Problem 147 - Algebra and Trigonometry Real Mathematics, Real People

Question

The percent \(p\) (in decimal form) of the United States population who own a smartphone is given by $$p=\frac{1}{1+e^{-(t-93) / 22.5}}$$ where \(t\) is the number of months after smartphones were available on the market. Find the number of months \(t\) when the percent of the population owning smartphones is (a) \(50 \%\) and (b) \(80 \%\).

Step-by-Step Solution

Answer

(a) 93 months and (b) 171.35 months

Step 1: Express the Percent as a Decimal

The first thing we do here is express the percentage as a decimal: \(50\% = 0.5\) and \(80\% = 0.8\). So we set \(p = 0.5\) in (a) and \(p = 0.8\) in (b) and continue.

Step 2: Procedure for solving for \(t\) when \(p = 0.5\) and \(p = 0.8\)

We start by substituting \(p = 0.5\) in the equation, and then \(p = 0.8\). Next, we solve for \(t\) by isolating \(t\). Doing this requires inverse functions, so we need to use the logarithmic properties of the constant \(e\) to simplify the exponential.

Step 3: Computations for \(p = 0.5\)

We rewrite \(0.5 = \frac{1}{1+e^{-(t-93) / 22.5}}\) as \(2 = 1+e^{-(t-93) / 22.5}\), which simplifies further to \(1 = e^{-(t-93) / 22.5}\). By taking the natural logarithm on both sides, we get: \(-\frac{(t-93)}{22.5} = \ln(1)\). Simplifying gives that \(t = 93\) months.

Step 4: Computations for \(p = 0.8\)

We rewrite \(0.8 = \frac{1}{1+e^{-(t-93) / 22.5}}\) as \(1.25 = 1+e^{-(t-93) / 22.5}\), which simplifies further to \(0.25 = e^{-(t-93) / 22.5}\). By taking the natural logarithm on both sides, we get: \(-\frac{(t-93)}{22.5} = \ln(0.25)\). Simplifying gives that \(t \approx 171.35\) months.

Key Concepts

Natural Logarithm

The natural logarithm is a key concept in mathematics that often appears in problems involving exponential growth and decay. Its symbol is \( \ln \), and it is specifically the logarithm to the base \( e \), where \( e \approx 2.71828 \). Here, \( e \) is a mathematical constant that arises naturally in various calculations, such as in compound interest and population growth models.In our exercise, we simplify the exponential equation using natural logarithms. When we have an equation like \( e^x = y \), we can take the natural logarithm of both sides to solve for \( x \), yielding \( x = \ln(y) \). This property is essential because it allows us to transition from an exponential form, where our variable is in the exponent, to a linear form, which is much simpler to solve.

Solving Exponential Equations

Exponential equations can appear complex because the variable you're solving for is in the exponent. The general strategy involves isolating the exponential term before applying logarithmic functions to simplify.For instance, if we have an equation like the one from our problem: \( \frac{1}{1+e^{-(t-93) / 22.5}} = p \), the first step is to express the equation with the exponential isolated, such as bringing all non-exponential terms to one side.After rearranging terms, you get a simplified form such as \( e^{-(t-93) / 22.5} = x \), where \( x \) is a numerical value based on your specific \( p \). At this point, we use logarithms (natural logs, specifically) to solve for the variable in the exponent, thus reducing the problem to something more manageable.

Percentage Conversion

Understanding how to convert percentages to decimals and vice versa is a vital skill in various mathematical contexts, including problem-solving with exponential functions. Converting a percentage, say \( 50\% \) or \( 80\% \), into a decimal is straightforward: simply divide by 100.For example:

• \( 50\% \) becomes \( 0.5 \)
• \( 80\% \) becomes \( 0.8 \)

This conversion is critical because mathematical equations, especially those involving growth and decay, often require inputs as decimals rather than percentages. Working with decimals rather than percentages simplifies equations and calculations, making it easier to manipulate mathematical expressions.

Inverse Functions

Inverse functions are a fundamental concept in algebra and calculus, used extensively in solving equations where the variable is trapped inside a more complex function, like an exponential one.An inverse function essentially "undoes" another function. The classic example involves exponential and logarithmic functions, which are inverses. If you have \( e^x \), the inverse of this function is \( \ln(x) \). This relationship is powerful because it allows us to transition between exponential and logarithmic forms.In practice, when faced with an exponential equation in the form \( e^{f(t)} = g \), you'd apply the natural logarithm to both sides, using the property \( \ln(e^x) = x \), to isolate \( f(t) \). This is exactly what's done in our exercise to solve for the time \( t \) corresponding to specific population percentages having smartphones. By recognizing and using inverse functions, we simplify complex functions, making them solvable.