Problem 90 - Intermediate Algebra

Question

For Problems \(81-97\), solve each of the equations. $$ \log _{6} x+\log _{6}(x+5)=2 $$

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Apply Logarithm Product Rule

Start by applying the logarithm product rule to combine the logs. Recall that \( \log_b(a) + \log_b(c) = \log_b(a \cdot c) \). Here, we have \( \log_6 x + \log_6(x+5) = \log_6(x(x+5)) \). Thus, the equation simplifies to \( \log_6(x^2 + 5x) = 2 \).

Step 2: Exponentiate to Remove Logarithm

To solve for \(x\), remove the logarithm by rewriting the equation in exponential form. If \( \log_b(a) = c \), then \( a = b^c \). Thus, \( x^2 + 5x = 6^2 \), which further simplifies to \( x^2 + 5x = 36 \).

Step 3: Rearrange to Form a Quadratic Equation

Rearrange the equation \( x^2 + 5x = 36 \) to set it to zero, resulting in \( x^2 + 5x - 36 = 0 \). This forms a quadratic equation.

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \( x^2 + 5x - 36 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -36 \). Calculate the discriminant \( b^2 - 4ac = 25 + 144 = 169 \).

Step 5: Calculate the Roots

Substitute into the quadratic formula: \( x = \frac{-5 \pm \sqrt{169}}{2} \), which simplifies to \( x = \frac{-5 \pm 13}{2} \). Therefore, the roots are \( x = 4 \) and \( x = -9 \).

Step 6: Verify Solutions

Check for extraneous solutions by substituting back into the original equation: \(\log_6 4 + \log_6(4+5) = \log_6 4 + \log_6 9 = \log_6(36) = 2\), which satisfies the equation. However, \(x = -9\) is not valid because logarithms of negative numbers are undefined.

Key Concepts

Logarithm Product Rule

When working with logarithmic equations, the logarithm product rule is a fundamental concept. This rule helps to consolidate multiple logarithmic expressions by combining them into one. The logarithm product rule is expressed as

• \( \log_b(a) + \log_b(c) = \log_b(a \cdot c) \)

If you ever encounter a situation where you have two logs being summed together, then you can simplify the equation by combining them into a single logarithm. In our given problem, \( \log_6 x + \log_6(x+5) \) was simplified to \( \log_6(x(x+5)) \) using this rule. This transformation makes it easier to handle even complex equations.

Keep in mind that this rule only applies when the logarithms share the same base. Thus, it is a crucial step that sets the stage for further simplifications in solving logarithmic equations.

Quadratic Equations

Once we've applied the logarithm product rule, often you'll end up with an equation that needs rearranging. In our case, this results in a quadratic equation. Quadratic equations are any polynomial equations of the form

• \( ax^2 + bx + c = 0 \)

where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations can be solved through various methods: factorization, completing the square, or using the quadratic formula.

In our problem, the quadratic formula was used:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

which led to the solutions \(x = 4\) and \(x = -9\). This method is particularly useful when the quadratic does not factor neatly, providing a straightforward way to find the roots.

Extraneous Solutions

An important consideration when solving equations, especially those involving logarithms, is the possibility of extraneous solutions. An extraneous solution is a solution derived from the algebraic manipulation of an equation, which does not actually satisfy the original equation.

After solving the quadratic equation, it is crucial to verify if the solutions are valid. Substituting back into the original logarithmic expression helps in this verification process. In our problem, after solving, we obtained \(x = 4\) and \(x = -9\).

The solution \(x = 4\) verified correctly, whereas \(x = -9\) did not—since logarithms of negative values are undefined. Therefore, \(x = -9\) is identified as an extraneous solution. Logarithmic functions are only defined for positive numbers, hence double-checking is key to ensure the validity of the solutions.