Problem 70 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{3}(a+3)+\log _{3}(a-3)=\log _{3} 16\)

Step-by-Step Solution

Answer

The solution is \(a = 5\).

Step 1: Apply the Product Property of Logarithms

The problem gives us two logarithms added together with the same base: \( \log_{3}(a+3) + \log_{3}(a-3) \). We can use the property \( \log_{b}(M) + \log_{b}(N) = \log_{b}(MN) \). Here, it simplifies to \( \log_{3}((a+3)(a-3)) \).

Step 2: Simplify the Expression

Applying the product property gives \( \log_{3}((a+3)(a-3)) = \log_{3} 16 \).

Step 3: Set the Arguments Equal

Since the bases and the resulting expressions must be equal for the logs to be equal, we can equate the arguments: \((a+3)(a-3) = 16\).

Step 4: Expand and Simplify the Quadratic Equation

Expanding the expression \((a+3)(a-3)\) using the difference of squares gets \(a^2 - 9 = 16\). Simplify to \(a^2 - 25 = 0\).

Step 5: Solve the Quadratic Equation

The equation \(a^2 - 25 = 0\) can be factored into \((a-5)(a+5) = 0\). Solving these gives two possible solutions: \(a = 5\) and \(a = -5\).

Step 6: Check Solution Validity

Plug back solutions into the original arguments of the logarithm to verify. If \(a = 5\), the arguments are 8 and 2, positive numbers, so it's valid. If \(a = -5\), the arguments become -2 and -8, which are not valid as logs cannot have negative arguments.

Key Concepts

Product Property of Logarithms

The product property of logarithms is a useful tool when dealing with logarithmic equations. This property states that the sum of two logarithms with the same base can be condensed into a single logarithm. It is expressed as \( \log_{b}(M) + \log_{b}(N) = \log_{b}(MN) \). This means you multiply the arguments of the logarithms, turning two logs into one.

For instance, the equation \( \log_{3}(a+3) + \log_{3}(a-3) \) simplifies to \( \log_{3}((a+3)(a-3)) \) using this property. This property simplifies the equations and makes solving for the unknown variable much easier. It allows us to equate the arguments within the logs, provided the bases are the same. This step reduces the complexity of the equation dramatically, paving the way for further simplification.

Difference of Squares

The difference of squares is a valuable algebraic identity used to simplify expressions. It is applicable when you have two terms that are each squares, separated by a subtraction sign. Mathematically, it is represented as \( x^2 - y^2 = (x-y)(x+y) \). This identity helps in breaking down complex expressions into a product of two binomials.

In the context of our problem, after applying the product property, we have \( (a+3)(a-3) \). This expression is a classic example of the difference of squares, simplifying to \( a^2 - 9 \). Recognizing and applying this pattern decreases the challenge of dealing with polynomials, converting a multiplication problem into addition and subtraction components. This concept is essential to solve quadratic equations efficiently by simplifying the steps needed to reach the solution.

Quadratic Equations

Quadratic equations generally appear in the form \( ax^2 + bx + c = 0 \). Solving them might seem intimidating, but they can be simplified and solved using various techniques, such as factoring, completing the square, or applying the quadratic formula.

In the solution provided, after using the product property and the difference of squares, the equation simplifies to \( a^2 - 25 = 0 \). This can be factored into \( (a-5)(a+5) = 0 \). By factoring the quadratic, we determine the values of \( a \) that satisfy the equation. In this case, the solutions \( a = 5 \) and \( a = -5 \) emerge naturally from the factored form.

The process involves simple algebraic steps and offers two potential solutions that need verification. Checking each solution ensures that they fit the original equation constraints, such as considering the valid domain of logarithm expressions. This last step is crucial to ascertain which solutions are valid, especially given logarithmic limitations.