Solving Logarithmic Equations A Step By Step Guide

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Introduction

Logarithmic equations, a fundamental concept in mathematics, often pose challenges for students and enthusiasts alike. In this comprehensive guide, we will delve into the intricacies of solving logarithmic equations, focusing on a specific example to illustrate the step-by-step process. Our main goal is to provide a clear and concise explanation that empowers readers to confidently tackle these equations. Logarithmic equations are equations in which the logarithm of an expression containing a variable appears. To solve these equations, we must use the properties of logarithms to isolate the variable. However, it's critical to remember that not all solutions obtained algebraically are valid. We need to check our solutions against the original equation to eliminate extraneous solutions, which arise due to the domain restrictions of logarithmic functions. These functions are only defined for positive arguments, so any solution that makes the argument of a logarithm negative or zero must be discarded. This article will walk you through a detailed example, explaining each step in a manner that is easy to understand and apply to other logarithmic problems. By understanding the core principles and methodologies outlined here, you'll be well-equipped to solve a wide range of logarithmic equations accurately and efficiently.

Understanding Logarithms

Before we dive into solving the equation, let's quickly recap what logarithms are. A logarithm is the inverse operation to exponentiation. In simple terms, the logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. Mathematically, if by=xb^y = x, then log⁡b(x)=y\log_b(x) = y. Here, 'b' is the base of the logarithm, 'x' is the argument, and 'y' is the logarithm itself. Understanding the basic properties of logarithms is crucial for solving logarithmic equations. Some key properties include the product rule, quotient rule, and power rule. The product rule states that the logarithm of the product of two numbers is the sum of their logarithms: log⁡b(mn)=log⁡b(m)+log⁡b(n)\log_b(mn) = \log_b(m) + \log_b(n). The quotient rule states that the logarithm of the quotient of two numbers is the difference of their logarithms: log⁡b(mn)=log⁡b(m)−log⁡b(n)\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n). The power rule states that the logarithm of a number raised to a power is the product of the power and the logarithm of the number: log⁡b(mp)=plog⁡b(m)\log_b(m^p) = p \log_b(m). These properties allow us to manipulate logarithmic expressions, combining or separating them as needed to simplify equations. Furthermore, the definition of a logarithm implies that the argument of a logarithm must be positive. This is a critical consideration when solving equations, as we must always check our solutions to ensure they do not result in taking the logarithm of a non-positive number. Neglecting this step can lead to incorrect solutions and a misunderstanding of the true nature of the logarithmic function. By mastering these fundamental concepts and properties, you can approach logarithmic equations with greater confidence and accuracy.

The Problem: log⁡4(x2+1)=log⁡4(−2x)\log_4(x^2+1)=\log_4(-2x)

Now, let's address the problem at hand: log⁡4(x2+1)=log⁡4(−2x)\log_4(x^2+1)=\log_4(-2x). This equation presents a classic scenario where we can leverage the properties of logarithms to find the solution. The equation involves logarithms with the same base (base 4), which simplifies our task significantly. The presence of logarithmic terms on both sides suggests that we can equate the arguments of the logarithms, provided the arguments are within the domain of the logarithmic function. The first step in solving this equation is to recognize that if log⁡b(A)=log⁡b(B)\log_b(A) = \log_b(B), then A = B, provided A and B are both positive. This principle is fundamental because the logarithm function is one-to-one for a given base. Applying this principle to our equation, we can set the arguments equal to each other: x2+1=−2xx^2 + 1 = -2x. This transforms our logarithmic equation into a quadratic equation, which is a more familiar form to solve. However, it is crucial to remember that this step is only valid if both x2+1x^2 + 1 and −2x-2x are positive. This condition is essential for the existence of the logarithms in the original equation. Ignoring this condition can lead to extraneous solutions, which are solutions that satisfy the transformed equation but not the original logarithmic equation. By paying careful attention to the domain restrictions of logarithms, we can ensure that we only accept solutions that are mathematically valid. The next step involves solving the quadratic equation, which will give us potential values for 'x'. We will then need to check these values in the original logarithmic equation to confirm their validity. This meticulous approach is the key to correctly solving logarithmic equations and avoiding common pitfalls.

