Solving Logarithmic Equations A Step By Step Guide

Introduction

In this comprehensive guide, we will delve into the intricacies of solving logarithmic equations, specifically focusing on the equation ${\log _4(x^2-6x+9) = \log _4(-x+3)}$. Logarithmic equations often present a challenge due to the nature of logarithms and the need to carefully consider domain restrictions. However, by breaking down the problem into manageable steps, we can systematically arrive at the solution. Our primary focus will be on understanding the properties of logarithms, applying algebraic techniques, and, most importantly, verifying the validity of our solutions within the domain of the logarithmic functions involved. Remember, a thorough understanding of these concepts is crucial not only for solving this particular equation but also for tackling a wide range of logarithmic problems. We will explore the domain restrictions imposed by logarithmic functions, the algebraic manipulations required to simplify the equation, and the verification process to ensure our solutions are valid. So, let's embark on this journey to unravel the solution to this intriguing logarithmic equation.

Understanding the Basics of Logarithmic Equations

Before we dive into the specifics of our equation, let's take a moment to solidify our understanding of logarithmic equations in general. A logarithm is essentially the inverse operation of exponentiation. The expression ${\log_b a = c}$ means that ${b^c = a}$, where ${b}$ is the base, ${a}$ is the argument, and ${c}$ is the exponent. The key to solving logarithmic equations lies in understanding this relationship and the properties that govern logarithms.

One of the most critical aspects of working with logarithms is the domain restriction. The argument of a logarithm must always be positive. This is because we can only raise a positive base to a power and obtain a positive result. Therefore, when dealing with equations involving logarithms, we must always ensure that the expressions inside the logarithms are greater than zero. This constraint will play a vital role in verifying the solutions we obtain later in the process. Another fundamental property that we'll use is that if ${\log_b a = \log_b c}$, then ${a = c}$, provided that ${a}$, ${c}$, and ${b}$ satisfy the domain restrictions of the logarithmic functions. This property allows us to eliminate the logarithms and work with a simpler algebraic equation. Furthermore, understanding the properties of exponents is crucial as they are intimately linked to logarithms. For example, the power rule of logarithms, which states that ${\log_b(a^k) = k \log_b(a)}$, can be useful in simplifying certain logarithmic expressions. Keeping these fundamental principles in mind will enable us to approach logarithmic equations with confidence and clarity.

Solving the Equation ${\log _4(x^2-6x+9) = \log _4(-x+3)}$

Now, let's tackle the specific equation at hand: ${\log _4(x^2-6x+9) = \log _4(-x+3)}$. The first step in solving this equation is to recognize that we have logarithms with the same base on both sides. This allows us to apply the property that if ${\log_b a = \log_b c}$, then ${a = c}$. Therefore, we can equate the arguments of the logarithms:

${x^2 - 6x + 9 = -x + 3}$

This simplifies the problem from a logarithmic equation to a quadratic equation. The next step is to rearrange the terms and set the equation to zero:

${x^2 - 6x + 9 + x - 3 = 0}$

${x^2 - 5x + 6 = 0}$

Now, we have a standard quadratic equation that we can solve by factoring. We are looking for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Therefore, we can factor the quadratic as:

${(x - 2)(x - 3) = 0}$

This gives us two potential solutions for ${x}$: ${x = 2}$ and ${x = 3}$. However, it is crucial to remember that we must verify these solutions against the domain restrictions of the original logarithmic equation. We'll delve into the verification process in the next section.

Verifying the Solutions and Checking for Extraneous Roots

As we've identified two potential solutions, ${x = 2}$ and ${x = 3}$, it is now paramount to verify whether these solutions are valid within the context of the original equation, ${\log _4(x^2-6x+9) = \log _4(-x+3)}$. This verification process is essential because logarithmic functions have domain restrictions. Specifically, the argument of a logarithm must be strictly greater than zero. In other words, we need to ensure that both ${x^2 - 6x + 9 > 0}$ and ${-x + 3 > 0}$ for our solutions to be valid. Let's examine each potential solution individually.

For ${x = 2}$:

• Let's substitute ${x = 2}$ into the first argument: ${x^2 - 6x + 9 = (2)^2 - 6(2) + 9 = 4 - 12 + 9 = 1}$. Since 1 is greater than 0, this condition is satisfied.
• Now, let's substitute ${x = 2}$ into the second argument: ${-x + 3 = -(2) + 3 = 1}$. Again, 1 is greater than 0, so this condition is also satisfied.

