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A town has a population of 7000 and grows at
3.5% every year. To the nearest tenth of a year, how long will it be until the population will reach 9400 ?
Answer:

A town has a population of $7000$ and grows at $3.5 \%$ every year. To the nearest tenth of a year, how long will it be until the population will reach $9400$ ? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A town has a population of

7000

and grows at

3.5 \%

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

9400

?

\newline

Answer:

Determine growth type: Determine the type of growth. The town's population grows at a constant percentage rate each year, which indicates exponential growth.
Identify key values: Identify the initial population ( $P_0$ ), the growth rate ( $r$ ), and the final population ( $P$ ). $\newline$ $P_0 = 7000$ , $r = 3.5\%$ , and $P = 9400$ .
Convert rate to decimal: Convert the growth rate from a percentage to a decimal. $r = 3.5\% = 0.035$ .
Set up growth formula: Set up the exponential growth formula. The formula for exponential growth is $P = P_0 \times (1 + r)^t$ , where $P$ is the final population, $P_0$ is the initial population, $r$ is the growth rate, and $t$ is the time in years.
Substitute values and solve: Substitute the known values into the formula and solve for $t$ . $9400 = 7000 \times (1 + 0.035)^t$ .
Isolate exponential part: Divide both sides by $7000$ to isolate the exponential part of the equation. $\newline$ $\frac{9400}{7000} = (1 + 0.035)^t$ .
Calculate left side: Calculate the left side of the equation. $9400 / 7000 \approx 1.3429$ .
Take natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(1.3429) = \ln((1 + 0.035)^t)$ .
Use logarithm property: Use the property of logarithms that $\ln(a^b) = b \cdot \ln(a)$ . $\ln(1.3429) = t \cdot \ln(1.035)$ .
Solve for t: Divide both sides by $\ln(1.035)$ to solve for t. $\newline$ $t = \frac{\ln(1.3429)}{\ln(1.035)}$ .
Calculate final value: Calculate the value of $t$ using a calculator. $t \approx \frac{\ln(1.3429)}{\ln(1.035)} \approx 8.8$ .

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