# The initial population of a town is 2,600, and it grows with a doubling time of 10 years. What...

## Question:

The initial population of a town is 2,600, and it grows with a doubling time of 10 years. What will the population be in 12 years?

## Exponential Growth Equations:

When something grows in pattern such that its growth increases with time, we call this an exponential growth. An exponential growth equation takes the form of {eq}P_t = P_0e^{rt} {/eq}.

The exponential growth equation takes the form:

\begin{align} P_t = P_0e^{rt} \end{align}

where:

{eq}P_t {/eq} is the population at any Time {eq}t {/eq},

{eq}P_0 {/eq} is the initial population,

{eq}r {/eq} is the growth rate and,

{eq}t {/eq} is the time.

In our question, we have:

{eq}\begin{align} P_0 = 2,600\\[0.3cm] P_10 = 5,200\\[0.3cm] t = 10\\[0.3cm] r = ? \end{align} {/eq}

Therefore:

\begin{align} 5,200 = 2,600e^{10r} \end{align}

\begin{align} 2= e^{10r} \end{align}

Introducing natural logarithms on both sides:

\begin{align} & \ln 2= \ln e^{10r}\\[0.3cm] & ln 2 = 10r \ln e\\[0.3cm] & \ln 2 = 10r \end{align}

Solving for {eq}r {/eq}:

\begin{align} & r = \dfrac{\ln 2}{10}\\[0.3cm] r\approx 0.069315 \end{align}

Therefore, the growth equation is:

\begin{align} P_t = P_0e^{0.069315t} \end{align}

Thus, the population after {eq}12 {/eq} years is:

\begin{align} & P_{12}= 2,600e^{0.069315\times 12}\\[0.3cm] & P_{12}= 2,600e^{0.83178}\\[0.3cm] & \boxed{\color{blue}{P_{12} \approx 5,973}} \end{align} 