Therefore, ${x = 2}$ is a valid solution.

Now, let's consider ${x = 3}$:

• Substituting ${x = 3}$ into the first argument: ${x^2 - 6x + 9 = (3)^2 - 6(3) + 9 = 9 - 18 + 9 = 0}$. Notice that the result is 0, which is not strictly greater than 0. Thus, this condition is not satisfied.

Since one of the domain restrictions is violated, ${x = 3}$ is an extraneous root and is not a valid solution to the equation. This step highlights the critical importance of checking for extraneous roots in logarithmic equations. If we had not performed this verification, we would have incorrectly included ${x = 3}$ in our solution set. Extraneous roots are values that emerge during the solving process but do not actually satisfy the original equation due to domain restrictions or other constraints. The process of checking for extraneous roots is not merely a formality; it is a fundamental step in ensuring the accuracy of our solution.

Final Solution

After carefully solving the equation ${\log _4(x^2-6x+9) = \log _4(-x+3)}$ and, crucially, verifying our solutions against the domain restrictions, we have arrived at the final answer. We initially obtained two potential solutions: ${x = 2}$ and ${x = 3}$. However, upon substituting these values back into the original equation and checking the arguments of the logarithms, we discovered that ${x = 3}$ leads to a zero argument, which violates the domain of logarithmic functions. Therefore, ${x = 3}$ is an extraneous root and must be discarded.

This leaves us with only one valid solution: ${x = 2}$. This value satisfies both the equation and the domain restrictions, making it the definitive solution to the problem. Therefore:

${x = 2}$

In summary, the solution to the equation ${\log _4(x^2-6x+9) = \log _4(-x+3)}$ is ${x = 2}$. This process underscores the importance of a meticulous approach when solving logarithmic equations, emphasizing not only the algebraic manipulations but also the critical step of verifying solutions to avoid extraneous roots. The ability to accurately solve logarithmic equations is a valuable skill in various mathematical and scientific contexts. By understanding the properties of logarithms, applying algebraic techniques, and rigorously verifying solutions, we can confidently tackle even the most challenging logarithmic problems. The key takeaway is that the process is just as important as the answer, and a thorough understanding of the underlying principles will serve you well in future mathematical endeavors. Always remember to check your work and verify your solutions, especially when dealing with functions that have domain restrictions, such as logarithms.

Conclusion

In this detailed exploration, we have successfully navigated the intricacies of solving the logarithmic equation ${\log _4(x^2-6x+9) = \log _4(-x+3)}$. We began by establishing a solid foundation in the fundamentals of logarithmic equations, emphasizing the critical role of domain restrictions and the fundamental properties that govern logarithmic functions. We then methodically applied these principles to simplify the given equation, transforming it from a logarithmic problem into a more manageable quadratic equation. By factoring the quadratic, we identified two potential solutions: ${x = 2}$ and ${x = 3}$.

However, the process did not conclude there. A crucial step in solving logarithmic equations is the verification of solutions. We diligently checked each potential solution against the domain restrictions of the original logarithmic equation. This step revealed that while ${x = 2}$ satisfied all conditions and was indeed a valid solution, ${x = 3}$ resulted in a zero argument for one of the logarithms, thereby violating the domain restriction. Consequently, ${x = 3}$ was identified as an extraneous root and discarded.

Our final solution, therefore, is ${x = 2}$. This journey underscores the importance of a comprehensive approach when solving mathematical problems, particularly those involving logarithmic functions. It highlights the need to not only apply algebraic techniques correctly but also to maintain a vigilant awareness of the underlying principles and restrictions that govern the functions involved. The ability to solve logarithmic equations accurately is a valuable asset in mathematics and its applications. By mastering the concepts and techniques discussed in this guide, you will be well-equipped to tackle a wide range of logarithmic problems with confidence and precision. Remember, practice makes perfect, and the more you engage with these concepts, the more proficient you will become. The key is to always approach each problem with a clear understanding of the fundamentals and a meticulous attention to detail, ensuring that every step is justified and every solution is verified. This approach will not only lead to correct answers but also foster a deeper appreciation for the elegance and power of mathematics.