Problem 75 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{2}(x-6)+\log _{2}(x-4)-\log _{2} x=2 $$

Step-by-Step Solution

Answer

The solution to the given equation is \(x = 12\).

Step 1: Simplify using a logarithmic property

Use the property of logarithms, \(\log_b A + \log_b B = \log_b (A*B)\), and \(\log_b A - \log_b B = \log_b (A/B)\) to write the expression as a single logarithm: \(\log_2 \frac{(x-6)(x-4)}{x} = 2\).

Step 2: Get rid of the logarithm

To isolate \(x\), we can get rid of the log by using the property that if \(\log_b A = c\), then \(A = b^c\). So, \(\frac{(x-6)(x-4)}{x} = 2^2\). This simplifies to \((x-6)(x-4) = 4x\).

Step 3: Simplify to get a quadratic equation

Expanding we get, \(x^2 - 10x + 24 = 4x\), which simplifies to \(x^2 - 14x + 24 = 0\).

Step 4: Solve for \(x\)

To get \(x\), we can use the quadratic formula, \(x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\). Using this formula we get \(x = 7 ± \sqrt{25} = 7 ± 5\).

Step 5: Check the solutions

Testing both potential solutions \(x = 12\) and \(x = 2\), we see that only \(x = 12\) is part of the domain \(x > 6\), therefore \(x = 2\) needs to be discarded.

Key Concepts

Properties of Logarithms

Logarithms have several unique properties that simplify complex expressions. These properties are essential for manipulating and solving logarithmic equations. One key property is the product rule:

• \( \log_b A + \log_b B = \log_b (AB) \)

This rule allows you to combine two logarithms with the same base that are added, turning them into a single logarithm.
Another important rule is the quotient rule:

• \( \log_b A - \log_b B = \log_b \left( \frac{A}{B} \right) \)

This helps when you need to subtract two logarithms, combining them into one by dividing their arguments.
These properties are especially useful in rewriting logarithmic expressions in simpler forms, making them easier to work with or solve.
In the given logarithmic equation from the exercise, these properties were used to combine multiple log terms into one:

• \( \log_2 \frac{(x-6)(x-4)}{x} = 2 \)

By understanding and mastering these properties, solving complex and intertwined logarithmic equations becomes more straightforward.

Quadratic Equations

Quadratic equations are polynomials of degree two. They have the general form:

• \( ax^2 + bx + c = 0 \)

A quadratic equation may have zero, one, or two real solutions that can be found using various methods such as factoring, completing the square, or applying the quadratic formula.
The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is often used when the quadratic equation does not factor easily. It gives the solutions for any quadratic equation.
In our example, after using logarithm properties and simplifying the expression, the question was transformed into the quadratic equation \( x^2 - 14x + 24 = 0 \).
To solve this equation, the quadratic formula was applied, resulting in two potential solutions. By solving it, we determine the values for \( x \) where our original equation is valid.

Domain of a Function

The domain of a function is critical because it defines the set of input values for which the function is valid.
For logarithmic functions, the input must be greater than zero. This is because the logarithm of a non-positive number is undefined in real numbers.
In our exercise, we want each argument in the logarithmic expressions to be positive. Therefore, we must ensure:

• \( x - 6 > 0 \Rightarrow x > 6 \)

• \( x - 4 > 0 \Rightarrow x > 4 \)

• \( x > 0 \)

Combining these constraints, we find the domain is \( x > 6 \). This means that any potential solutions to the logarithmic equation must satisfy this domain condition.
Therefore, when solving the equation, it is crucial to test the solutions against these conditions to ensure they are valid. For example, among the solutions obtained, only \( x = 12 \) meets the domain requirement, while \( x = 2 \) does not, and therefore, \( x = 2 \) must be discarded.