Problem 351 - Precalculus

Question

Solve the equation for \(x,\) if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. \(\log _{9}(3-x)=\log _{9}(4 x-8)\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{11}{5}\).

Step 1: Set the arguments of the logarithms equal

Since both sides of the equation have the same base (9), we can equate the arguments of the logarithms. This gives us the equation: \[ 3 - x = 4x - 8 \]

Step 2: Solve the equation for x

Rearrange the equation from Step 1. Start by adding \(x\) to both sides:\[ 3 = 5x - 8 \]Then, add 8 to both sides to isolate the term with \(x\):\[ 11 = 5x \]Now, divide by 5 to solve for \(x\):\[ x = \frac{11}{5} \]

Step 3: Verify solution by substituting back

Substitute \(x = \frac{11}{5}\) back into the original expressions for both sides of the equation:Left side: \( \log_{9}(3 - \frac{11}{5}) = \log_{9}(\frac{4}{5})\)Right side: \(\log_{9}(4 \times \frac{11}{5} - 8) = \log_{9}(\frac{4}{5})\)Both sides are equal, confirming the solution is correct.

Step 4: Graph both sides to verify point of intersection

Graph the functions \( y = \log_{9}(3-x) \) and \( y = \log_{9}(4x-8) \) on the same set of axes. Observe where the graphs intersect.The point of intersection should occur at \( x = \frac{11}{5} \), verifying the solution visually.

Key Concepts

Solving Equations

The process of solving equations involves finding the value of the variable that makes the equation true. In this case, we have an equation that involves logarithms, specifically with base 9. The principle here is straightforward: if two logarithms with the same base are equal, then their arguments must be equal too.

• For example, given the equation \(\log _{9}(3-x)=\log _{9}(4 x-8)\), we can directly equate the arguments:
• This leads to the linear equation: \(3 - x = 4x - 8\).

Once we have this linear equation, the next step is to isolate \(x\). We do this by simplifying:

• Add \(x\) to both sides: \(3 = 5x - 8\).
• Add 8 to both sides: \(11 = 5x\).
• Divide by 5: \(x = \frac{11}{5}\).

This provides us with the solution \(x = \frac{11}{5}\). It's crucial to check this solution by plugging it back into the original equation to ensure both sides are equal. This confirms the accuracy of our solution.

Graphing Functions

Graphing is a powerful tool for visualizing the solutions of equations. In the context of this exercise, graphing helps to verify our solution by showing the point of intersection between two functions. Here, we graph the two logarithmic expressions:

• \(y = \log_{9}(3-x)\)
• \(y = \log_{9}(4x-8)\)

Plotting these on the same axes allows us to see where the two graphs intersect, which should happen at \(x = \frac{11}{5}\).To graph these functions:

• Identify the domain. For each function, the argument of the logarithm must be positive. Thus, \(3-x > 0\) and \(4x-8 > 0\).
• Use points within these domains to plot both sides.
• The intersection point confirms the solution obtained algebraically.

The intersection not only verifies our algebraic solution but also provides a visual understanding of why \(x = \frac{11}{5}\) is the point where both logarithmic functions are equal.

Logarithmic Functions

Understanding logarithmic functions is key when working with logarithmic equations. These functions are the inverse of exponential functions and are defined by bases, such as base 9 in our example. The logarithm of a number is the power to which the base must be raised to obtain that number.In the equation \(\log _{9}(3-x)=\log _{9}(4x-8)\), the base 9 logarithm means we compare the exponents that make the expressions inside the logarithms equal.

• If \(\log_b(A) = \log_b(B)\), then \(A = B\), assuming \(A > 0\) and \(B > 0\).
• Logarithms require their arguments to be positive, which defines their domain of definition.

The unique properties of logarithms, such as changing base, and their graphical characteristics, like being defined only for positive arguments, make them fascinating yet challenging. Graping them offers a concrete perspective, revealing how solutions are influenced by logarithmic behavior.