Question

Population Growth The population of Glenbrook is \(375,000\) and is increasing at the rate of 2.25\(\%\) per year. Predict when the population will be 1 million. In Exercises 49 and \(50,\) let \(x=0\) represent \(1990, x=1\) represent \(1991,\) and so forth.

Short answer

The population of Glenbrook is predicted to reach 1 million by the year 2025.

Step-by-step solution

Identify Initial Population, Growth Rate and Target Population

Initial Population (P₀) = 375,000, Growth Rate (r) = 2.25% or 0.0225, and Target Population (P) = 1,000,000.

Subtitute Variables into the Exponential Growth Formula

Substitute these values into the population growth formula P=P₀·(1+r)ᵗ, which then becomes 1,000,000 = 375,000 · (1 + 0.0225)ᵗ.

Solve for t

First, divide both sides of the equation by 375,000, which results in 2.67 = (1+0.0225)ᵗ. Then take the natural logarithm (ln) of both sides and apply the power rule of logarithms, which shifts the t in the equation, resulting in ln(2.67) = t · ln(1.0225). Finally, to isolate t, divide both sides of the equation by ln(1.0225). Now the equation is t = ln(2.67)/ln(1.0225).

Calculate t

Computing the right side of the equation using a calculator gives t ≈ 34.88. This value must be rounded up to the nearest whole number, as it refers to years, so t = 35.

Determine the Year

Since the problem deems that x=0 represents the year 1990, t represents the number of years from 1990. Thus, the year is 1990 + t = 1990 + 35 = 2025.