Math / Algebra

QuestionTwo towns experience changes in population. Equations modelling the population of each town, where $P$ is population and $t$ is the number of years after January 1, 2021, are shown below. $\begin{array}{c} \text { Town A } \rightarrow P_{A}=7000(0.93)^{t} \\ \text { Town B } \rightarrow P_{B}=2500(1.12)^{t} \end{array}$
The number of years, to the nearest tenth, that it will take for the population of the two towns to be the same. $t=\square \text { years }$

Studdy Solution

STEP 1

1. The population of Town A is decreasing exponentially, while the population of Town B is increasing exponentially.
2. We need to find the time $t$ when the populations of both towns are equal.

STEP 2

1. Set the population equations equal to each other.
2. Solve the resulting equation for $t$ .
3. Round the solution to the nearest tenth.

STEP 3

Set the population equations of the two towns equal to each other:
$7000(0.93)^t = 2500(1.12)^t$

STEP 4

To solve for $t$ , we need to isolate $t$ . Start by dividing both sides by $2500$ :
$\frac{7000}{2500}(0.93)^t = (1.12)^t$
Simplify the fraction:
$2.8(0.93)^t = (1.12)^t$

STEP 5

Take the natural logarithm of both sides to bring down the exponents:
$\ln(2.8) + t \ln(0.93) = t \ln(1.12)$
Rearrange the equation to solve for $t$ :
$\ln(2.8) = t \ln(1.12) - t \ln(0.93)$
Factor out $t$ from the right side:
$\ln(2.8) = t (\ln(1.12) - \ln(0.93))$
Solve for $t$ :
$t = \frac{\ln(2.8)}{\ln(1.12) - \ln(0.93)}$

STEP 6

Calculate the value of $t$ using a calculator and round to the nearest tenth:
$t \approx \frac{\ln(2.8)}{\ln(1.12) - \ln(0.93)} \approx 14.3$
The number of years it will take for the populations to be the same is:
$\boxed{14.3} \text{ years}$

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Studdy Solution

STEP 1

1. The population of Town A is decreasing exponentially, while the population of Town B is increasing exponentially.
2. We need to find the time $t$ when the populations of both towns are equal.

STEP 2

1. Set the population equations equal to each other.
2. Solve the resulting equation for $t$ .
3. Round the solution to the nearest tenth.

STEP 3

Set the population equations of the two towns equal to each other:
$7000(0.93)^t = 2500(1.12)^t$

STEP 4

To solve for $t$ , we need to isolate $t$ . Start by dividing both sides by $2500$ :
$\frac{7000}{2500}(0.93)^t = (1.12)^t$
Simplify the fraction:
$2.8(0.93)^t = (1.12)^t$

STEP 5

STEP 6

Calculate the value of $t$ using a calculator and round to the nearest tenth:
$t \approx \frac{\ln(2.8)}{\ln(1.12) - \ln(0.93)} \approx 14.3$
The number of years it will take for the populations to be the same is:
$\boxed{14.3} \text{ years}$

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Studdy Solution

STEP 1

1. The population of Town A is decreasing exponentially, while the population of Town B is increasing exponentially.2. We need to find the time t t t when the populations of both towns are equal.

STEP 2

1. Set the population equations equal to each other.2. Solve the resulting equation for t t t.3. Round the solution to the nearest tenth.

STEP 3

Set the population equations of the two towns equal to each other:7000(0.93)t=2500(1.12)t 7000(0.93)^t = 2500(1.12)^t 7000(0.93)t=2500(1.12)t

STEP 4

To solve for t t t, we need to isolate t t t. Start by dividing both sides by 2500 2500 2500:70002500(0.93)t=(1.12)t \frac{7000}{2500}(0.93)^t = (1.12)^t 25007000​(0.93)t=(1.12)tSimplify the fraction:2.8(0.93)t=(1.12)t 2.8(0.93)^t = (1.12)^t 2.8(0.93)t=(1.12)t

STEP 5

STEP 6

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The population of Town A is decreasing exponentially, while the population of Town B is increasing exponentially.
2. We need to find the time $t$ when the populations of both towns are equal.

1. Set the population equations equal to each other.
2. Solve the resulting equation for $t$ .
3. Round the solution to the nearest tenth.

Set the population equations of the two towns equal to each other:
$7000(0.93)^t = 2500(1.12)^t$

To solve for $t$ , we need to isolate $t$ . Start by dividing both sides by $2500$ :
$\frac{7000}{2500}(0.93)^t = (1.12)^t$
Simplify the fraction:
$2.8(0.93)^t = (1.12)^t$