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Difficulty: MediumExponential Functions and Equations

The population of a species of fish in a lake can be modeled by the function P(t)=P0⋅2tdP(t) = P_0 \cdot 2^{\frac{t}{d}}, where P0P_0 is the initial population when the population was first measured, tt represents the time in years since it was first measured, and dd is a constant representing the doubling time in years. If the population of the fish doubles every 6 years, and the population after 18 years is 3,200, what was the initial population of the fish?

Answer: 400

Answer

The initial population of the fish was 400.
By substituting the doubling time d=6d = 6 and the final population of 3,200 at t=18t = 18 into the exponential model P(t)=P0⋅2tdP(t) = P_0 \cdot 2^{\frac{t}{d}}, we get 3,200=P0⋅21863,200 = P_0 \cdot 2^{\frac{18}{6}}. Simplifying the exponent gives 3,200=P0⋅233,200 = P_0 \cdot 2^3, which simplifies further to 3,200=8P03,200 = 8P_0. Dividing both sides by 8 yields P0=400P_0 = 400.

Step-by-Step Solution

1
Identify the values for the known variables from the word problem.
d=6d = 6 years and at t=18t = 18 years, P(18)=3,200P(18) = 3,200.
To substitute these values into the exponential growth function model.
2
Substitute the known values into the exponential function P(t)=P0⋅2tdP(t) = P_0 \cdot 2^{\frac{t}{d}}.
3,200=P0⋅21863,200 = P_0 \cdot 2^{\frac{18}{6}}
To set up an equation to solve for the unknown parameter P0P_0.
3
Simplify the exponent and calculate the growth factor.
3,200=P0⋅23⇒3,200=8P03,200 = P_0 \cdot 2^3 \Rightarrow 3,200 = 8P_0
Reducing the fractional exponent simplifies the equation.
4
Solve for the initial population P0P_0 by dividing both sides of the equation by 8.
P0=400P_0 = 400
Isolating P0P_0 gives the initial population of the fish.

Key Concept

Using an exponential function to model real-world growth and solving for the initial value.
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Topics

Exponential Functions and Equations

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