Question

A population of wolves had 120 wolves in the year 2020 and was growing at a rate of 3% per year. (a)Write a formula for the exponential function f(t), where f(t) is the size of the wolf population t years after 2020.(b) Find the solution to the equation 60(2.1)^x=1100.

Answer 1

Given:

The initial number of wolves is a = 120.

The growth rate is r = 3%.

The objective is,

a) To write the formula for exponential function for the population of the wolf after t years.

b) To find the solution to the equation,

$60(2.1)^x=1100\text{ . . . . .. (1)}$

Explanation:

a)

The general formula for the exponential growth of population is,

$y=a(1+r)^x\text{ . . . . . .(2)}$

Here, x represents the number of years which is t.

On plugging the given values in equation (2)

$\begin{gathered} f(t)=120(1+\frac{3}{100})^t \\ f(t)=120(1+0.03)^t \\ f(t)=120(1.03)^t \end{gathered}$

b)

The solution for given equation (1) will be,

$\begin{gathered} 60(2.1)^x=1100 \\ (2.1)^x=\frac{1100}{60} \\ (2.1)^x=\frac{110}{6} \end{gathered}$

Multiply the natural logarithm on both sides of the equation,

$\begin{gathered} ln(2.1)^x=\ln (\frac{110}{6}) \\ x=\frac{\ln (\frac{110}{6})}{\ln (2.1)} \\ x=3.92044\ldots. \\ x\approx3.92 \end{gathered}$

Hence,

a) The exponential function is f(t) = 120(1.03)^t.

b) The solution of the equation is 3.92.