Problem 77 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The exact solution is \( x = 10 \).

Step 1: Apply the Logarithm Subtraction Rule

Use the property of logarithms that \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \). This allows us to combine the left-hand side of the equation into a single logarithm: \( \log\left( \frac{x-6}{x-2} \right) = \log\left( \frac{5}{x} \right) \).

Step 2: Set the Arguments Equal

Since the logarithm is the same on both sides of the equation, we can set the arguments equal to each other: \( \frac{x-6}{x-2} = \frac{5}{x} \).

Step 3: Cross-Multiply to Eliminate Fractions

To solve the equation \( \frac{x-6}{x-2} = \frac{5}{x} \), cross-multiply to get rid of the fractions: \( x(x-6) = 5(x-2) \). This simplifies to \( x^2 - 6x = 5x - 10 \).

Step 4: Rearrange and Simplify the Equation

Bring all terms to one side to form a quadratic equation: \( x^2 - 11x + 10 = 0 \).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -11 \), and \( c = 10 \). Calculate the discriminant: \( b^2 - 4ac = 121 - 40 = 81 \). The solutions are \( x = \frac{11 \pm \sqrt{81}}{2} \), which simplifies to \( x = 10 \) or \( x = 1 \).

Step 6: Check Solutions for Validity

We need to check whether the solutions fall within the domain of the original logarithmic expressions. For \( \log(x-6) \) and \( \log(x-2) \) to be defined, both \( x-6 > 0 \) and \( x-2 > 0 \) must hold, hence \( x > 6 \). Therefore, only \( x = 10 \) is valid.

Key Concepts

Understanding the Logarithm Subtraction Rule

The logarithm subtraction rule is a fundamental property of logarithms which states that the difference between two logarithms with the same base can be expressed as the logarithm of the division of their arguments. This can be written mathematically as:

• \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \).

In the original exercise, we apply this rule to combine the terms on the left side of the equation:

• \( \log(x-6) - \log(x-2) = \log\left(\frac{x-6}{x-2}\right) \).

This transformation simplifies complex logarithmic expressions and makes it easier to solve equations. Essentially, the logarithm subtraction rule helps us convert a subtraction of logs into a simpler single logarithm expression. Once the equation is in this form, we can then equate it to the right-hand side and solve it using simpler algebraic techniques.

Solving with Quadratic Equations

A quadratic equation is a second-degree polynomial equation in a single variable \( x \), with the standard form:

• \( ax^2 + bx + c = 0 \).

In the problem at hand, we derive a quadratic equation from the logarithmic expressions once simplified and equate the arguments. After cross-multiplying, it results in:

• \( x^2 - 11x + 10 = 0 \).

Quadratic equations can be solved using various methods including factoring, completing the square, or the quadratic formula. Here, the quadratic formula is used:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

By substituting \( a = 1 \), \( b = -11 \), and \( c = 10 \), we find the solutions. The discriminant \( b^2 - 4ac \) determines the nature of the roots. If it is positive, as in our exercise (discriminant of 81), there are two distinct real solutions. It’s crucial to check these solutions in the context of the original problem, especially since this involves logarithms.

Determining the Domain of Logarithms

The domain of logarithms is an essential consideration when solving logarithmic equations. A logarithmic function \( \log(b) \) is defined only for positive real numbers. Hence, any expression within a logarithm must be positive. In math terms, for \( \log(x-c) \), it follows that \( x-c > 0 \), meaning \( x > c \).
With our given equation, we must consider both \( \log(x-6) \) and \( \log(x-2) \). Logarithmic terms \( x-6 > 0 \) and \( x-2 > 0 \) imply that:

• \( x > 6 \).

Consequently, when we solve the quadratic equation and find potential solutions, we must check that these solutions satisfy the derived domain constraints. If a solution, like \( x = 1 \) in this case, does not meet the criteria of being greater than 6, it must be discarded. Therefore, among all possible solutions, only the ones that satisfy the domain conditions of the original logarithmic equation are valid.