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A fish population is modeled by P(t) = 1200 * (1.15)^t, where t is the number of years since 2020. a) Predict the population in 2025. b) Show this is equivalent to the log model log P = log 1200 + t l

Mathematics

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2 parts
a

2414 fish\text{2414 fish}

c

10.21 years\text{10.21 years}

Graph of P(t) = 1200·1.15^t

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Key points of P(t) = 1200·1.15^t
Pointty
y-intercept01200

Step-by-step solution

9 steps

Part a· 2 steps

To calculate the predicted fish population in the year 2025 using the index model, we first determine the number of years tt from the initial year 2020.

  1. Calculate the number of years tt.

    The initial year is 2020, and the target year is 2025.

    t=2025−2020=5 yearst = 2025 - 2020 = 5 \text{ years}
  2. Substitute t=5t=5 into the index model P(t)=1200×(1.15)tP(t) = 1200 \times (1.15)^t.

    P(5)=1200×(1.15)5P(5) = 1200 \times (1.15)^5 P(5)=1200×2.0113571875P(5) = 1200 \times 2.0113571875 P(5)=2413.628625P(5) = 2413.628625

    Rounding to the nearest whole fish, the predicted population is 2414 fish.

    2414 fish\boxed{\text{2414 fish}}

Part b· 3 steps

To show that the logarithmic model is equivalent to the index model, we start with the index model and apply the logarithm.

  1. Start with the index model.

    P=1200×(1.15)tP = 1200 \times (1.15)^t
  2. Apply log⁡10\log_{10} to both sides of the equation.

    log⁡10P=log⁡10(1200×(1.15)t)\log_{10} P = \log_{10} (1200 \times (1.15)^t)
  3. Use the logarithm properties log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log B and log⁡(AB)=Blog⁡A\log(A^B) = B \log A.

    log⁡10P=log⁡101200+log⁡10(1.15)t\log_{10} P = \log_{10} 1200 + \log_{10} (1.15)^t log⁡10P=log⁡101200+tlog⁡101.15\log_{10} P = \log_{10} 1200 + t \log_{10} 1.15

    This matches the given logarithmic model, thus showing their equivalence.

Part c

10.21 years\text{10.21 years}

To determine the approximate number of years (tt) it will take for the fish population to reach 5,000, we use the index model and solve for tt using logarithms.

  1. Set P(t)=5000P(t) = 5000 in the index model.

    5000=1200×(1.15)t5000 = 1200 \times (1.15)^t
  2. Isolate the exponential term.

    50001200=(1.15)t\frac{5000}{1200} = (1.15)^t 5012=(1.15)t\frac{50}{12} = (1.15)^t 256=(1.15)t\frac{25}{6} = (1.15)^t
  3. Take the logarithm (base 10 or natural logarithm) of both sides.

    log⁡10(256)=log⁡10((1.15)t)\log_{10}\left(\frac{25}{6}\right) = \log_{10}((1.15)^t)
  4. Use the logarithm property log⁡(AB)=Blog⁡A\log(A^B) = B \log A to solve for tt.

    log⁡10(256)=tlog⁡10(1.15)\log_{10}\left(\frac{25}{6}\right) = t \log_{10}(1.15) t=log⁡10(256)log⁡10(1.15)t = \frac{\log_{10}\left(\frac{25}{6}\right)}{\log_{10}(1.15)} t≈0.6197890.060698t \approx \frac{0.619789}{0.060698} t≈10.21t \approx 10.21

    The approximate number of years is 10.21 years.

    10.21 years\boxed{\text{10.21 years}}
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