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The population of a town, in thousands, is modeled by the equation p(t)=50*(1.06)^(t) where t is time measured in years. About how many years will it take for the population to double? Round to the nearest year. Explain how you know

The population of a town, in thousands, is modeled by the equation p(t)=50⋅(1.06)t p(t)=50 \cdot (1.06)^{t} where t t is time measured in years.\newlineAbout how many years will it take for the population to double? Round to the nearest year. Explain how you know

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Q. The population of a town, in thousands, is modeled by the equation p(t)=50⋅(1.06)t p(t)=50 \cdot (1.06)^{t} where t t is time measured in years.\newlineAbout how many years will it take for the population to double? Round to the nearest year. Explain how you know
  1. Identify Population and Condition: Identify the initial population and the population doubling condition.\newlineThe initial population is given by p(0)=50p(0) = 50 (since 1.060=11.06^0 = 1). To double, the population needs to reach 100100 thousand.
  2. Set Up Equation for Doubling: Set up the equation to solve for the time it takes for the population to double.\newlineWe need to find tt such that p(t)=100p(t) = 100.\newlineSo, we set up the equation 50×(1.06)t=10050 \times (1.06)^t = 100.
  3. Simplify Equation for t: Simplify the equation to solve for t.\newlineDivide both sides of the equation by 5050 to isolate (1.06)t(1.06)^t on one side:\newline(1.06)t=10050(1.06)^t = \frac{100}{50}\newline(1.06)t=2(1.06)^t = 2
  4. Use Logarithms to Solve: Use logarithms to solve for tt.\newlineTake the natural logarithm (ln⁡)(\ln) of both sides to remove the exponent on the left side:\newlineln⁡((1.06)t)=ln⁡(2)\ln((1.06)^t) = \ln(2)\newlinet⋅ln⁡(1.06)=ln⁡(2)t \cdot \ln(1.06) = \ln(2)
  5. Isolate and Solve for t: Isolate tt and solve for its value.\newlineDivide both sides by ln⁡(1.06)\ln(1.06) to solve for tt:\newlinet=ln⁡(2)ln⁡(1.06)t = \frac{\ln(2)}{\ln(1.06)}
  6. Calculate t Value: Calculate the value of t using a calculator.\newlinet≈ln⁡(2)/ln⁡(1.06)t \approx \ln(2) / \ln(1.06)\newlinet≈0.69314718056/0.05826890812t \approx 0.69314718056 / 0.05826890812\newlinet≈11.895661236t \approx 11.895661236
  7. Round to Nearest Year: Round the answer to the nearest year.\newlineSince we are looking for an approximation to the nearest year, we round 11.89566123611.895661236 to 1212.

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