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A town has a population of 13000 and grows at
2% every year. To the nearest tenth of a year, how long will it be until the population will reach 21200 ?
Answer:

A town has a population of $13000$ and grows at $2 \%$ every year. To the nearest tenth of a year, how long will it be until the population will reach $21200$ ? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A town has a population of

13000

and grows at

2 \%

every year. To the nearest tenth of a year, how long will it be until the population will reach

21200

?

\newline

Answer:

Determine growth type: Determine the type of growth. The town's population grows by a fixed percentage each year. This indicates exponential growth.
Identify parameters: Identify the initial population ( $P_0$ ), growth rate ( $r$ ), and final population ( $P$ ). $\newline$ $P_0 = 13000$ $\newline$ $r = 2\%$ per year or $0.02$ per year when converted to a decimal $\newline$ $P = 21200$
Use exponential growth formula: Use the formula for exponential growth: $P = P_0 \times (1 + r)^t$ , where $P$ is the final population, $P_0$ is the initial population, $r$ is the growth rate, and $t$ is the time in years. $\newline$ We need to solve for $t$ . $\newline$ $21200 = 13000 \times (1 + 0.02)^t$
Isolate growth factor: Divide both sides by the initial population to isolate the growth factor on one side. $\newline$ $\frac{21200}{13000} = (1 + 0.02)^t$ $\newline$ $1.63076923077 \approx (1.02)^t$
Take natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(1.63076923077) = \ln((1.02)^t)$ $\ln(1.63076923077) = t \cdot \ln(1.02)$
Divide by $\ln(1.02)$ : Divide both sides by $\ln(1.02)$ to solve for $t$ .
$t = \frac{\ln(1.63076923077)}{\ln(1.02)}$
$t \approx \frac{\ln(1.63076923077)}{\ln(1.02)}$
Calculate t value: Calculate the value of t using a calculator. $\newline$ $t \approx \frac{\ln(1.63076923077)}{\ln(1.02)}$ $\newline$ $t \approx 22.0240937876$
Round to nearest tenth: Round the answer to the nearest tenth of a year. $\newline$ $t \approx 22.0$ years

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