Problem 49 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log (x-3)+\log (x+2)=\log (4 x)$$

Step-by-Step Solution

Answer

The exact solution is \( x = 6 \).

Step 1: Use the Product Rule of Logarithms

The equation given is \( \log (x-3) + \log (x+2) = \log (4x) \). Apply the product rule of logarithms which states that \( \log a + \log b = \log (a \cdot b) \). So, \( \log (x-3) + \log (x+2) = \log ((x-3)(x+2)) \). Now the equation becomes \( \log ((x-3)(x+2)) = \log (4x) \).

Step 2: Set the Arguments Equal

Since the logs are equal (assuming all values are in the domain), we can set the arguments equal: \( (x-3)(x+2) = 4x \). Expand the left side: \( x^2 + 2x - 3x - 6 = 4x \), which simplifies to \( x^2 - x - 6 = 4x \).

Step 3: Solve the Quadratic Equation

Rearrange the equation to \( x^2 - x - 6 = 4x \) into standard quadratic form: \( x^2 - x - 6 - 4x = 0 \) simplifies to \( x^2 - 5x - 6 = 0 \). To factor, find two numbers that multiply to -6 and add to -5: \( x^2 - 5x - 6 = (x - 6)(x + 1) = 0 \). This gives the solutions \( x - 6 = 0 \) or \( x + 1 = 0 \), so \( x = 6 \) or \( x = -1 \).

Step 4: Check for Extraneous Solutions

Substitute \( x = 6 \) back into the arguments of the original logarithmic expressions: \( x - 3 = 3 \) and \( x + 2 = 8 \), both are positive which ensures they are in the domain of the logarithm. Now check \( x = -1 \): \( x - 3 = -4 \) and \( x + 2 = 1 \). Since \( x - 3 = -4 \) is negative, \( x = -1 \) is outside the domain of the logarithm and is not a valid solution.

Key Concepts

Product Rule of Logarithms

Understanding the product rule of logarithms is crucial when solving logarithmic equations. This rule is grounded in the property of logarithms: when you have the addition of two logarithmic expressions with the same base, you can combine them into a single log by multiplying their arguments. This is shown as follows:

• If you have \( \log_a b + \log_a c = \log_a (b \cdot c) \), then the arguments \( b \) and \( c \) are multiplied together.

In our exercise, we applied this rule to combine \( \log(x-3) + \log(x+2) \) into a single logarithm: \( \log((x-3)(x+2)) \). This simplification makes it possible to equate the arguments directly, because now we are only dealing with one logarithmic expression on each side. Consequently, this step forms the bridge from a logarithmic equation to a polynomial, thus allowing us to use further algebraic techniques to solve.
Always remember that the product rule only works for logarithms that share the same base. This foundational property streamlines working with complex expressions and is a key tool in solving logarithmic equations efficiently.

Quadratic Equations

Once the product rule of logarithms has been applied, and the equation is simplified, it's typical for these types of problems to turn into polynomial equations, such as quadratics. In this exercise, the logarithmic equation simplifies into the quadratic equation \( (x-3)(x+2) = 4x \).
This requires us to expand and rearrange it into standard quadratic form: \( x^2 - 5x - 6 = 0 \). Here, the task becomes one of finding values of \( x \) that satisfy this equation. Factoring is one of the most straightforward methods for solving quadratic equations unless the quadratic is a perfect square or requires completing the square.

• Look for two numbers that multiply to give the constant term (-6) and add to give the coefficient of the x term (-5).
• In this case, \( -6 \) and \( +1 \) work perfectly because they satisfy these requirements.

Breaking the quadratic into \( (x-6)(x+1) = 0 \) gives us the roots, leading us to solutions \( x = 6 \) and \( x = -1 \). The discovery of solutions in this form needs further verification for validity, which is influenced by the domain of our original logarithmic expressions.

Extraneous Solutions

Extraneous solutions are those that appear valid during the solving process but do not satisfy the original equation due to domain restrictions or other factors. It is especially common to encounter such solutions in logarithmic equations.
When solving \( \log (x-3)+\log (x+2)=\log (4 x) \), we found potential solutions \( x = 6 \) and \( x = -1 \). But to determine if these are truly solutions, they need to fit into the domain of the original logarithmic expressions.

• Logarithms are only defined for positive arguments, meaning \( x - 3 \) and \( x + 2 \) must be positive.
• Checking these conditions, \( x = 6 \) works because both arguments \( x-3=3 \) and \( x+2=8 \) are positive.
• However, for \( x = -1 \), \( x-3 \) results in -4, which is negative, thus invalidating this solution.

This validation step is crucial to avoid adopting incorrect solutions. By verifying the solutions against the original equation's domain, we prevent logical errors in mathematical processes, affirming the accuracy and reliability of our solution.