Solving Logarithmic Equations Find The Solution To Log3(x² + 6x) = Log3(2x + 12)

In this comprehensive article, we will delve into the step-by-step solution of the logarithmic equation ${ \log_3(x^2 + 6x) = \log_3(2x + 12) }$. This type of problem is common in algebra and calculus, and mastering the techniques to solve it is crucial for students and enthusiasts alike. We will not only find the solution but also discuss the underlying principles and potential pitfalls in solving logarithmic equations. Our goal is to provide a clear, detailed explanation that will help you understand the process thoroughly. The question requires us to identify the correct value of ${ x }$ from the given options that satisfies the equation. The options are A. ${ x = -6 }$, B. ${ x = -2 }$, C. ${ x = 0 }$, D. ${ x = 2 }$, and E. ${ x = 6 }$. Let's embark on this mathematical journey together.

Understanding Logarithmic Equations

Before we dive into solving the specific equation, it's essential to understand the fundamentals of logarithmic equations. A logarithmic equation is an equation that involves a logarithm of an expression containing a variable. The key to solving these equations is to use the properties of logarithms to simplify the equation and isolate the variable. One of the most important properties we will use is that if ${ \log_b(A) = \log_b(B) }$, then ${ A = B }$, provided that ${ A }$ and ${ B }$ are positive and ${ b > 0 }$, ${ b \neq 1 }$. This property allows us to eliminate the logarithms and work with a simpler algebraic equation. However, it is crucial to check our solutions at the end to ensure they are valid within the domain of the logarithmic functions involved. Logarithmic functions are only defined for positive arguments, so any solution that results in a negative or zero argument for any logarithm in the original equation must be discarded. This step is often overlooked but is vital for obtaining the correct solution. Failing to check for extraneous solutions can lead to choosing an incorrect answer, particularly in multiple-choice scenarios like the one presented. Moreover, understanding the domain restrictions of logarithmic functions is not just important for solving equations but also for graphing logarithmic functions and understanding their behavior.

Step-by-Step Solution

Let's solve the equation ${ \log_3(x^2 + 6x) = \log_3(2x + 12) }$ step by step. First, we apply the property that if ${ \log_b(A) = \log_b(B) }$, then ${ A = B }$. This gives us:

${ x^2 + 6x = 2x + 12 }$

Next, we rearrange the equation to form a quadratic equation by moving all terms to one side:

${ x^2 + 6x - 2x - 12 = 0 }$

${ x^2 + 4x - 12 = 0 }$

Now, we factor the quadratic equation. We are looking for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2. So, we can factor the quadratic as:

${ (x + 6)(x - 2) = 0 }$

Setting each factor equal to zero gives us two potential solutions:

${ x + 6 = 0 \quad \text{or} \quad x - 2 = 0 }$

${ x = -6 \quad \text{or} \quad x = 2 }$

Checking for Extraneous Solutions

Now, we must check these solutions in the original equation to ensure they are valid. Remember, the argument of a logarithm must be positive.

First, let's check ${ x = -6 }$:

For the left side of the equation, we have:

${ x^2 + 6x = (-6)^2 + 6(-6) = 36 - 36 = 0 }$

Since the argument of the logarithm becomes 0, ${ x = -6 }$ is not a valid solution because logarithms are not defined for non-positive arguments. This highlights the critical importance of checking for extraneous solutions in logarithmic equations.

Now, let's check ${ x = 2 }$:

For the left side of the equation, we have:

${ x^2 + 6x = (2)^2 + 6(2) = 4 + 12 = 16 }$

For the right side of the equation, we have:

${ 2x + 12 = 2(2) + 12 = 4 + 12 = 16 }$

Since both arguments are positive when ${ x = 2 }$, this is a valid solution. This step demonstrates how substituting the potential solution back into the original equation helps us confirm its validity.

The Final Answer

Therefore, the only solution to the equation ${ \log_3(x^2 + 6x) = \log_3(2x + 12) }$ is ${ x = 2 }$. This corresponds to option D. We have shown through a detailed, step-by-step process how to arrive at this solution, emphasizing the importance of understanding logarithmic properties and checking for extraneous solutions. This comprehensive approach ensures accuracy and builds a solid understanding of the underlying mathematical principles. In summary, solving logarithmic equations requires careful application of logarithmic properties, algebraic manipulation, and, most importantly, verification of solutions to avoid extraneous results. This process not only solves the specific problem at hand but also enhances overall problem-solving skills in mathematics.

Why Other Options Are Incorrect

It's also beneficial to understand why the other options are incorrect. This reinforces the importance of the steps we took to arrive at the correct answer.

• Option A: ${ x = -6 }$: As we demonstrated earlier, substituting ${ x = -6 }$ into the original equation results in the argument of the logarithm being zero, which is undefined. Therefore, ${ x = -6 }$ is an extraneous solution.

• Option B: ${ x = -2 }$: If we substitute ${ x = -2 }$ into the original equation, we get:

  For the left side: ${ x^2 + 6x = (-2)^2 + 6(-2) = 4 - 12 = -8 }$. Since the argument is negative, ${ x = -2 }$ is not a valid solution.

• Option C: ${ x = 0 }$: Substituting ${ x = 0 }$ gives:

  For the left side: ${ x^2 + 6x = (0)^2 + 6(0) = 0 }$. Again, the argument is zero, making ${ x = 0 }$ an extraneous solution.

• Option E: ${ x = 6 }$: While ${ x = 6 }$ might seem like a plausible solution, substituting it into the equation yields:

  For the left side: ${ x^2 + 6x = (6)^2 + 6(6) = 36 + 36 = 72 }$ For the right side: ${ 2x + 12 = 2(6) + 12 = 12 + 12 = 24 }$

  Since ${ \log_3(72) \neq \log_3(24) }$, ${ x = 6 }$ is not a solution.

This analysis of each option provides a comprehensive understanding of why only ${ x = 2 }$ satisfies the original equation. Understanding why incorrect options are wrong is as important as knowing why the correct answer is right, as it deepens the understanding of the problem-solving process.

Conclusion

In conclusion, the only solution to the logarithmic equation ${ \log_3(x^2 + 6x) = \log_3(2x + 12) }$ is ${ x = 2 }$. This solution was obtained by applying the properties of logarithms, solving the resulting quadratic equation, and, crucially, checking for extraneous solutions. The detailed step-by-step process outlined in this article provides a clear methodology for solving similar logarithmic equations. Remember, solving mathematical problems, especially those involving logarithms, requires not only a solid understanding of the underlying principles but also careful attention to detail and a systematic approach. By mastering these techniques, you can confidently tackle a wide range of mathematical challenges.

Key takeaways from this exercise include:

1. Understanding the properties of logarithms is essential for simplifying and solving logarithmic equations.
2. Always rearrange the equation into a standard form, such as a quadratic equation, to facilitate solving.
3. Factoring or using the quadratic formula are common methods for solving quadratic equations.
4. Checking for extraneous solutions is a critical step in solving logarithmic equations, as logarithms are only defined for positive arguments.
5. Substituting potential solutions back into the original equation is the best way to verify their validity.

By following these steps and understanding the underlying concepts, you can confidently solve logarithmic equations and avoid common pitfalls. This comprehensive guide aims to provide you with the necessary tools and knowledge to excel in this area of mathematics.