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After the closing of the mill, the town of Sawyerville experienced a decline in its population. The relationship between the elapsed time, t, in years, since the closing of the mill, and the town's population, P(t), is modeled by the following function. P(t)=12,000*2^(-(t)/( 15)) In how many years will Sawyerville's population be 9000 ? Round your answer, if necessary, to the nearest hundredth. years

After the closing of the mill, the town of Sawyerville experienced a decline in its population.\newlineThe relationship between the elapsed time, t t , in years, since the closing of the mill, and the town's population, P(t) P(t) , is modeled by the following function.\newlineP(t)=12,000⋅2−t15 P(t)=12,000 \cdot 2^{-\frac{t}{15}} \newlineIn how many years will Sawyerville's population be 90009000 ? Round your answer, if necessary, to the nearest hundredth.\newlineyears

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Q. After the closing of the mill, the town of Sawyerville experienced a decline in its population.\newlineThe relationship between the elapsed time, t t , in years, since the closing of the mill, and the town's population, P(t) P(t) , is modeled by the following function.\newlineP(t)=12,000⋅2−t15 P(t)=12,000 \cdot 2^{-\frac{t}{15}} \newlineIn how many years will Sawyerville's population be 90009000 ? Round your answer, if necessary, to the nearest hundredth.\newlineyears
  1. Write population function and target: Write down the given population function and the target population.\newlineThe population function is P(t)=12,000×2−(t15)P(t) = 12,000 \times 2^{-(\frac{t}{15})}, and we want to find the time tt when the population P(t)P(t) is 90009000.
  2. Solve for t: Set the population function equal to 90009000 and solve for t.\newline90009000 = 1212,000000 \times 22^{-\left(\frac{t}{1515}\right)}
  3. Isolate exponential term: Divide both sides of the equation by 12,00012,000 to isolate the exponential term.\newline9,00012,000=2−(t15)\frac{9,000}{12,000} = 2^{-\left(\frac{t}{15}\right)}
  4. Simplify left side: Simplify the left side of the equation. 900012,000=34\frac{9000}{12,000} = \frac{3}{4}
  5. Convert fraction to decimal: Convert the fraction to a decimal to make it easier to work with. 34=0.75\frac{3}{4} = 0.75
  6. Take natural logarithm: Now we have the equation 0.75=2−(t15)0.75 = 2^{-\left(\frac{t}{15}\right)}. Take the natural logarithm (ln) of both sides to remove the exponent.\newlineln⁡(0.75)=ln⁡(2−(t15))\ln(0.75) = \ln\left(2^{-\left(\frac{t}{15}\right)}\right)
  7. Simplify using logarithm property: Use the property of logarithms that ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b \cdot \ln(a) to simplify the right side of the equation.\newlineln⁡(0.75)=−t15⋅ln⁡(2)\ln(0.75) = -\frac{t}{15} \cdot \ln(2)
  8. Divide by −ln⁡(2)-\ln(2): Divide both sides by −ln⁡(2)-\ln(2) to solve for tt.t=15⋅ln⁡(0.75)−ln⁡(2)t = \frac{15 \cdot \ln(0.75)}{-\ln(2)}
  9. Calculate t using calculator: Calculate the value of t using a calculator.\newlinet≈15×ln⁡(0.75)/−ln⁡(2)t \approx 15 \times \ln(0.75) / -\ln(2)\newlinet≈15×(−0.28768)/−0.693147t \approx 15 \times (-0.28768) / -0.693147\newlinet≈15×0.28768/0.693147t \approx 15 \times 0.28768 / 0.693147\newlinet≈4.3452/0.693147t \approx 4.3452 / 0.693147\newlinet≈6.271t \approx 6.271
  10. Round to nearest hundredth: Round the answer to the nearest hundredth. t≈6.27t \approx 6.27 years

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