Problem 51 - Precalculus Mathematics for Calculus

Question

Solve the logarithmic equation for \(x .\) $$\log _{2} x+\log _{2}(x-3)=2$$

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Use the Product Property of Logarithms

Start with the equation \( \log_{2} x + \log_{2} (x-3) = 2 \). Use the product property of logarithms to combine the terms: \( \log_{2} (x(x-3)) = 2 \). This simplifies the equation to \( \log_{2} (x^2 - 3x) = 2 \).

Step 2: Convert the Logarithmic Equation to an Exponential Equation

Recall that if \( \log_{b} y = c \), then \( b^c = y \). Here, \( \log_{2} (x^2 - 3x) = 2 \) means \( 2^2 = x^2 - 3x \), which simplifies to \( 4 = x^2 - 3x \).

Step 3: Rearrange into Quadratic Form

Bring all terms to one side to create a quadratic equation: \( x^2 - 3x - 4 = 0 \).

Step 4: Factor the Quadratic Expression

Factor the quadratic equation \( x^2 - 3x - 4 = 0 \) into \( (x - 4)(x + 1) = 0 \).

Step 5: Solve for x

Set each factor equal to zero and solve: \( x - 4 = 0 \) gives \( x = 4 \), and \( x + 1 = 0 \) gives \( x = -1 \).

Step 6: Check for Logarithm Domain

Substitute \( x = 4 \) and \( x = -1 \) back into the original equation to check for validity. \( x = -1 \) is not valid since \( \log_2(-1) \) and \( \log_2(-4) \) are undefined. Therefore, \( x = 4 \) is the only valid solution.

Key Concepts

Product Property of Logarithms

The product property of logarithms is a fundamental rule that helps simplify expressions involving logarithms. It states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. This can be expressed as:

• \( \log_b (mn) = \log_b m + \log_b n \)

In our exercise, the terms \( \log_2 x + \log_2 (x-3) \) utilize this property to combine into one single term: \( \log_2 (x(x-3)) \). This simplification is crucial because it transforms the original equation into an easier form to manipulate. Evolving from two terms into one, makes the equation more manageable and sets the stage for further solving steps.

Exponential Equations

Once we have simplified our logarithmic equation using the product property, the next step is to convert it into an exponential equation. This process relies on understanding that logarithms and exponentials are inverse functions. If you have an equation of the form \( \log_b y = c \), it can be rewritten in exponential form as:

• \( b^c = y \)

In the exercise, \( \log_2 (x^2 - 3x) = 2 \) becomes \( 2^2 = x^2 - 3x \). Simplifying the exponential expression \( 2^2 \), which equals \( 4 \), transforms the logarithmic equation into \( 4 = x^2 - 3x \). This step is vital because it eliminates the logarithmic expression altogether, allowing us to handle the equation using algebraic techniques.

Solving Quadratic Equations

After transforming the equation into an exponential form and subsequently to a polynomial, we often find ourselves with a quadratic equation. The standard form of such an equation is \( ax^2 + bx + c = 0 \). Solving it involves finding the values of \( x \) that satisfy the equation.
In the exercise, the derived quadratic equation is \( x^2 - 3x - 4 = 0 \). To solve this, we use factoring:

• Express \( x^2 - 3x - 4 \) as \( (x - 4)(x + 1) = 0 \).
• Set each factor equal to zero: \( x - 4 = 0 \) and \( x + 1 = 0 \).
• This gives solutions \( x = 4 \) and \( x = -1 \).

Factoring is a common method for solving quadratic equations, especially when they can be easily broken down into integers, as it provides straightforward solutions for \( x \).

Logarithm Domain Checking

The final step in solving any logarithmic equation is to ensure that your solutions are valid, particularly regarding the domain of a logarithm. Logarithms are defined only for positive real numbers. Thus, any solution resulting in a negative or zero input under a logarithm is not acceptable.
In our equation, after identifying \( x = 4 \) and \( x = -1 \) as potential solutions, a domain check is necessary.

• Substituting \( x = 4 \) gives valid logarithmic inputs \( \log_2 (4) \) and \( \log_2 (1) \).
• However, \( x = -1 \) results in \( \log_2 (-1) \) and \( \log_2 (-4) \), which are not defined, indicating the solution is invalid.

Hence, \( x = 4 \) is the only solution that fits within the correct domain for logarithmic functions. This highlights the importance of considering domain restrictions in problems involving logarithmic equations.