[tex]1.1P_0=P_0e^{r \times 2}[/tex]

This implies that:

[tex]1.1=e^{2r }[/tex]

Take natural log to get:

[tex] ln(1.1) = 2r[/tex]

[tex]r = \frac{ ln(1.1) }{2} [/tex]

[tex]r = 0.0477[/tex]

Therefore the annual percentage rate is

[tex]4.8\%[/tex]

The annual percentage rate according to the exponential growth function is 5%

An exponential growth is in the form:

y = abˣ;

where y, x are variables, a is the initial value of y and b > 1

Let y represent the population after x years.

Let us assume that initially the population is c. hence, a = c.

The population reaches 1.1 times its previous size in 2 years:

[tex]1.1c=c(b)^2\\\\1.1=b^2\\\\2ln(b)=ln(1.1)\\\\ln(b)=0.0477\\\\b=1.05[/tex]

Hence the annual percentage rate = 1.05 - 1 = 5%

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