Problem 53 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x.\) \(\log _{9}(x-5)+\log _{9}(x+3)=1\)

Step-by-Step Solution

Answer

The solution is \(x=6\).

Step 1: Use the Product Rule for Logarithms

The given equation is \(\log_{9}(x-5) + \log_{9}(x+3) = 1\). According to the product rule, \(\log_b(A) + \log_b(B) = \log_b(AB)\). Therefore, combine the logarithms: \(\log_{9}((x-5)(x+3)) = 1\).

Step 2: Exponentiate Both Sides

To eliminate the logarithm, rewrite the equation in exponential form. This gives \((x-5)(x+3) = 9^1\). So, \((x-5)(x+3) = 9\).

Step 3: Expand the Equation

Expand \((x-5)(x+3)\) to form a quadratic equation. \((x-5)(x+3) = x^2 + 3x - 5x - 15 = x^2 - 2x - 15\). Thus, the equation becomes \(x^2 - 2x - 15 = 9\).

Step 4: Rearrange the Quadratic Equation

Move all terms to one side of the equation to set it to zero: \(x^2 - 2x - 15 - 9 = 0\). Simplify to get \(x^2 - 2x - 24 = 0\).

Step 5: Factor the Quadratic Equation

Factorize \(x^2 - 2x - 24\). Look for two numbers that multiply to \(-24\) and add to \(-2\), which are \(-6\) and \(4\). Hence, the factorization is \((x-6)(x+4)=0\).

Step 6: Solve for x

Set each factor equal to zero: \(x-6=0\) or \(x+4=0\). Solving these gives \(x=6\) or \(x=-4\).

Step 7: Verify the Solutions

Check the solutions in the original equation to ensure they are valid. Substituting \(x=6\) into the original equation, \(\log_{9}(6-5) + \log_{9}(6+3) = \log_{9}(1) + \log_{9}(9) = 0 + 1 = 1\). Now check \(x=-4\): \(\log_{9}(-4-5) + \log_{9}(-4+3)\), which is undefined as the argument of a logarithm cannot be negative. Thus, \(x=-4\) is not a valid solution.

Key Concepts

Product Rule for Logarithms

When dealing with logarithmic expressions, the product rule is a vital tool. It states that when you add two logarithms with the same base, you can combine them into a single logarithm of the product of their arguments:

• Mathematically, it is expressed as: \( \log_b(A) + \log_b(B) = \log_b(AB) \).
• This rule is extremely useful for simplifying episodes like in the original problem where you had \( \log_9(x-5) + \log_9(x+3) = 1 \).
• By applying the product rule, it simplifies to \( \log_9((x-5)(x+3)) = 1 \), which further makes the equation easier to handle.

Imagine you're stacking blocks. Instead of counting them one by one, if they're all the same type, you can consider them as a whole block. This is the essence of the product rule, making complex additions straightforward.

Exponential Form

Converting a logarithmic equation into an exponential form is analogous to turning a complex magic formula into something more understandable. The basic idea is that the expression \( \log_b(x) = y \) can be rewritten as an exponential equation \( b^y = x \). This transformation gives you a clear path to solve real-world equations.

• In the context of the solved equation, \( \log_9((x-5)(x+3)) = 1 \), let's convert it. In exponential form, it becomes \( 9^1 = (x-5)(x+3) \), simplifying your work significantly.
• This step removes the logarithm and paves the way for straightforward algebraic manipulation.
• By switching forms, you're peeling away the layers of complexity, allowing you to see the core relational structure of the numbers involved.

Think of it like translating a book into your native language - it breaks down barriers and makes the content relatable and solvable.

Quadratic Equations

Once converted into exponential form, our equation becomes \( (x-5)(x+3) = 9 \). The next step is expanding this product to create a quadratic equation. Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). Understanding their structure, solving them is achievable through different methods, with factoring being one of the most common.

• The expansion gives us \( x^2 - 2x - 15 = 9 \). Rearranging terms, we have: \( x^2 - 2x - 24 = 0 \).
• Factoring this results in \( (x - 6)(x + 4) = 0 \), paving the path to potential solutions for \( x \).
• Setting each factor to zero, \( x = 6 \) and \( x = -4 \) are revealed as possible solutions.

Once these solutions are unearthed, it's necessary to verify them, as shown in the exercise, ensuring they satisfy the original conditions of the problem. Just as a detective verifies the evidence, it's crucial to confirm that these solutions don't break any mathematical rules, especially when dealing with phenomena like logarithms where negative inputs aren't permissible.