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The population of a town is $500,000$ . If the population growth rate is $\newline$ $6\%$ yearly, when will the population exceed $600,000$ ? (Express the answer correct to the nearest integer.) $\newline$ A. $2$ $\newline$ B. $3$ $\newline$ C. $4$ $\newline$ D. $5$

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Exponential growth and decay: word problems

Full solution

Q. The population of a town is

500,000

. If the population growth rate is

\newline

6\%

yearly, when will the population exceed

600,000

? (Express the answer correct to the nearest integer.)

\newline

A.

2

\newline

B.

3

\newline

C.

4

\newline

D.

5

Identify Population and Growth Rate: Identify the initial population and the growth rate. $\newline$ The initial population $P_0$ is $500,000$ and the growth rate $r$ is $6\%$ per year.
Set Up Exponential Growth Formula: Set up the exponential growth formula. $\newline$ The formula for exponential growth is $P(t) = P_0 \times (1 + r)^t$ , where $P(t)$ is the population at time $t$ , $P_0$ is the initial population, $r$ is the growth rate, and $t$ is the time in years.
Determine Population Threshold: Determine the population at which we want to find the time. We want to find the time when the population exceeds $600,000$ .
Set Up Inequality for t: Set up the inequality to solve for t. $\newline$ $600,000 < 500,000 \times (1 + 0.06)^t$
Isolate Exponential Expression: Divide both sides of the inequality by $500,000$ to isolate the exponential expression. $\newline$ $\frac{600,000}{500,000} < (1 + 0.06)^t$ $\newline$ $1.2 < (1.06)^t$
Use Logarithms to Solve: Use logarithms to solve for $t$ . Take the natural logarithm $(\ln)$ of both sides to get: $\ln(1.2) < t \times \ln(1.06)$
Divide by $\ln(1.06)$ for $t$ : Divide both sides by $\ln(1.06)$ to solve for $t$ .
$t > \frac{\ln(1.2)}{\ln(1.06)}$
Calculate t Value: Calculate the value of t using a calculator. $\newline$ $t > \frac{\ln(1.2)}{\ln(1.06)}$ $\newline$ $t > \frac{0.1823215567939546}{0.05826890812397534}$ $\newline$ $t > 3.129496404455733$
Round $t$ to Nearest Integer: Round $t$ to the nearest integer since we cannot have a fraction of a year. $t \approx 4$ (since the population must exceed $600,000$ , we round up to the next whole year)

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