Solving the Equation Step-by-Step

  1. Equate the arguments: Since the bases of the logarithms are the same, we can equate the expressions inside the logarithms: x2+1=−2xx^2 + 1 = -2x.
  2. Rearrange into a quadratic equation: To solve for x, we need to rearrange the equation into a standard quadratic form: x2+2x+1=0x^2 + 2x + 1 = 0. This is achieved by adding 2x to both sides of the equation. Recognizing the quadratic form is crucial because it allows us to apply standard techniques for solving such equations, such as factoring or using the quadratic formula. The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants. In our case, a = 1, b = 2, and c = 1. By bringing all terms to one side, we create a scenario where we can analyze the equation as a whole and identify potential solutions. The next step is to find the roots of this quadratic equation, which are the values of 'x' that make the equation true. These roots are potential solutions to our original logarithmic equation. However, it's important to emphasize that not all roots of the quadratic equation will necessarily be solutions to the logarithmic equation. This is why checking our solutions in the original equation is a critical step in the process. By carefully rearranging the equation into standard form, we set the stage for applying effective solving methods and moving closer to the correct solution.
  3. Factor the quadratic equation: The left side of the equation is a perfect square trinomial: (x+1)2=0(x + 1)^2 = 0. This factorization significantly simplifies the process of finding the roots of the equation. Recognizing perfect square trinomials is a valuable skill in algebra, as it allows for quick and efficient factorization. In general, a perfect square trinomial is a trinomial that can be factored into the square of a binomial. In our case, x2+2x+1x^2 + 2x + 1 fits this pattern perfectly. The factored form, (x+1)2=0(x + 1)^2 = 0, tells us that the expression (x + 1) multiplied by itself equals zero. This can only happen if (x + 1) itself is equal to zero. This observation leads us directly to the solution for 'x'. The simplicity of this step highlights the power of factorization as a problem-solving technique. By rewriting the quadratic equation in its factored form, we transform a potentially complex problem into a straightforward one. The next step is to solve for 'x' by setting the factor equal to zero, which will give us the potential solution to our equation. However, as always, we must remember to check this solution against the original equation to ensure its validity. This rigorous approach is essential for accurate and reliable problem-solving in mathematics. By mastering factorization techniques, we can tackle a wide range of algebraic problems with greater ease and efficiency.
  4. Solve for x: Taking the square root of both sides, we get x+1=0x + 1 = 0, so x=−1x = -1. This step is a direct consequence of the factored form of the quadratic equation. When we have an expression squared equal to zero, the only way this can be true is if the expression itself is zero. Therefore, (x+1)2=0(x + 1)^2 = 0 implies that x+1=0x + 1 = 0. Solving this simple linear equation for 'x' is a straightforward process. We subtract 1 from both sides to isolate 'x', resulting in the solution x=−1x = -1. This value is a potential solution to our original logarithmic equation. However, it is crucial to emphasize the word "potential" here. We have found a value of 'x' that satisfies the quadratic equation we derived from the logarithmic equation, but we have not yet verified whether it satisfies the logarithmic equation itself. This verification step is essential because logarithmic functions have domain restrictions. Specifically, the argument of a logarithm must be positive. If substituting x=−1x = -1 into the original logarithmic equation results in taking the logarithm of a non-positive number, then x=−1x = -1 is an extraneous solution and must be discarded. Therefore, the next critical step is to check this potential solution in the original equation to ensure its validity. This meticulous approach is a hallmark of careful and accurate mathematical problem-solving.
  5. Check the solution: Substitute x=−1x = -1 into the original equation: log⁡4((−1)2+1)=log⁡4(−2(−1))\log_4((-1)^2+1)=\log_4(-2(-1)). This gives us log⁡4(2)=log⁡4(2)\log_4(2) = \log_4(2), which is true. However, we must also check if the arguments of the logarithms are positive. (−1)2+1=2(-1)^2 + 1 = 2 which is positive, and −2(−1)=2-2(-1) = 2 which is also positive. This check is a crucial step in solving logarithmic equations. While we have found a value of 'x' that satisfies the equation obtained by equating the arguments, it is imperative to verify that this value also satisfies the original logarithmic equation. This verification process involves substituting the potential solution back into the original equation and checking two key conditions. First, we need to ensure that the equation holds true after the substitution. In our case, substituting x=−1x = -1 leads to log⁡4(2)=log⁡4(2)\log_4(2) = \log_4(2), which is indeed a true statement. However, this is not the only condition we need to check. The second, and equally important, condition is that the arguments of all logarithms in the original equation must be positive after the substitution. This condition arises from the fundamental definition of logarithms, which are only defined for positive arguments. If substituting the value of 'x' leads to taking the logarithm of a negative number or zero, then that value is an extraneous solution and must be rejected. In our case, we need to check both x2+1x^2 + 1 and −2x-2x for positivity when x=−1x = -1. We find that (−1)2+1=2(-1)^2 + 1 = 2 and −2(−1)=2-2(-1) = 2, both of which are positive. Therefore, the solution x=−1x = -1 satisfies both conditions and is a valid solution to the original logarithmic equation. This thorough checking process ensures that we arrive at the correct answer and avoid the pitfalls of extraneous solutions.

Conclusion

Therefore, the solution to the equation log⁡4(x2+1)=log⁡4(−2x)\log_4(x^2+1)=\log_4(-2x) is x=−1x = -1. Option B is the correct answer. Solving logarithmic equations requires a systematic approach. It is essential to understand the properties of logarithms and to check solutions to avoid extraneous results. The process we followed demonstrates the importance of each step, from equating arguments to verifying the final solution. By mastering these techniques, you can confidently tackle a wide range of logarithmic problems. The key takeaways from this example include: 1) Equating the arguments of logarithms with the same base is a valid step, but only when the arguments are within the domain of the logarithmic function. 2) Always check your solutions in the original equation to eliminate extraneous solutions. 3) A thorough understanding of the properties of logarithms is crucial for solving logarithmic equations effectively. 4) Practice is essential for mastering these concepts and building confidence in your problem-solving abilities. By applying these principles, you can successfully navigate the challenges of logarithmic equations and achieve accurate results. Logarithmic equations are a fundamental topic in mathematics, and mastering them opens the door to more advanced concepts and applications. From calculus to complex analysis, logarithms play a crucial role in various areas of mathematics and its applications in science and engineering. Therefore, investing time and effort in understanding and solving logarithmic equations is a valuable endeavor for any student or professional in a technical field.