Solving Logarithmic Equations Log Base 2 Of (x+6) = 2 - Log Base 2 Of (x+3)

In the realm of mathematics, solving logarithmic equations can seem daunting at first, but with a systematic approach, it becomes a manageable task. This comprehensive guide aims to equip you with the necessary tools and techniques to confidently tackle logarithmic equations. We will delve into the fundamental properties of logarithms, explore various solution strategies, and illustrate these concepts with practical examples. The focus will be on a specific equation, $\log _2(x+6)=2-\log _2(x+3)$, demonstrating a detailed, step-by-step solution to not only find the value(s) of x but also to understand the underlying principles that govern logarithmic functions.

Before we dive into the solution, let's establish the groundwork by reviewing the core properties of logarithms. A logarithm is essentially the inverse operation of exponentiation. The expression $\log_b a = c$ is equivalent to $b^c = a$, where b is the base, a is the argument, and c is the logarithm. The base b must be a positive number not equal to 1, and the argument a must be positive. These conditions are crucial, as they define the domain of logarithmic functions and influence the solutions we obtain. Understanding these fundamental rules is key to navigating the intricacies of logarithmic equations and avoiding common pitfalls, such as overlooking extraneous solutions. These are values that emerge during the solving process but do not satisfy the original equation. As we embark on this journey, keep in mind that precision and adherence to these basic rules are paramount in the quest to accurately solve logarithmic equations. Logarithmic equations appear in various scientific and engineering contexts, including calculating the pH of solutions in chemistry, modeling population growth in biology, and analyzing signal processing in electrical engineering. Understanding how to solve these equations is therefore essential for students and professionals in these fields. We will cover the strategies to ensure that the reader gains a strong grasp of logarithmic equation solving. The process involves algebraic manipulation and careful evaluation to eliminate any inconsistencies. The initial step involves transforming the equation into a manageable form, usually by condensing multiple logarithmic terms into a single logarithm, and then applying the exponential form to eliminate the logarithm altogether.

Understanding Logarithmic Properties and Transformations

To effectively solve the given logarithmic equation, $\log _2(x+6)=2-\log _2(x+3)$, a solid understanding of logarithmic properties is crucial. These properties allow us to manipulate and simplify logarithmic expressions, paving the way for isolating the variable x. One of the most fundamental properties we'll utilize is the product rule of logarithms, which states that the logarithm of the product of two numbers is equal to the sum of the logarithms of the individual numbers. Mathematically, this is expressed as $\log_b(mn) = \log_b(m) + \log_b(n)$. Conversely, the quotient rule states that the logarithm of the quotient of two numbers is equal to the difference of the logarithms: $\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)$. Another important property is the power rule, which tells us that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number: $\log_b(m^p) = p \log_b(m)$. These properties, along with the definition of a logarithm, form the cornerstone of our problem-solving approach. Another key aspect to grasp is the inverse relationship between logarithms and exponentiation. The equation $\log_b a = c$ is equivalent to $b^c = a$. This equivalence allows us to transition between logarithmic and exponential forms, which is often necessary to solve logarithmic equations. In our specific equation, we will leverage the properties of logarithms to combine logarithmic terms and subsequently convert the equation into an exponential form. This transformation is a pivotal step, as it eliminates the logarithmic functions and allows us to deal with a more familiar algebraic equation. Additionally, we must always be mindful of the domain of logarithmic functions. The argument of a logarithm must be strictly positive. This means that when solving for x, we need to check if the solutions obtained satisfy the conditions $x+6 > 0$ and $x+3 > 0$. Solutions that do not meet these criteria are extraneous and must be discarded. This step is a critical safeguard against errors and ensures that our final solutions are valid within the context of the logarithmic functions involved.

