Problem 82 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 2 \log _{3} x-\log _{3}(x-4)=2+\log _{3} 2 $$

Step-by-Step Solution

Answer

The solutions are \(x = 12\) and \(x = 6\).

Step 1: Simplify Logarithmic Equation

Start with the given equation: \(2 \log_{3} x - \log_{3} (x-4) = 2 + \log_{3} 2\). Use the property of logarithms that states \(a \log_{b} c = \log_{b} c^a\) to rewrite the first term as \(\log_{3} x^2\). This gives us: \[\log_{3} x^2 - \log_{3} (x-4) = 2 + \log_{3} 2\].

Step 2: Combine Logarithms

Use the property of logarithms that states \(\log_{b} A - \log_{b} B = \log_{b} \left(\frac{A}{B}\right)\). Substitute into the equation: \[\log_{3} \left(\frac{x^2}{x-4}\right) = 2 + \log_{3} 2\].

Step 3: Simplify the Right-Hand Side

On the right-hand side, use the property that \(c + \log_{b} d = \log_{b} (b^c \cdot d)\). Transform \(2 + \log_{3} 2\) into \(\log_{3} (9 \times 2)\). Simplifying gives: \[\log_{3} 18\]. The equation is now: \[\log_{3} \left(\frac{x^2}{x-4}\right) = \log_{3} 18\].

Step 4: Remove Logarithms

Since the logarithms have the same base, if \(\log_{b} A = \log_{b} B\), then \(A = B\). Set the arguments equal: \[\frac{x^2}{x-4} = 18\].

Step 5: Solve the Rational Equation

Multiply both sides of the equation by \((x-4)\) to eliminate the fraction: \[x^2 = 18(x - 4)\]. Simplify and rearrange to form a quadratic equation: \[x^2 - 18x + 72 = 0\].

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -18\), \(c = 72\). Calculate: \[x = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 72}}{2}\]. This simplifies to \[x = \frac{18 \pm \sqrt{36}}{2}\], giving solutions \(x = 12\) or \(x = 6\).

Step 7: Verify Solutions

Check both solutions in the context of the original logarithmic equation to ensure they do not create negative arguments. \(x = 12\) and \(x = 6\) both result in positive arguments, so both are valid solutions.

Key Concepts

Properties of Logarithms

Logarithms come with a variety of properties that allow us to manipulate and solve logarithmic equations. Understanding these properties can simplify complex expressions:

• Power Rule: One of the most used properties in our case is the power rule. It states that multiplying a logarithm by a constant can be rewritten as raising the logarithmic argument to that constant's power: \(a \log_{b} c = \log_{b} c^a\).
• Quotient Rule: Another useful property is the quotient rule, which allows us to express the difference between two logarithms as a single logarithmic term: \(\log_{b} A - \log_{b} B = \log_{b} \left(\frac{A}{B}\right)\).
• Product Rule: Although not explicitly used in our example, the product rule states that the sum of two logarithms can be combined as a product: \(\log_{b} A + \log_{b} B = \log_{b} (AB)\).

These properties help us transform and eventually solve logarithmic equations efficiently.
In the given exercise, the power and quotient rules played a significant role in simplifying the equation before equating the arguments of the logarithms.

Quadratic Equations

Quadratic equations often appear after transforming and simplifying logarithmic equations. A standard quadratic equation is written in the form:
\(ax^2 + bx + c = 0\).
The key characteristics of quadratic equations involve:

• Being polynomial equations of degree two, meaning the highest exponent of the variable (usually \(x\)) is two.
• They can be solved using various methods, such as factoring, completing the square, graphing, or using the quadratic formula.
• The solutions to these equations are called the "roots" and can be real or complex numbers.

In our exercise, through the properties of logarithms, the problem simplifies to a quadratic equation which was in the format \(x^2 - 18x + 72 = 0\). This equation was then solved using the quadratic formula, leading us to find the potential solutions for \(x\).

Exact Solution

An exact solution refers to a precise answer without any rounding or estimation. When using the quadratic formula:

• We got the general form \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• By substituting the values into the quadratic formula with \(b = -18\), \(a = 1\), and \(c = 72\), the solutions became \(x = 12\) and \(x = 6\).
• An exact solution is valuable because it represents the complete and definitive answer derived directly from the original problem without approximation.

Check your solutions by plugging them back into the initial equation to ensure they do not result in undefined or negative logarithmic values. In this case, both values provided valid solutions for the problem.

Approximation

Sometimes, solutions to equations are given in approximate forms especially when finding roots of quadratic equations, where the discriminant might not be a perfect square. Approximation involves providing a numerical solution, often to a specified number of decimal places, particularly useful in real-world applications.

• However, in this particular problem, the quadratic formula yielded exact integer solutions, meaning no approximation was needed for this exercise.
• Approximations often occur in logarithmic problems that do not result in whole numbers, thus requiring calculation tools or methods such as a calculator to derive values close to the actual roots.
• Understanding when to approximate helps students distinguish between when an exact or approximate answer is more suitable based on the problem context or requirements.

In conclusion, the distinction between requiring an exact solution and when an approximation suffices is crucial, particularly in mathematical studies involving more complex equations.