Problem 38 - College Algebra with Corequisite Support

Question

Use the product rule for logarithms to find all \(x\) values such that \(\log _{12}(2 x+6)+\log _{12}(x+2)=2 .\) Show the steps for solving.

Step-by-Step Solution

Answer

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

Step 1: Apply the Product Rule

First, use the product rule for logarithms, which states \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \). Therefore, the expression can be rewritten as \( \log_{12}((2x+6)(x+2)) = 2 \).

Step 2: Simplify the Logarithmic Equation

Now rewrite the right-hand side of the equation in exponential form. Since \( \log_b(a) = c \) implies \( a = b^c \), it follows that \( (2x+6)(x+2) = 12^2 \). Calculate \( 12^2 = 144 \), so the equation becomes \( (2x+6)(x+2) = 144 \).

Step 3: Expand the Quadratic Equation

Expand the left-hand side: \( (2x+6)(x+2) \) gives \( 2x^2 + 4x + 6x + 12 = 2x^2 + 10x + 12 \). The equation is now \( 2x^2 + 10x + 12 = 144 \).

Step 4: Rearrange into a Standard Quadratic Equation

Subtract 144 from both sides to set the quadratic equation to zero: \( 2x^2 + 10x + 12 - 144 = 0 \), which simplifies to \( 2x^2 + 10x - 132 = 0 \).

Step 5: Simplify and Solve the Quadratic

Divide the entire equation by 2 to simplify: \( x^2 + 5x - 66 = 0 \). Solve for \( x \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 5, \) and \( c = -66 \).

Step 6: Calculate the Discriminant and Find Roots

Calculate the discriminant: \( b^2 - 4ac = 5^2 - 4(1)(-66) = 25 + 264 = 289 \). The roots are \( x = \frac{-5 \pm \sqrt{289}}{2} = \frac{-5 \pm 17}{2} \).

Step 7: Evaluate the Roots

Calculate the potential solutions: \( x = \frac{-5 + 17}{2} = 6 \) and \( x = \frac{-5 - 17}{2} = -11 \).

Step 8: Validate the Solutions

Check for domain restrictions: \( 2x+6 > 0 \) and \( x+2 > 0 \). This implies \( x > -3 \) and \( x > -2 \). The solution \( x = -11 \) does not satisfy these, so it is extraneous. Therefore, only \( x = 6 \) is valid.

Key Concepts

Product Rule

The product rule is a handy tool for combining separate logarithmic expressions into a single one. This rule is expressed as follows: If you have

• \( \log_b(m) \)
• \( \log_b(n) \)

then you can combine them using the product rule:

• \( \log_b(m \cdot n) \).

This makes it easier to simplify expressions with multiple logarithms. In the given exercise, we used this to combine the logs, changing

• \( \log_{12}(2x+6)+\log_{12}(x+2) \)
into
• \( \log_{12}((2x+6)(x+2)) \).

This simplification is so powerful because it reduces the complexity of the equation, making it easier to solve.

Exponential Form

Rewriting equations in exponential form is crucial when dealing with logarithms. The relation

• \( \log_b(a) = c \)

can be transformed into an exponential equation:

• \( a = b^c \).

This form is useful because it turns the problem into a more familiar territory - multiplication and powers - that we might find easier to handle. In our specific application, transforming

• \( \log_{12}((2x+6)(x+2)) = 2 \)
into
• \( (2x+6)(x+2) = 12^2 \)

allowed us to calculate that

• \(12^2 = 144\).

This step is key to establishing a solid foundation for further simplification.

Quadratic Equation

Once in exponential form, solving for \( x \) becomes a task of dealing with a quadratic equation. A quadratic equation generally looks like

• \( ax^2 + bx + c = 0 \).

Our goal is to rearrange everything into this format. Expanding the expression

• \((2x+6)(x+2)\)
results in a quadratic equation
• \(2x^2 + 10x + 12 = 144\).
By further simplifying, we get
• \(x^2 + 5x - 66 = 0\).
At this point, we can solve it using simple algebraic techniques like factoring or the quadratic formula.

Discriminant

The discriminant is an important part of the quadratic formula

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

The discriminant

• \( b^2 - 4ac \)

determines the number and type of solutions we can expect:

• If it's positive, we get two real solutions.
• If zero, exactly one real solution.
• If negative, no real solutions, only complex ones.

In our exercise,

• \( b = 5, a = 1, c = -66 \).

The discriminant is

• \( 5^2 - 4(1)(-66) = 289 \).

So, this indicates two real solutions exist. Plugging this into the quadratic formula yields the potential values for \( x \). However, always remember to verify these in the context of the problem to ensure they meet all conditions.