Solving Logarithmic Equations Log_3(v^2+v-5) = Log_3(-3v-8)

Introduction

In mathematics, solving equations is a fundamental skill, and logarithmic equations are a significant part of this domain. Logarithmic equations appear in various fields, including physics, engineering, and computer science, making it crucial to understand how to solve them. This article delves into the process of solving the logarithmic equation ${\log _3(v^2+v-5) = \log _3(-3v-8)}$. We will break down each step, ensuring clarity and a thorough understanding. Our primary keyword, logarithmic equations, will be emphasized throughout this guide. Grasping the concepts presented here will empower you to tackle similar problems with confidence.

Logarithmic equations are mathematical expressions where the logarithm of one or more terms is involved. To solve such equations effectively, one needs to be adept at applying the properties of logarithms and algebraic manipulations. The equation we are addressing, ${\log _3(v^2+v-5) = \log _3(-3v-8)}$, is a classic example where understanding these properties is key. The base of the logarithm here is 3, and we need to find the value(s) of ${v}$ that satisfy the equation. The process involves equating the arguments of the logarithms, forming a quadratic equation, and then solving for ${v}$. However, it is crucial to check the solutions to ensure they do not result in taking the logarithm of a negative number or zero, as this is undefined. This step-by-step approach will help in accurately finding the solution.

Logarithmic equations often require careful attention to detail, especially when dealing with the domain of logarithmic functions. Remember, the argument of a logarithm must always be positive. This constraint is essential when verifying potential solutions. For example, if a solution for ${v}$ makes the expression inside the logarithm negative, it must be discarded. The given equation, ${\log _3(v^2+v-5) = \log _3(-3v-8)}$, exemplifies this requirement perfectly. We begin by equating the expressions inside the logarithms: ${v^2 + v - 5 = -3v - 8}$. This leads to a quadratic equation, which can be solved using various methods such as factoring, completing the square, or using the quadratic formula. After obtaining the potential solutions, we must substitute them back into the original logarithmic equation to confirm their validity. This verification step is vital to avoid extraneous solutions, which are values that satisfy the transformed equation but not the original logarithmic equation. Thus, understanding and applying this verification process is a critical component of solving logarithmic equations successfully.

Step-by-Step Solution

1. Equate the Arguments

The fundamental property of logarithms states that if ${\log_b(x) = \log_b(y)}$, then ${x = y}$, provided that both ${x}$ and ${y}$ are positive and ${b}$ is a positive base not equal to 1. Applying this to our equation, ${\log _3(v^2+v-5) = \log _3(-3v-8)}$, we equate the arguments:

${v^2 + v - 5 = -3v - 8}$

This step is the cornerstone of solving this logarithmic equation. By equating the arguments, we eliminate the logarithmic function and transform the problem into a more manageable algebraic equation. The condition that the arguments must be positive is crucial and will be revisited later when we verify the solutions. The equation ${v^2 + v - 5 = -3v - 8}$ is a quadratic equation, which we can solve using standard techniques. The next step involves rearranging the terms to bring all terms to one side, setting the equation to zero, and then proceeding to factor or use the quadratic formula. This algebraic manipulation is a common strategy in solving various types of equations, and it is essential to be proficient in these techniques to solve logarithmic equations and other mathematical problems effectively.

2. Form a Quadratic Equation

To solve for ${v}$, we rearrange the equation to form a standard quadratic equation, which is in the form ${ax^2 + bx + c = 0}$. In our case, we add ${3v}$ and ${8}$ to both sides of the equation ${v^2 + v - 5 = -3v - 8}$:

${v^2 + v - 5 + 3v + 8 = 0}$

Combining like terms, we get:

${v^2 + 4v + 3 = 0}$

Now we have a quadratic equation in the standard form. Solving quadratic equations is a fundamental skill in algebra, and mastering different methods to solve them is crucial. The quadratic equation we have obtained, ${v^2 + 4v + 3 = 0}$, can be solved by factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method if the quadratic expression can be factored easily. The equation ${v^2 + 4v + 3 = 0}$ is a relatively simple quadratic equation, making it an excellent example for illustrating the process of solving logarithmic equations after the logarithmic part has been dealt with. The next step will involve factoring this quadratic equation to find potential solutions for ${v}$.

