# The population of a certain city was 174,000 in 2006, and the observed doubling time for the...

## Question:

The population of a certain city was {eq}\,\textrm{174,000}\, {/eq} in {eq}\,2006 {/eq}, and the observed doubling time for the population is {eq}\,19\, {/eq} years.

a. Find an exponential model {eq}\; n(t) = n_{0}2^{t/a}\; {/eq} for the population {eq}t {/eq} years after {eq}2006 {/eq}.

b. Find an exponential model {eq}\,\; n(t) = n_{0}e^{rt}\;\, {/eq} for the population {eq}t {/eq} years after {eq}2006 {/eq}.

## Exponential Models

As we see in this question, there can be different types of exponential models. Such models have specific parameters whose values have to be found from the data given.

## Answer and Explanation:

a) To complete the exponential model, we have to find the value of {eq}a {/eq}. This can be done as follows using the doubling time of {eq}19 {/eq} years.

$$\begin{align} 174000*2^{19/a}&=2*174000\\ 2^{19/a}&=2\\ \frac {19}a\ln 2&=\ln 2\\ \therefore a&=19 \end{align}\\ $$

The model is with the initial population of 174000 in 2006 is:

$$n(t) =174000 2^{19/t} $$

b) Here we have to find {eq}r {/eq}. This can be found as follows.

$$\begin{align} 174000*e^{19r}&=2*174000\\ e^{19r}&=2\\ 19r&=\ln 2\\ r&\approx 0.03648 \end{align}\\ $$

The model is;

$$n(t) = 174000e^{0.03648t} $$

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10