Step-by-Step Solution: Solving the Logarithmic Equation

Let's embark on the journey of solving the logarithmic equation $\log _2(x+6)=2-\log _2(x+3)$ step-by-step, applying the principles and properties we've discussed. The first crucial step is to consolidate the logarithmic terms on one side of the equation. This simplifies the equation and sets the stage for further manipulation. By adding $\log _2(x+3)$ to both sides, we achieve this consolidation: $\log _2(x+6) + \log _2(x+3) = 2$. Now, we can apply the product rule of logarithms, which states that $\log_b(m) + \log_b(n) = \log_b(mn)$. Using this property, we combine the two logarithmic terms on the left side: $\log _2((x+6)(x+3)) = 2$. This step is a pivotal transformation, condensing two logarithmic terms into a single, more manageable term. Next, we leverage the fundamental relationship between logarithms and exponentials. To eliminate the logarithm, we convert the equation from logarithmic form to exponential form. Recall that $\log_b a = c$ is equivalent to $b^c = a$. Applying this to our equation, we get: $(x+6)(x+3) = 2^2$. This transformation liberates x from the confines of the logarithmic function, allowing us to work with a simpler algebraic equation. Now, we expand the left side and simplify the equation: $x^2 + 9x + 18 = 4$. To solve for x, we need to rearrange the equation into a standard quadratic form: $x^2 + 9x + 14 = 0$. This quadratic equation can be solved using various methods, such as factoring, completing the square, or the quadratic formula. In this case, factoring is the most straightforward approach. We look for two numbers that multiply to 14 and add up to 9. These numbers are 2 and 7. Thus, we can factor the quadratic equation as: $(x+2)(x+7) = 0$. Setting each factor equal to zero yields two potential solutions: $x = -2$ and $x = -7$. However, we are not done yet. It is imperative to check these solutions against the original equation to ensure they are valid and not extraneous. Remember, the argument of a logarithm must be positive.

Verifying Solutions and Identifying Extraneous Roots

The final and arguably most critical step in solving logarithmic equations is the verification of solutions. We must meticulously check each potential solution in the original equation to ensure it doesn't lead to any undefined logarithmic terms. In our case, we obtained two potential solutions: $x = -2$ and $x = -7$. Let's substitute each of these values back into the original equation, $\log _2(x+6)=2-\log _2(x+3)$, and see if they hold true.

First, consider $x = -2$. Substituting this value into the equation, we get: $\log _2(-2+6)=2-\log _2(-2+3)$, which simplifies to $\log _2(4)=2-\log _2(1)$. Since $\log _2(4) = 2$ and $\log _2(1) = 0$, the equation becomes $2 = 2 - 0$, which is a true statement. Therefore, $x = -2$ is a valid solution.

Now, let's check $x = -7$. Substituting this value into the equation, we get: $\log _2(-7+6)=2-\log _2(-7+3)$, which simplifies to $\log _2(-1)=2-\log _2(-4)$. Here, we encounter a significant issue. The arguments of the logarithms, -1 and -4, are both negative. Logarithms are only defined for positive arguments. Therefore, $\log _2(-1)$ and $\log _2(-4)$ are undefined. This means that $x = -7$ is not a valid solution; it is an extraneous root. Extraneous roots often arise when manipulating logarithmic equations, especially when squaring both sides or combining logarithmic terms. These operations can introduce solutions that do not satisfy the original equation's domain restrictions. In summary, the only valid solution to the equation $\log _2(x+6)=2-\log _2(x+3)$ is $x = -2$. The other potential solution, $x = -7$, is an extraneous root and must be discarded. This meticulous process of verification is essential for ensuring the accuracy of our solutions and avoiding common mistakes in logarithmic equation solving.

Conclusion: Mastering Logarithmic Equations

In conclusion, solving logarithmic equations requires a blend of algebraic manipulation, understanding logarithmic properties, and careful verification of solutions. By following a systematic approach, we can confidently tackle these types of problems. We started with the logarithmic equation $\log _2(x+6)=2-\log _2(x+3)$ and demonstrated a step-by-step solution. This involved consolidating logarithmic terms using the product rule, converting the equation to exponential form, solving the resulting quadratic equation, and, most importantly, verifying the solutions to eliminate extraneous roots.

The key takeaways from this exploration are the importance of logarithmic properties, the inverse relationship between logarithms and exponentials, and the necessity of checking for extraneous solutions. The product rule of logarithms allowed us to combine logarithmic terms, simplifying the equation. The conversion to exponential form enabled us to eliminate the logarithms and work with a more familiar algebraic structure. However, the most crucial aspect was the verification step, where we identified and discarded an extraneous root. This highlights the fact that not all solutions obtained during the solving process are necessarily valid solutions to the original equation.

Mastering logarithmic equations is not just about finding the correct numerical answers; it's about developing a deep understanding of the underlying concepts and principles. This understanding empowers us to approach more complex problems with confidence and accuracy. The techniques and strategies discussed in this guide can be applied to a wide range of logarithmic equations, making it a valuable resource for students, educators, and anyone interested in mathematics. Logarithmic equations find applications in various fields, including physics, engineering, computer science, and finance. Understanding how to solve them is therefore a valuable skill in many disciplines. Through practice and consistent application of these principles, anyone can master the art of solving logarithmic equations.

Therefore, the solution to the equation $\log _2(x+6)=2-\log _2(x+3)$ is $x=-2$.