Problem 55 - A Graphical Approach to College Algebra

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{5}(x+2)+\log _{5}(x-2)=1$$

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Use the Logarithm Product Rule

Recall the product rule for logarithms: \( \log_b(M) + \log_b(N) = \log_b(M \cdot N) \). Using this rule, combine the given logarithmic expression: \( \log_{5}(x+2) + \log_{5}(x-2) = \log_{5}((x+2)(x-2)) \). This simplifies the equation to: \( \log_{5}(x^2 - 4) = 1 \).

Step 2: Convert the Logarithmic Equation to Exponential Form

Recall the definition of a logarithm: if \( \log_b(M) = N \), then \( M = b^N \). Apply this definition to our equation: \( \log_{5}(x^2 - 4) = 1 \) becomes \( x^2 - 4 = 5^1 \). Simplify the right side to get: \( x^2 - 4 = 5 \).

Step 3: Solve for \( x \)

Now, solve the quadratic equation: \( x^2 - 4 = 5 \). Add 4 to both sides to isolate \( x^2 \): \( x^2 = 9 \). Take the square root of both sides: \( x = \pm 3 \). This gives us potential solutions \( x = 3 \) and \( x = -3 \).

Step 4: Verify the Solutions

Check each potential solution to ensure they are valid: first verify \( x = 3 \). Substitute back into the original equation: \( \log_{5}(3+2) + \log_{5}(3-2) = \log_{5}(5) + \log_{5}(1) = 1 + 0 = 1 \), which is true. Then verify \( x = -3 \). Substitute back: \( \log_{5}(-3+2) + \log_{5}(-3-2) \), both terms are undefined as the arguments are negative or zero in logarithms. Hence, \( x = -3 \) is not a valid solution.

Key Concepts

Product Rule for Logarithms

The product rule for logarithms is a powerful tool that allows us to simplify expressions where two logs with the same base are added together. This rule states that \( \log_b(M) + \log_b(N) = \log_b(M \cdot N) \). Hence, instead of dealing with two separate logarithms, we can condense them into a single log.

This not only makes solving equations easier, but it also helps in factoring out complex expressions through multiplication inside a single logarithm. In our specific case with the expression \( \log_{5}(x+2) + \log_{5}(x-2) = \log_{5}((x+2)(x-2)) \), we used the product rule to combine the terms:

• The two logs have the same base (5).
• The arguments \((x+2)\) and \((x-2)\) are multiplied together.
• The result is a simplified log: \( \log_{5}(x^2 - 4) \).

This step simplifies our work by allowing us to focus on solving fewer terms in subsequent steps.

Exponential Form Conversion

Converting logarithmic equations to exponential form is a crucial step in solving them. This process relies on the understanding that the logarithm function is the inverse of exponentiation. If you know that \( \log_b(M) = N \), this directly implies that \( M = b^N \).

The transformative property here is that it shifts the equation from a logarithmic form, where we often have complicated terms, to a simpler exponential form. In this scenario, from the equation \( \log_{5}(x^2 - 4) = 1 \), we apply the exponential conversion:

• The base is 5, exponent is 1, and thus the expression becomes \( x^2 - 4 = 5^1 \).
• Compute the exponent to simplify it: \( x^2 - 4 = 5 \).

By converting to exponential form, it's easier to isolate \( x \) and solve for it, leading us to the next step of solving it as a quadratic equation.

Quadratic Equations

Quadratic equations are a foundational component in algebra, characterized by terms squared as the highest degree. A standard form is \( ax^2 + bx + c = 0 \). Solving them often requires isolating \( x \) and simplifying further.

When we arrived at the equation \( x^2 - 4 = 5 \), it was transformed into a more standard quadratic equation \( x^2 - 9 = 0 \) by adding 4 to both sides:

• This equation resembles \( x^2 - c = 0 \), where \( c = 9 \).
• We isolated \( x^2 \) to the left to simplify the process of finding its roots.

Taking the square root of both sides gives solutions \( x = \pm 3 \).

Don't forget to verify potential solutions in the original problem to ensure their validity, especially considering the nature of logarithms which only take positive arguments.