Problem 38 - College Algebra

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 solution is \(x = 6\).

Step 1: Identify the relevant logarithm rule

The problem is asking us to find the values of \(x\) for which two logarithmic expressions add up to a constant. To do this, we'll use the property of logarithms known as the product rule: \(\log_b(M) + \log_b(N) = \log_b(M \cdot N)\).

Step 2: Apply the product rule to combine logs

Using the product rule from the first step, we combine the two logarithmic expressions: \( \log_{12}(2x + 6) + \log_{12}(x + 2) = \log_{12}((2x + 6)(x + 2)) \).

Step 3: Simplify the equation

Set the expression from Step 2 equal to the given constant on the right side of the equation: \( \log_{12}((2x + 6)(x + 2)) = 2 \). Solve for the argument of the logarithm: \((2x + 6)(x + 2) = 12^2 \).

Step 4: Expand and simplify the polynomial

Expand \((2x + 6)(x + 2)\):\[2x(x + 2) + 6(x + 2) = 2x^2 + 4x + 6x + 12 = 2x^2 + 10x + 12\]. Then equate to the square of 12: \(2x^2 + 10x + 12 = 144\).

Step 5: Transform into standard quadratic form

Rearrange to form a standard quadratic equation: \(2x^2 + 10x + 12 - 144 = 0\), simplifying to \(2x^2 + 10x - 132 = 0\).

Step 6: Factor the quadratic equation

Divide through by 2 to simplify: \(x^2 + 5x - 66 = 0\). Look for two numbers that multiply to -66 and add to 5, which are 11 and -6. Therefore, the polynomial factors to \((x + 11)(x - 6) = 0\).

Step 7: Solve for roots

Set each factor equal to zero: 1. \(x + 11 = 0\) gives \(x = -11\)2. \(x - 6 = 0\) gives \(x = 6\)

Step 8: Verify the domain constraints

Since logarithms are only defined for positive arguments, we need to ensure that \(2x + 6 > 0\) and \(x + 2 > 0\). Solving these inequalities gives \(x > -3\) and \(x > -2\). The intersection of these constraints means \(x > -2\). Thus, only \(x = 6\) is valid because \(x = -11\) is outside this range.

Key Concepts

product rule for logarithms

When working with logarithms, one helpful rule to know is the product rule. This rule simplifies expressions involving the addition of logarithms with the same base. The product rule states: for any two positive numbers, \(M\) and \(N\), and a logarithm base \(b\), \(\log_b(M) + \log_b(N) = \log_b(M \cdot N)\). This rule allows us to combine two separate logarithmic terms into one.
In the exercise provided, we used the product rule to transform the sum of two logs, \(\log_{12}(2x + 6) + \log_{12}(x + 2)\), into a single log: \(\log_{12}((2x + 6)(x + 2))\).
By doing this, we simplify the equation significantly, making it possible to solve for the variable \(x\) in subsequent steps.

quadratic equation and solving techniques

A quadratic equation is a polynomial equation of degree 2, typically expressed in the form \(ax^2 + bx + c = 0\). In our exercise, after simplifying the logarithmic expression, we encountered the quadratic equation \(2x^2 + 10x - 132 = 0\).
To solve this, we transformed it into its standard form by dividing by 2: \(x^2 + 5x - 66 = 0\).
Solving quadratic equations can be done through various methods such as factoring, completing the square, or using the quadratic formula.

• Factoring: We look for two numbers that multiply to \(-66\) and add up to \(5\), which are \(11\) and \(-6\).
  This gives us the factored form \((x + 11)(x - 6) = 0\).


• Finding roots: Set each factor equal to zero gives us the roots \(x = -11\) and \(x = 6\).

Quadratic equations are a key step in effectively solving many algebraic problems and understanding their properties helps to solve them with ease.

importance of domain constraints in logarithmic equations

When dealing with logarithms, it is crucial to consider domain constraints because logarithms are only defined for positive real numbers. This means the argument inside a logarithm must be positive, which imposes restrictions on acceptable values for the variable \(x\).
In this exercise, the expressions inside the logarithms, \(2x + 6\) and \(x + 2\), must both be greater than zero. Solving these inequalities:

• \(2x + 6 > 0\) simplifies to \(x > -3\).
• \(x + 2 > 0\) simplifies to \(x > -2\).

Combining these inequalities, we find that \(x\) must be greater than \(-2\).
Thus, the root \(x = -11\) is not valid because it does not satisfy the domain constraints.
Only \(x = 6\) remains as a feasible solution, highlighting how domain constraints affect which solutions are valid in logarithmic equations. Always check domain constraints to ensure a solution is applicable in the context of the problem.