Problem 49 - 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: Apply Logarithm Property

The equation is given as \( \log_{9}(x-5) + \log_{9}(x+3) = 1 \). We can use the property of logarithms that states \( \log_{a}(b) + \log_{a}(c) = \log_{a}(b \times c) \). This allows us to combine the logs into one: \( \log_{9}((x-5)(x+3)) = 1 \).

Step 2: Convert Logarithmic Equation to Exponential Form

The equation \( \log_{9}((x-5)(x+3)) = 1 \) means that the base 9 raised to the power of 1 equals \((x-5)(x+3)\). So we have the equation \( 9^1 = (x-5)(x+3) \). Simplify this to obtain \( (x-5)(x+3) = 9 \).

Step 3: Expand and Simplify

Expand the left side to get \( x^2 + 3x - 5x - 15 = 9 \). Combine like terms to get \( x^2 - 2x - 15 = 9 \).

Step 4: Set Equation to Zero

Subtract 9 from both sides to set the equation to zero: \( x^2 - 2x - 15 - 9 = 0 \), which simplifies to \( x^2 - 2x - 24 = 0 \).

Step 5: Factor the Quadratic Equation

Find two numbers that multiply to \(-24\) and add up to \(-2\). These numbers are \(-6\) and \(4\). Thus, the equation factors as \((x-6)(x+4) = 0\).

Step 6: Solve for x

Set each factor equal to zero and solve: \(x - 6 = 0\) or \(x + 4 = 0\). Therefore, \(x = 6\) or \(x = -4\).

Step 7: Check for Extraneous Solutions

Since log terms, where the argument is negative or zero, are undefined in the real number system, we need to check that \(x\) values result in valid arguments inside the logarithms. For \(x = 6\), \(x-5 = 1\) and \(x+3 = 9\), both valid. For \(x = -4\), \(x-5 = -9\) is not valid. Therefore, we discard \(x = -4\).

Key Concepts

Properties of Logarithms

When solving logarithmic equations, understanding the properties of logarithms is crucial. Two important properties can simplify the process:

• The product property: This states that the sum of two logarithms with the same base is equivalent to the logarithm of their product. For example, \(\log_{a}(b) + \log_{a}(c) = \log_{a}(b \times c)\).
• The power property: This rule tells us that a logarithm of a power can be expressed as a product, like \(\log_{a}(b^m) = m \cdot \log_{a}(b)\).

In the provided exercise, we applied the product property to combine \(\log_{9}(x-5)\) and \(\log_{9}(x+3)\) into one single logarithm, \(\log_{9}((x-5)(x+3))\). This simplifies our problem, making it easier to solve.

Exponential Form

Once we have the equation \(\log_{9}((x-5)(x+3)) = 1\), the next step is to convert it to its exponential form. This transformation leverages the definition of a logarithm, which asserts that if \(\log_{a}(b) = c\), then \(a^c = b\).

For our problem, the logarithmic equation \(\log_{9}((x-5)(x+3)) = 1\) translates to the exponential form \(9^1 = (x-5)(x+3)\). This conversion is vital, as it strips away the logarithmic function, allowing direct manipulation of the equation.

With this exponential form, we're prepared to simplify and solve the equation, stepping away from any complexities tied to logarithms.

Quadratic Equations

After converting the logarithmic equation to exponential form, you're typically left with an algebraic equation to solve. Here, it simplifies into a quadratic form, \((x-5)(x+3) = 9\), which upon expansion results in \(x^2 - 2x - 24 = 0\).

Quadratic equations like this one can usually be solved by factoring, completing the square, or using the quadratic formula. However, for this problem, factoring is the most straightforward approach.

The key is identifying which numbers add up to \(-2\) and multiply to \(-24\). Factors of \(-24\) such as \(-6\) and \(4\) satisfy this condition, enabling us to factor the equation as \((x-6)(x+4) = 0\). Solving these factors provides potential solutions for \(x\).

Extraneous Solutions

Extraneous solutions are values that emerge during the solving process but don't satisfy the original equation. It's vital to check each potential solution in the context of the original problem, especially for logarithmic equations.

Logarithms require positive arguments, as they are undefined for negative numbers or zero in real numbers. For our solutions, \(x = 6\) and \(x = -4\), careful consideration is necessary:

• For \(x = 6\): Substitution results in \(x-5 = 1\) and \(x+3 = 9\); both are positive, making it a valid solution.
• For \(x = -4\): Substitution gives \(x-5 = -9\), a negative value, which is invalid.

Hence, we discard \(x = -4\) because it yields an invalid logarithmic argument. Only \(x = 6\) will be the solution.