Problem 99 - Introductory and Intermediate Algebra for College Students 4th

Question

The formula \(A=36.1 e^{0.0126 t}\) models the population of California, \(A,\) in millions, \(t\) years after 2005 a. What was the population of California in \(2005 ?\) b. When will the population of California reach 40 million?

Step-by-Step Solution

Answer

a. The population of California in 2005 was approximately 36.1 million. \n b. The population of California will reach 40 million around year 2012.

Step 1: Calculating the population in 2005

For the year 2005, \(t=0\). So, replace \(t\) in the given formula \(A=36.1 e^{0.0126 t}\) with \(0\) to find the initial population:\n \(A=36.1 e^{0.0126*0} = 36.1 e^{0}= 36.1 millions.\)

Step 2: Determining when the population will reach 40 million

To find when the population will reach 40 million, set \(A\) equal to 40 in the given function: \(40=36.1 e^{0.0126 t}\). We need to solve this equation for \(t\). Firstly, let's isolate \(e\) to the power of \(t\) by dividing both sides by 36.1: \(\frac{40}{36.1}=e^{0.0126t}\), which when simplified gives us: \(\approx 1.108= e^{0.0126t}\). Now, We know that the inverse of exponentiation is taking the logarithm. So, take a natural logarithm on both sides of the equation: \( ln(1.108) = ln(e^{0.0126t})\). Using the logarithm identity \(ln(e^x) = x\), the formula simplifies to: \(ln(1.108) = 0.0126t\). Solve this equation for \(t\) to get the result, \(t = \frac{ln(1.108)}{0.0126}\) which gives approx \(t≈6.95\).

Step 3: Interpret the result

The result is in years after 2005. When we add this to 2005, we find that the population of California will reach 40 million around year 2012.

Key Concepts

Exponential Growth

Exponential growth is a process that increases quantity over time. It is characterized by the presence of a constant proportional growth rate, meaning the quantity grows exponentially, rather than linearly. The formula given in the exercise, A=36.1 e^{0.0126 t}, is a classic example of an exponential growth model, where A represents the population of California in millions, and t represents time in years. This model indicates that the population grows by a factor of e^{0.0126} every year. With exponential growth, small changes in the growth rate can lead to huge differences in outcomes over time.

Understanding this model is crucial for planning resources and infrastructure to meet future demands. When plotted on a graph, exponential growth produces a J-shaped curve that gets steeper over time. This is important to note because it helps us understand how populations can quickly grow to large sizes, highlighting the potential for overpopulation or resource depletion if growth is unchecked.

Natural Logarithm

The natural logarithm, often represented as ln, is the inverse operation of raising e to a power x, where e is the base of the natural logarithm and approximately equal to 2.71828. The relationship between a logarithm and exponentiation is central in solving exponential equations.

In the exercise, the natural logarithm makes it possible to solve for t when the population, A, reaches 40 million. The calculation ln(1.108) reverses the exponential function, reducing the equation to a linear form that can be solved for t. It is this quality that allows us to work with exponential growth models algebraically, transforming the multiplicative rates of change into additive rates, which are much easier to manipulate and understand.

Solving Exponential Equations

Solving exponential equations is a key skill in algebra, particularly when dealing with growth models. An exponential equation typically involves a variable in the exponent, as seen in the population growth model with e^{0.0126 t}. To solve these types of equations, one common method is to isolate the exponential expression and then apply the natural logarithm to both sides of the equation, as logarithms can convert the exponentiation of variables into multiplication, enabling us to solve for the unknown.

In the step-by-step solution of the exercise, after isolating e^{0.0126 t}, we use the property that ln(e^x) = x to deduce that 0.0126 t is equal to ln(1.108). From this point, the variable t can be easily isolated to find the number of years it will take for the population to reach a certain threshold. This approach is critical in various fields that use mathematical modeling, from ecological studies to financial forecasting.