Problem 62 - A Graphical Approach to College Algebra

Question

Solve each equation in part (a) analytically. Support your answer with a calculator graph. Then use the graph to solve the associated inequalities in parts (b) and (c). (a) \(\left(\frac{2}{3}\right)^{x-1}=\left(\frac{81}{16}\right)^{x+1}\) (b) \(\left(\frac{2}{3}\right)^{x-1} \leq\left(\frac{81}{16}\right)^{x+1}\) (c) \(\left(\frac{2}{3}\right)^{x-1} \geq\left(\frac{81}{16}\right)^{x+1}\)

Step-by-Step Solution

Answer

(a) Solution: \(x = \frac{3}{5}\); (b) \(x \geq \frac{3}{5}\); (c) \(x \leq \frac{3}{5}\).

Step 1: Set Up the Equation

We are given the equation \( \left(\frac{2}{3}\right)^{x-1} = \left(\frac{81}{16}\right)^{x+1} \). To solve this analytically, we first equate the exponents, assuming both sides can be written with the same base.

Step 2: Convert Bases

Notice \( \frac{81}{16} = \left(\frac{3}{2}\right)^4 \). Therefore, we can rewrite the right-hand side as \( \left(\left(\frac{3}{2}\right)^4\right)^{x+1} = \left(\frac{3}{2}\right)^{4(x+1)} \). Now take the logarithm on both sides using base \(\frac{2}{3}\).

Step 3: Solve for x

Setting the exponents equal gives us: \( x-1 = 4(x+1) \log_{\left(\frac{2}{3}\right)}\left(\frac{3}{2}\right) \).Solve for \( x \):\[ x - 1 = 4(x+1)\cdot (-1)\] which simplifies to \[ x - 1 = -4x - 4 \] Rearranging gives:\[ x + 4x = 4 - 1 \]\[ 5x = 3 \]\[ x = \frac{3}{5} \].

Step 4: Verify Solution with a Graph

Graph the equations \( y_1 = \left(\frac{2}{3}\right)^{x-1} \) and \( y_2 = \left(\frac{81}{16}\right)^{x+1} \) on a calculator. Verify that they intersect at \( x = \frac{3}{5} \).

Step 5: Solve Inequality (b)

The inequality \( \left(\frac{2}{3}\right)^{x-1} \leq \left(\frac{81}{16}\right)^{x+1} \) implies that the graph of \( y_1 \) is below or on \( y_2 \). Include the solution to the equation for equality, so: For this inequality, \( x \geq \frac{3}{5} \).

Step 6: Solve Inequality (c)

For the inequality \( \left(\frac{2}{3}\right)^{x-1} \geq \left(\frac{81}{16}\right)^{x+1} \), the graph of \( y_1 \) should be above or on \( y_2 \). Hence, the solution is \( x \leq \frac{3}{5} \).

Key Concepts

Inequality Solutions

Inequality solutions involve finding the ranges or values for which the inequality holds true. In our exercise, we have two inequalities:

• \(\left(\frac{2}{3}\right)^{x-1} \leq \left(\frac{81}{16}\right)^{x+1}\)
• \(\left(\frac{2}{3}\right)^{x-1} \geq \left(\frac{81}{16}\right)^{x+1}\)

To solve these, first determine where the equality \(\left(\frac{2}{3}\right)^{x-1} = \left(\frac{81}{16}\right)^{x+1}\) occurs. By solving, we found \(x = \frac{3}{5}\). The solution \(x = \frac{3}{5}\) represents the only intersection point where both expressions are equal.

For inequality (b), \(\left(\frac{2}{3}\right)^{x-1} \leq \left(\frac{81}{16}\right)^{x+1}\), this means finding where the first expression is less than or equals to the second. Graphically, it represents the points where the first curve is below or exactly on the second curve, giving us the solution \(x \geq \frac{3}{5}\).Similarly, for inequality (c), \(\left(\frac{2}{3}\right)^{x-1} \geq \left(\frac{81}{16}\right)^{x+1}\), it implies finding \(x\) where the first expression overlaps or rises above the second curve.

Therefore, the solution is \(x \leq \frac{3}{5}\). Inequalities are crucial in understanding range and limitations in mathematical models.

Graphical Verification

Graphical verification is a powerful strategy used to confirm analytical solutions. In this exercise, we graph the two functions:

• \(y_1 = \left(\frac{2}{3}\right)^{x-1}\)
• \(y_2 = \left(\frac{81}{16}\right)^{x+1}\)

Using a calculator or graphing software, plot these functions to observe their interaction.

The goal is to verify the point of intersection, previously found to be \(x = \frac{3}{5}\), which confirms the solution for the original equation \( \left(\frac{2}{3}\right)^{x-1} = \left(\frac{81}{16}\right)^{x+1}\).
This step visualizes where the two expressions are equal, and inspecting beyond this point can aid in determining where inequalities become effective.For inequality solutions, you identify ranges on the graph:

• At \(x = \frac{3}{5}\), the graphs intersect.
• On the domain \(x > \frac{3}{5}\), \(y_1\) falls below \(y_2\).
• On the domain \(x < \frac{3}{5}\), \(y_1\) remains above \(y_2\).

By examining these segments visually, it confirms and complements analytical findings, enhancing understanding of the function's behavior over different ranges.

Base Conversion

Base conversion is pivotal in making equations solvable by using the common base for comparison. Originally, our equation was \( \left(\frac{2}{3}\right)^{x-1} = \left(\frac{81}{16}\right)^{x+1} \). To facilitate solving this, the right side could be expressed with the base \(\frac{3}{2}\) since \(\frac{81}{16} = \left(\frac{3}{2}\right)^4\).

This conversion is crucial because logarithmic solutions to equations require expressions to have the same base. Once in the form \( \left(\frac{2}{3}\right)^{x-1} = \left(\frac{3}{2}\right)^{4(x+1)} \), logarithms help transition this equality into an easy-to-solve algebraic form:- Apply logarithms to both sides, converting exponents to productsThis simplifies comparisons between the two expressions as it aligns them through equivalent bases.

Base conversion elucidates the forms of exponential equations, simplifying their handling and graphical interpretation, and allows for straightforward algebraic manipulations leading to solutions.