Problem 75 - College Algebra Essentials

Question

Solve each logarithmic equation. 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 equation is \( x = 6 \).

Step 1: Consolidate the Logarithmic Expressions

We can use the properties of logarithms to simplify the given equation. The sum of logarithms is equal to the logarithm of the product of the two numbers: \[ \log _{2}(x-6) + \log _{2}(x-4) = \log _{2}((x-6) \cdot (x-4)) \]. Also, the difference of logarithms equals the logarithm of the quotient: \[ \log _{2}((x-6) \cdot (x-4)) - \log _{2} x = \log _{2}\left( \frac{(x-6) \cdot (x-4)}{x} \right) \]. Thus, the original equation is simplified to \[ \log _{2} \left( \frac{(x-6) \cdot (x-4)}{x} \right) = 2 \].

Step 2: Solve the Equation

By the definition of logarithmic equations, we can say that \[ \frac{(x-6) \cdot (x-4)}{x} = 2^{2} = 4 \]. Simplifying this expression will give us a quadratic equation: \( x^{2} - 10x + 24 = 4x \). Continue to simplify this quadratic equation by subtracting 4x from both sides to isolate terms involving x on one side to obtain: \( x^{2} - 14x + 24 = 0 \).

Step 3: Solve for \( x \) by Factoring

Next, we must solve for \( x \) in the quadratic equation \( x^{2} - 14x + 24 = 0 \) by factoring. The factors of the equation are \((x - 6) \) and \((x - 4) \). So, \( (x - 6)(x - 4) = 0 \). The solutions to this equation are: \( x = 6 \) or \( x = 4 \).

Step 4: Check the Domain of the Original Logarithmic Expressions

The values of \( x \) must be greater than 6 to ensure that \( x-6 \) and \( x - 4 \) are positive. Moreover, \( x \) must be more than 0 as per the original logarithmic expression \( \log _{2} x \). So, the only suitable solution after checking the domain will be \( x = 6 \).

Step 5: Give a Decimal Approximation

Providing a decimal approximation is unnecessary in this case as \( x = 6 \) is already in decimal form.