Math  /  Algebra

QuestionTwo towns experience changes in population. Equations modelling the population of each town, where PP is population and tt is the number of years after January 1,2021 , are shown below.  Town A PA=7000(0.93)t Town B PB=2500(1.12)t\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=t= \square years

Studdy Solution

STEP 1

1. The populations of Town A and Town B are modeled by the given exponential equations.
2. We need to find the time 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 .
3. Round the solution to the nearest tenth.

STEP 3

Set the population equations for Town A and Town B equal to each other:
7000(0.93)t=2500(1.12)t 7000(0.93)^t = 2500(1.12)^t

STEP 4

To solve for t t , first divide both sides by 2500 to simplify the equation:
70002500(0.93)t=(1.12)t \frac{7000}{2500}(0.93)^t = (1.12)^t
Simplify the fraction:
2.8(0.93)t=(1.12)t 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)+tln(0.93)=tln(1.12) \ln(2.8) + t\ln(0.93) = t\ln(1.12)

STEP 6

Rearrange the equation to solve for t t :
ln(2.8)=tln(1.12)tln(0.93) \ln(2.8) = t\ln(1.12) - t\ln(0.93)
Factor out t t :
ln(2.8)=t(ln(1.12)ln(0.93)) \ln(2.8) = t(\ln(1.12) - \ln(0.93))
Solve for t t :
t=ln(2.8)ln(1.12)ln(0.93) t = \frac{\ln(2.8)}{\ln(1.12) - \ln(0.93)}

STEP 7

Calculate the value of t t using a calculator and round to the nearest tenth:
tln(2.8)ln(1.12)ln(0.93)12.3 t \approx \frac{\ln(2.8)}{\ln(1.12) - \ln(0.93)} \approx 12.3
The number of years it will take for the populations to be the same is approximately:
12.3 \boxed{12.3}

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