Problem 26 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$ \log _{6}(x+5)+\log _{6} x=2 $$

Step-by-Step Solution

Answer

The solution is \(x = 4\).

Step 1: Apply Logarithm Property

The equation is given as \( \log _{6}(x+5) + \log _{6} x = 2 \). Use the logarithmic property \( \log_b A + \log_b B = \log_b (A \times B) \). This combines the two logarithms into one: \( \log_6((x+5)x) = 2 \).

Step 2: Eliminate the Logarithm

To eliminate the logarithm, we rewrite the equation in exponential form. This gives \((x+5)x = 6^2\). Calculating \(6^2\), we get 36, so now our equation is \((x+5)x = 36\).

Step 3: Solve the Quadratic Equation

The expression \((x+5)x\) expands to \(x^2 + 5x\). Our equation is now \(x^2 + 5x = 36\). Rearrange this to form a standard quadratic equation: \(x^2 + 5x - 36 = 0\).

Step 4: Factor the Quadratic Equation

To factor \(x^2 + 5x - 36\), we need two numbers that multiply to \(-36\) and add to \(5\). These numbers are \(9\) and \(-4\). Thus, the equation factors to \((x+9)(x-4) = 0\).

Step 5: Solve for x

Set each factor to zero: \(x+9 = 0\) or \(x-4 = 0\). Solving these gives \(x = -9\) and \(x = 4\).

Step 6: Check for Valid Solutions

Insert the solutions back into the original equation to check for validity: \(x = -9\) is not valid since \(\log_6(-9+5)\) is undefined. For \(x = 4\), both logarithms are defined, making \(x = 4\) the valid solution.

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools in simplifying and solving logarithmic equations. The property used in this exercise is the product property of logarithms. It states that the sum of two logarithms with the same base can be rewritten as a single logarithm of the product of their arguments:

• \( \log_b A + \log_b B = \log_b (A \times B) \)

In this problem, the logarithmic property helps us combine the logs:

• Given \( \log_6(x+5) + \log_6 x = 2 \)
• We combine into \( \log_6((x+5)x) = 2 \)

This property simplifies the equation and makes it possible to solve for \(x\) more easily. Knowing and applying these properties allows for efficient manipulation of logarithmic terms.

Exponential Form

Changing a logarithmic expression to exponential form is a critical step in solving these types of equations. The idea is that if \( \log_b A = C \), it can be expressed in exponential form as:

• \( b^C = A \)

In the given exercise, after combining the logarithms, we have:

• \( \log_6((x+5)x) = 2 \)

This can be converted to exponential form:

• \( (x+5)x = 6^2 \)
• Calculating \(6^2\), we find \(36\)

The exponential form transitions the problem from a logarithmic equation to an algebraic equation, easily solved using familiar techniques.

Quadratic Equations

When the logarithmic equation is transformed to an algebraic one, it results in a quadratic equation. A quadratic equation has the standard form:

• \( ax^2 + bx + c = 0 \)

For this particular exercise, substituting into the equation of the form gives:

• \( x^2 + 5x = 36 \)

Rearranging terms, we get:

• \( x^2 + 5x - 36 = 0 \)

Quadratic equations can be solved through various methods like factoring, completing the square, or using the quadratic formula. In this exercise, the quadratic equation is solved by factoring.

Factoring

Factoring is a method used to solve quadratic equations. It involves rewriting the equation in terms of its roots or factors. Here, factors of the quadratic \(x^2 + 5x - 36\) are required such that they multiply to \(-36\) and add to \(5\). The steps are as follows:

• Identify two numbers that multiply to \(-36\) and sum to \(5\): \(9\) and \(-4\)
• Rewrite the quadratic equation as a product of its factors: \((x+9)(x-4) = 0\)

Now, by setting each factor equal to zero and solving, \(x+9 = 0\) or \(x-4 = 0\), we find the solutions: \(x = -9\) and \(x = 4\). A crucial final step is to check for valid solutions by substituting them back into the original equation. This ensures no errors due to the nature of logarithms' domains. Here, \(x = 4\) is valid, while \(x = -9\) is not, as negatives aren't permissible in logs with real number bases.