3. Solve the Quadratic Equation

We can solve the quadratic equation ${v^2 + 4v + 3 = 0}$ by factoring. We look for two numbers that multiply to 3 and add to 4. These numbers are 1 and 3. Thus, we can factor the quadratic equation as:

${(v + 1)(v + 3) = 0}$

Setting each factor equal to zero gives us the potential solutions for ${v}$:

${v + 1 = 0 \Rightarrow v = -1}$

${v + 3 = 0 \Rightarrow v = -3}$

Therefore, the potential solutions for ${v}$ are ${-1}$ and ${-3}$. Factoring is an efficient method for solving quadratic equations when the roots are integers or simple fractions. The ability to factor quadratic equations quickly can significantly speed up the process of solving logarithmic equations. After finding the potential solutions, the next critical step is to verify whether these solutions are valid by substituting them back into the original logarithmic equation. This verification process is essential because not all solutions obtained from the transformed equation may satisfy the original logarithmic equation due to the domain restrictions of logarithms. This step ensures that we avoid extraneous solutions and arrive at the correct answer.

4. Verify the Solutions

It is crucial to verify the solutions in the original logarithmic equation to ensure they are valid. The argument of a logarithm must be positive. We need to check ${v^2 + v - 5 > 0}$ and ${-3v - 8 > 0}$ for both ${v = -1}$ and ${v = -3}$.

For ${v = -1}$:

${v^2 + v - 5 = (-1)^2 + (-1) - 5 = 1 - 1 - 5 = -5}$ ${-3v - 8 = -3(-1) - 8 = 3 - 8 = -5}$

Since both arguments are negative, ${v = -1}$ is not a valid solution.

For ${v = -3}$:

${v^2 + v - 5 = (-3)^2 + (-3) - 5 = 9 - 3 - 5 = 1}$ ${-3v - 8 = -3(-3) - 8 = 9 - 8 = 1}$

Since both arguments are positive, ${v = -3}$ is a valid solution.

This verification step is the most crucial part of solving logarithmic equations. It highlights the importance of understanding the domain of logarithmic functions. Substituting the potential solutions back into the original equation and checking the arguments ensures that we only accept solutions that satisfy the initial conditions. The arguments of the logarithms must be positive for the logarithm to be defined. This verification process not only confirms the validity of the solutions but also deepens our understanding of the properties of logarithms. For the given equation, we found that ${v = -1}$ leads to negative arguments, making it an extraneous solution. On the other hand, ${v = -3}$ satisfies the conditions, making it the correct solution. This careful approach is essential for accurately solving logarithmic equations.

Final Answer

After verifying the solutions, we find that only ${v = -3}$ is a valid solution. Therefore, the solution to the logarithmic equation ${\log _3(v^2+v-5) = \log _3(-3v-8)}$ is:

${v = -3}$

The final answer is the culmination of a series of steps, each crucial for solving the logarithmic equation accurately. We started by equating the arguments of the logarithms, formed a quadratic equation, solved it by factoring, and then, most importantly, verified the solutions. This step-by-step approach ensures that we not only find the potential solutions but also confirm their validity within the context of logarithmic functions. The solution ${v = -3}$ satisfies the original equation, making the arguments of the logarithms positive, thus adhering to the domain restrictions. This complete process exemplifies the rigor required in solving mathematical problems and underscores the importance of each step in arriving at the correct answer. The final answer, ${v = -3}$, is the definitive solution to the given logarithmic equation, reflecting a thorough and accurate problem-solving approach.

Conclusion

In summary, solving logarithmic equations involves several key steps: equating the arguments, forming and solving a quadratic equation, and, most importantly, verifying the solutions. The equation ${\log _3(v^2+v-5) = \log _3(-3v-8)}$ serves as an excellent example to illustrate this process. By following these steps carefully, you can confidently solve similar logarithmic equations. Remember to always check your solutions to ensure they are valid within the domain of the logarithmic functions.

The process of solving logarithmic equations, as demonstrated through the example ${\log _3(v^2+v-5) = \log _3(-3v-8)}$, highlights the necessity of a systematic approach. Each step, from equating the arguments to verifying the solutions, plays a vital role in obtaining the correct answer. The algebraic manipulations involved, such as forming and solving quadratic equations, are fundamental skills that are applicable in various mathematical contexts. However, the verification step is particularly crucial in the context of logarithmic equations due to the domain restrictions of logarithmic functions. Failing to verify the solutions can lead to accepting extraneous solutions, which are not valid. Therefore, a comprehensive understanding of these steps and the ability to apply them accurately is essential for mastering the art of solving logarithmic equations. The confidence gained through this process extends beyond this specific type of equation and enhances overall problem-solving skills in mathematics.