Problem 115 - Precalculus

Question

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth. \(f(x)=a b^{x}+d .\) $$116=\frac{1}{4}\left(\frac{1}{8}\right)^{x}$$

Step-by-Step Solution

Answer

The solution is approximately \(x \approx -1.573\).

Step 1: Rearrange the Given Equation

The equation to solve is \(116=\frac{1}{4}\left(\frac{1}{8}\right)^{x}\). Rewriting this gives the function \(f(x) = \frac{1}{4}\left(\frac{1}{8}\right)^{x}\). You need to find the value of \(x\) for which \(f(x) = 116\).

Step 2: Set Up the Equation for Graphing

To simplify graphing, express the equation as \(f(x) = \frac{1}{4}(0.125)^x\). You will compare this function to \(f(x) = 116\) by setting them equal, as this is how you find where the graph intersects the line \(y = 116\).

Step 3: Graph the Functions

Using a graphing calculator, enter the function \(y_1 = \frac{1}{4}(0.125)^x\) and the line \(y_2 = 116\). This will show the curve of the function and a horizontal line intersecting it at the solution.

Step 4: Identify the Intersection Point

Locate the intersection point of \(y_1\) and \(y_2\) on the graphing calculator. You can use the 'Intersect' feature to find the \(x\) value where these two graphs meet.

Step 5: Approximate the Solution

The graphing calculator should provide a numerical approximation for \(x\). Round this solution to the nearest thousandth as the problem specifies.

Key Concepts

Exponential Equation

An exponential equation involves an expression with a variable in the exponent. In this exercise, the equation is given by \(116=\frac{1}{4}(\frac{1}{8})^{x}\). Let's break it down to understand it better. An exponential equation generally takes the form \(ab^x + d\), where:

• \(a\) is a constant.
• \(b\) is the base of the exponent.
• \(x\) is the variable and appears as the exponent.
• \(d\) is an additional constant term, but in our exercise \(d\) is zero.

In this context, the exponential part \((\frac{1}{8})^{x}\) shows that small changes in \(x\) can lead to rapid changes in the outcome due to the power function. Exponential equations are used to model growth or decay processes, such as population growth or radioactive decay. These equations are particularly interesting because they describe processes that are not just linear but have a rate of change that also changes.

Intersection Point

An intersection point in the context of this exercise is the \(x\) value where two graphs meet on a graphing calculator. We have two main functions here:

• The function \(y_1 = \frac{1}{4}(0.125)^x\) – this is derived from our original exponential equation.
• The line \(y_2 = 116\) – a horizontal line representing the constant value in the original equation.

When these two graphs intersect, the point at which they meet corresponds to the solution of the equation \(116=\frac{1}{4}(\frac{1}{8})^{x}\). At this point of intersection, the value of \(x\) satisfies both expressions, making it a critical component in solving the equation. Using the graphing calculator's "Intersect" feature allows us to visually identify this point and verify the solution.

Numerical Approximation

Numerical approximation is the method used to estimate the value of a solution when it's difficult to obtain an exact number analytically. In this scenario, once the intersection point is identified on the graph, the graphing calculator helps find a close approximation of the \(x\) value.Due to the nature of real-world problems and the limitations of calculators, the solutions obtained are often rounded to maintain precision within practical limits. In this exercise, the problem specifies rounding the solution to the nearest thousandth. This means that, while the value may not be exact, it provides a sufficiently accurate approximation useful for most practical purposes.Numerical approximations are powerful because they allow us to work with complex equations that might not have simple algebraic solutions, making them a vital tool in science and engineering.