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A town has a population of 9000 and grows at
5% every year. To the nearest year, how long will it be until the population will reach 18900 ?
Answer:

A town has a population of $9000$ and grows at $5 \%$ every year. To the nearest year, how long will it be until the population will reach $18900$ ? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A town has a population of

9000

and grows at

5 \%

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

18900

?

\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 key values: Identify the initial population $P_0$ , the growth rate $r$ , and the final population $P$ . $P_0 = 9000$ $r = 5\%$ or $0.05$ (as a decimal) $P = 18900$
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 $t$ .
Rearrange formula for t: Rearrange the formula to solve for t. $\newline$ $t = \frac{\log(\frac{P}{P_0})}{\log(1 + r)}$ $\newline$ Substitute the known values into the equation. $\newline$ $t = \frac{\log(\frac{18900}{9000})}{\log(1 + 0.05)}$
Substitute values into equation: Calculate the values. $\newline$ $t = \frac{\log(\frac{18900}{9000})}{\log(1.05)}$ $\newline$ $t = \frac{\log(2.1)}{\log(1.05)}$
Calculate values: Use a calculator to find the values of $\log(2.1)$ and $\log(1.05)$ .
$t \approx \frac{\log(2.1)}{\log(1.05)}$
$t \approx \frac{0.3222}{0.0212}$
$t \approx 15.1981$
Use calculator for logs: Round the result to the nearest year. $\newline$ $t \approx 15$ years

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