Problem 50 - College Algebra

Question

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth. \(f(x)=a b^{x}+d\). $$ -30=-4(2)^{x+2}+2 $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 1\), verified by calculation and graphing.

Step 1: Simplify the Equation

We begin by isolating terms. Start with the equation \(-30 = -4(2)^{x+2} + 2\).Subtract 2 from both sides:\[-30 - 2 = -4(2)^{x+2}\] which simplifies to \[-32 = -4(2)^{x+2}\].

Step 2: Divide by -4

Divide both sides by -4 to further isolate the exponential term:\[-32 \div -4 = (2)^{x+2}\] which simplifies to \[8 = (2)^{x+2}\].

Step 3: Solve for x in the Equation

Now solve the equation \(8 = 2^{x+2}\).Write 8 as a power of 2: \(8 = 2^3\), so\(2^3 = 2^{x+2}\).Equate the exponents: \(3 = x+2\).Subtract 2 from both sides:\(x = 1\).

Step 4: Verify Using a Graphing Calculator

Use a graphing calculator to plot the function \(y = -4(2)^{x+2} + 2\).Find where the graph intersects the line \(y = -30\).Confirm that the intersection occurs when \(x \approx 1\). Since we already found \(x = 1\) analytically, we expect this value, but check for precision.

Key Concepts

Exponential Equations

Exponential equations are equations where a variable appears in the exponent. This makes them quite different and often more challenging than other types of equations. In the exercise, we have an equation of the form \( f(x) = a \cdot b^{x} + d \), which is a typical representation of an exponential equation. Here, the variable \( x \) is in the exponent, making the function grow rapidly for large values of \( x \) if \( b > 1 \). The general form of an exponential equation can show

• Exponential growth if \( b > 1 \)
• Exponential decay if \( 0 < b < 1 \)

In our specific case, the equation is \(-30 = -4(2)^{x+2} + 2\). This involves exponential growth since the base 2 is greater than 1. Here are some basic properties of exponential equations:

• The base (\( b \)) determines the rate of growth or decay. A larger base results in faster growth.
• The exponent (\( x \)) directly influences the outcome, allowing exponential equations to model real-world phenomena like population growth and radioactive decay.

Understanding these properties helps you analyze the behavior of the equation as \( x \) changes. For example, when \( x \) increases, the value of \( f(x) \) grows rapidly, especially when the base is more than 1.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. In our example, the goal is to find \( x \) for the equation \(-30 = -4(2)^{x+2} + 2\). Solving such equations often requires a careful step-by-step approach as outlined. Here are the simplified steps to solve:

• Isolate the exponential term: Begin by moving the constant terms to one side, achieving \(-32 = -4(2)^{x+2}\).
• Divide and simplify: Divide both sides by -4 to get \( 8 = (2)^{x+2} \). This step simplifies the equation, making it easier to proceed.
• Use logarithms or match bases: Recognize that 8 can be expressed as a power of 2, specifically \( 2^3 \). Hence, you set the exponents equal to each other, giving the equation \( x + 2 = 3 \).

The final step is straightforward, subtract 2 from both sides resulting in \( x = 1 \). This method works well for equations where the exponential term can be easily transformed. When dealing with more complex bases or non-integer results, logarithms are often used to find approximate solutions.

Graphical Solution

Finding a graphical solution involves using technology to visualize the equation and find intersections that represent solutions. In this exercise, a graphing calculator is used, which is a powerful tool for approximating solutions, especially when equations are complex or when precision to several decimal places is required.Here is how you can use a graphing calculator for our example:

• First, graph the exponential function \( y = -4(2)^{x+2} + 2 \). This will give you the visual shape and behavior of the equation based on the variable \( x \).
• Then, plot the line \( y = -30 \) as a horizontal line.
• The solution to the equation is where these two graphs intersect. The x-coordinate of this intersection provides the approximate solution.

A graphing calculator not only allows you to find intersections but also to zoom in closely, verifying the precise value of \( x \). For instance, plotting these graphs confirms the analytical solution where the intersection is very close to \( x = 1 \). This method is crucial when you need an approximate solution that the analytical process alone cannot deliver, especially for complicated equations where solving by hand is impractical.