Problem 68 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{6}(x+5)+\log _{6} x=2 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 4\).

Step 1: Combine logarithms

Using the law of logarithms, combine \(\log _{6}(x+5)+\log _{6} x\) into one expression: \(\log _{6}((x+5)x)\).

Step 2: Convert to exponential form

Convert the logarithmic equation to an exponential equation, following the rule \(a = \log_b(c) \) translates to \(b^a = c\). Therefore, the equation transforms into \(6^2 = (x+5)x\).

Step 3: Simplify and solve for \(x\)

Expand and simplify the equation to its quadratic form: \(36 = x^2 + 5x\). Move all the terms to one side to obtain a quadratic equation and solve for \(x\), either using the quadratic formula \(x = \frac{-b ± \sqrt {b^2 - 4ac}}{2a}\) or factorisation if possible. The equation becomes: \(x^2 + 5x - 36 = 0\). Solving this, we find two potential solutions for \(x\) : \(x = -9, 4\).

Step 4: Check and reject invalid solutions

Check the solutions by substituting each into the original logarithmic expressions. Since logarithm because is only defined for positive values, we find that the solution \(x = -9\) does not satisfy the original equation while \(x = 4\) does. Thus, the solution \(x = -9\) is rejected.

Key Concepts

Logarithm Properties

In solving logarithmic equations, understanding the properties of logarithms is crucial. These properties help simplify and solve complex expressions. One of the most useful properties is the product rule of logarithms. This rule states that the sum of two logarithms with the same base is equivalent to the logarithm of the product of their arguments:

• \( \log_b(M) + \log_b(N) = \log_b(M \times N) \)

In our problem, this rule was applied to combine \( \log_{6}(x+5) \) and \( \log_{6}x \), resulting in \( \log_{6}((x+5)x) \). Another important property is the change from logarithmic to exponential form, which involves rewriting \( \log_b(c) = a \) as \( b^a = c \). This step is crucial for converting logarithmic expressions into equations more easily solvable for \(x\). By mastering these properties, solving logarithmic equations becomes straightforward and more intuitive.

Domain of Logarithmic Functions

Logarithms are defined only for positive arguments, meaning the expression inside the logarithm must be greater than zero. This characteristic defines the domain for any logarithmic function. When solving logarithmic equations, it's vital to consider this limitation.

• The domain of \( \log_6(x+5) \) is \( x+5 > 0 \), which simplifies to \( x > -5 \).
• The domain of \( \log_6x \) is simply \( x > 0 \).

Thus, the overall domain for the solution of the equation is the intersection of these conditions, specifically \( x > 0 \). During the solution process, if any potential solutions fall outside this domain, they must be rejected since they do not make sense within the context of the original logarithmic statement. This is why, after solving the quadratic equation, we discarded the negative solution \( x = -9 \), because it does not meet the domain criteria.

Quadratic Equations

Quadratic equations are commonly encountered when dealing with logarithmic expressions, especially after converting logarithmic to exponential forms. Solving a quadratic equation involves standard methods like factoring, completing the square, or applying the quadratic formula.

• A quadratic equation typically looks like \( ax^2 + bx + c = 0 \).
• In our example, we eventually arrived at \( x^2 + 5x - 36 = 0 \).
• This form allows us to solve for \( x \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our specific problem, this quadratic equation yields two potential solutions: \( x = -9 \) and \( x = 4 \). It is important to verify which solutions are valid by checking the domain of the original logarithmic expressions. Quadratic math techniques provide the backbone to many solutions, bridging the gap between logarithmic functions and their real-world applications.