Question

There are two parts to this question. A) The exponential growth function that models the data is A=

Answer 1

Answer:

$\begin{gathered} (a)A=5.86e^{0.01t} \\ (b)Year\;2100 \end{gathered}$

Explanation:

The exponential growth model is given as:

$A=A_oe^{kt}$

The population in Year 2000 (t=0) = 5.86 million.

$\begin{gathered} 5.86=A_0e^{k(0)} \\ \implies A_0=5.86 \end{gathered}$

The projected population in Year 2015 (t=15) = 7 million.

$\begin{gathered} A(15)=5.86e^{15k} \\ 7=5.86e^{15k} \end{gathered}$

We solve the equation for k.

$\begin{gathered} \text{ Divide both sides by 5.86} \\ \frac{7}{5.86}=e^{15k} \\ \text{ Take the natural logarithm \lparen ln\rparen of both sides:} \\ ln(\frac{7}{5.86})=ln(e^{15k}) \\ \ln(\frac{7}{5.86})=15k \\ \text{ DIvide both sides by 15} \\ k=\frac{1}{15}\ln(\frac{7}{5.86}) \\ k\approx0.01 \end{gathered}$

(a)The exponential growth function that models the data is:

$A=5.86e^{0.01t}$

(b)When the population is 16 million, i.e. A=16, we want to find the value of t.

$\begin{gathered} 16=5.86e^{0.01t} \\ \text{ Divide both sides by 5.86} \\ \frac{16}{5.86}=e^{0.01t} \\ \text{ Take the }\ln\text{ of both sides} \\ \ln(\frac{16}{5.86})=\ln(e^{0.01t}) \\ \ln(\frac{16}{5.86})=0.01t \\ \text{ Divide both sides by 0.01} \\ t=\frac{1}{0.01}\ln(\frac{16}{5.86}) \\ t\approx100.44 \end{gathered}$

Add 100 to the year 2000:

$2000+100=2100$

The population will be 16 million in the year 2100.