Problem 87 - College Algebra with Modeling and Visualization

Question

Population Growth The population \(P\) (in millions) of California \(x\) years after 2000 can be modeled by \(P=34 e^{0.013 x}\) A. Use properties of logarithms to solve this equation for \(x\) B. Use your equation to find \(x\) when \(P=38\). Interpret your answer.

Step-by-Step Solution

Answer

California's population reaches 38 million in approximately 2009.

Step 1: Set up the Equation

We are given the population model \( P = 34 e^{0.013 x} \). To solve for \( x \), we start by setting the equation equal to the given population value \( P \). This gives us \( 34 e^{0.013 x} = P \).

Step 2: Isolate the Exponential Expression

Divide both sides of the equation by 34 to isolate the exponential term:\[ e^{0.013 x} = \frac{P}{34} \]

Step 3: Apply the Natural Logarithm

Take the natural logarithm (\( \ln \)) of both sides to remove the exponential:\[ \ln(e^{0.013 x}) = \ln \left( \frac{P}{34} \right)\] Using the property \( \ln(e^y) = y \), this simplifies to:\[ 0.013 x = \ln \left( \frac{P}{34} \right)\]

Step 4: Solve for x

Divide both sides by 0.013 to solve for \( x \):\[ x = \frac{\ln \left( \frac{P}{34} \right)}{0.013} \]

Step 5: Substitute P = 38 and Calculate x

Substitute \( P = 38 \) into the equation:\[ x = \frac{\ln \left( \frac{38}{34} \right)}{0.013} \]Calculate the value of \( x \):\[ x \approx \frac{\ln(1.1176)}{0.013} \approx \frac{0.1116}{0.013} \approx 8.585\]

Step 6: Interpret the Result

The calculated value of \( x \) is approximately 8.59. This means that the population of California reaches 38 million about 8.59 years after 2000, which rounds to about the year 2009.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions used to model situations where a quantity grows or decays at a constant relative rate. In our context, we are examining the population growth of California represented by the function \( P = 34 e^{0.013x} \). In this equation:

• \( P \) represents the population in millions,
• \( 34 \) is the initial population in the year 2000,
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828,
• \( 0.013 \) is the growth rate expressed as a percentage.

The function \( e^{0.013x} \) indicates an exponential growth model. Over time, as \( x \) increases, the exponent causes the population to grow at a gradually accelerating pace. This is a characteristic of exponential functions, where growth becomes faster as there is more to build upon. Understanding exponential functions is crucial in real-life applications such as population growth, radioactive decay, and interest calculations in finance.

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are the inverse functions of exponential functions involving base \( e \). They are used to simplify equations where the variable appears in the exponent. In our problem, once we isolate the exponent \( 0.013x \) by dividing \( P \) by 34, we use the natural logarithm to "bring down" the exponent:

• \( \ln(e^{0.013x}) = \ln \left( \frac{P}{34} \right) \)

By the property of logarithms, \( \ln(e^y) = y \), which simplifies our equation to \( 0.013x = \ln \left( \frac{P}{34} \right) \). This step is critical because it transforms the exponential equation into a linear form, allowing us to easily solve for \( x \). The natural logarithm's base \( e \) aligns well with exponential growth models, making it a preferred choice for calculations involving exponential relationships.

Solving Equations

Solving equations involving exponential functions requires a step-by-step approach to isolate the variable of interest. Here's how we solve our problem:

• Start with the equation \( 34 e^{0.013x} = P \).
• Divide by 34 to isolate the exponential term: \( e^{0.013x} = \frac{P}{34} \).
• Apply the natural logarithm to both sides: \( \ln(e^{0.013x}) = \ln \left( \frac{P}{34} \right) \).
• Simplify using the logarithmic property: \( 0.013x = \ln \left( \frac{P}{34} \right) \).
• Solve for \( x \) by dividing by 0.013: \( x = \frac{\ln \left( \frac{P}{34} \right)}{0.013} \).

Finally, to find when the population reaches 38 million, substitute \( P = 38 \) and calculate \( x \). This results in \( x \approx 8.59 \), meaning around 8.59 years after 2000, the population will be 38 million, approximately the year 2009. Breaking down each step clarifies the process of moving from an exponential equation to finding a specific time frame for a projected population size.