Problem 87 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \frac{\log (5 x+6)}{2}=\log x $$

Step-by-Step Solution

Answer

The exact solution is \( x = 6 \) and it doesn't require approximation.

Step 1: Eliminate the Fraction

Multiply both sides of the equation by 2 to get rid of the fraction: \[ \log(5x+6) = 2 \log x \]

Step 2: Apply the Power Rule of Logarithms

Use the power rule, \(a \log b = \log(b^a)\), to rewrite the right side: \[ \log(5x+6) = \log(x^2) \]

Step 3: Use the Property of Logarithms

Since both sides of the equation are in logarithmic form, set the internal expressions equal to each other: \[ 5x + 6 = x^2 \]

Step 4: Rearrange into a Quadratic Equation

Rearrange the terms to form a quadratic equation: \[ x^2 - 5x - 6 = 0 \]

Step 5: Solve the Quadratic Equation

Factor the quadratic equation or use the quadratic formula. Factoring gives:\[ (x - 6)(x + 1) = 0 \]This gives the solutions:\[ x = 6 \quad \text{and} \quad x = -1 \]

Step 6: Discard Invalid Solutions

Since logarithms are not defined for non-positive numbers, discard any extraneous solutions. Here, \( x = -1 \) is not valid because \( \log x \) is undefined for \( x = -1 \), leaving only:\[ x = 6 \]

Step 7: Verify the Solution

Substitute \( x = 6 \) back into the original equation to verify:\[ \frac{\log(5 \times 6 + 6)}{2} = \log 6 \]Simplifying the left side:\[ \frac{\log 36}{2} = \log 6 \]This equality holds since both sides evaluate to the same logarithmic statement. Thus, \( x = 6 \) is a valid solution.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations are characterized by having a degree of 2, which means the highest power of the variable \( x \) is 2.
Quadratic equations can often be solved by several methods:

• Factoring: This involves expressing the equation as a product of two binomials. From the factorized form, the solutions (also called roots) can be easily found.
• The Quadratic Formula: This is a universal formula that provides the solutions for any quadratic equation, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), applicable when factoring is complex or impossible.
• Completing the Square: A method that involves arranging the equation into a perfect square trinomial. This can then be solved by taking square roots on both sides.

In our exercise, the quadratic equation \(x^2 - 5x - 6 = 0\) was solved by factoring, leading us to the solutions \(x = 6\) and \(x = -1\). Since quadratic equations usually have two solutions, it's important to consider the context and constraints of the problem to determine which solutions are valid.

Properties of Logarithms

Logarithms have several useful properties that make them invaluable tools in solving equations that involve exponential terms.

• Product Rule: \( \log_b(mn) = \log_b(m) + \log_b(n) \), which simplifies multiplication inside a logarithm into an addition of separate logarithms.
• Quotient Rule: \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \), allowing division within a logarithm to be expressed as subtraction.
• Power Rule: \( \log_b(m^n) = n \cdot \log_b(m) \), which simplifies powers within a logarithm to multiplication outside.

In the provided solution, the power rule was used to transform \(2 \log x\) into \(\log(x^2)\). This significantly simplifies the equation, making it easier to isolate and solve for the unknown variable. Understanding these properties can greatly enhance one's ability to manipulate and solve logarithmic expressions effectively.

Solving Equations

Solving equations involves finding the values of the variables that make the equation true. This process can include isolating the variable on one side and ensuring that valid operations are applied throughout.
In the context of the given logarithmic equation, the solution process involves transforming the equation using logarithmic properties and subsequently recognizing or rearranging into a familiar form, such as a quadratic equation.

• Transformation: This step changes the form of the equation to make it easier to solve. For example, transforming \( \log(5x+6) = \log(x^2) \) to \( 5x + 6 = x^2 \) by dropping the logarithm (because both sides have the same base) allows you to see the equation as a quadratic.
• Verification: Once a solution is obtained, it's crucial to verify its validity. Substitute back into the original equation to confirm whether the potential solution holds true throughout the entire equation.
• Consideration of Domain: Not all solutions may be valid in the original context, especially with logarithms, which only accept positive arguments.

Ultimately, solving equations is about carefully applying mathematical concepts to find solutions that are consistent and valid within the constraints of the problem. With practice, solving both simple and complex equations becomes more intuitive.