Extraneous Solution Of Logarithmic Equation Log₇(3x³+x) - Log₇(x) = 2

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In the realm of mathematics, solving equations often leads us to potential solutions. However, not all solutions obtained through algebraic manipulations are valid. Some might be extraneous solutions, which, while satisfying the transformed equation, do not satisfy the original equation. This phenomenon is particularly prevalent in equations involving logarithms, radicals, and rational expressions. This article delves into the intricacies of extraneous solutions, focusing on logarithmic equations, and provides a comprehensive guide on how to identify and eliminate them. We will use a specific example to illustrate the process, ensuring a clear understanding of the concepts involved.

Understanding Extraneous Solutions

Extraneous solutions are essentially false positives in the world of equation solving. They arise when we perform operations that alter the domain of the equation. For instance, squaring both sides of an equation can introduce solutions that were not present in the original equation. Similarly, in logarithmic equations, the domain is restricted to positive values, so any solution that results in a negative argument for the logarithm is an extraneous solution. To grasp the concept fully, let's delve deeper into the nature of logarithmic equations and the steps involved in solving them.

Logarithmic equations, at their core, involve the inverse relationship between logarithms and exponential functions. The equation $\log_b(a) = c$ is equivalent to $b^c = a$, where 'b' is the base, 'a' is the argument, and 'c' is the exponent. A crucial aspect to remember is that the argument 'a' must always be positive. This restriction stems from the definition of logarithms; we cannot raise a positive base to any power and obtain a non-positive result. This domain restriction is the primary reason why extraneous solutions often crop up in logarithmic equations. When solving these equations, we manipulate them algebraically, potentially overlooking this domain restriction. As a result, we might arrive at solutions that, when plugged back into the original equation, lead to the logarithm of a negative number or zero, which is undefined. Therefore, it is imperative to check all potential solutions against the original equation to weed out any extraneous ones.

Consider the simple logarithmic equation $\log_2(x) + \log_2(x-2) = 3$. To solve this, we can use the properties of logarithms to combine the terms on the left side: $\log_2[x(x-2)] = 3$. Converting this to exponential form, we get $x(x-2) = 2^3$, which simplifies to $x^2 - 2x = 8$. Rearranging the terms, we have a quadratic equation: $x^2 - 2x - 8 = 0$. Factoring this equation gives us $(x-4)(x+2) = 0$, leading to two potential solutions: $x = 4$ and $x = -2$. However, we must check these solutions in the original equation.

For $x = 4$, we have $\log_2(4) + \log_2(4-2) = \log_2(4) + \log_2(2) = 2 + 1 = 3$, which is a valid solution. But for $x = -2$, we encounter a problem. The original equation contains the terms $\log_2(x)$ and $\log_2(x-2)$. If we substitute $x = -2$, we get $\log_2(-2)$ and $\log_2(-4)$, both of which are undefined since we cannot take the logarithm of a negative number. Thus, $x = -2$ is an extraneous solution. This example illustrates the importance of verifying solutions in the original equation to eliminate any extraneous solutions that might arise during the algebraic manipulation process. In the subsequent sections, we will tackle a more complex logarithmic equation and demonstrate a systematic approach to identifying and discarding extraneous solutions.

Solving the Logarithmic Equation: A Step-by-Step Approach

Let's consider the logarithmic equation: $\log _7(3 x^3+x)-\log _7(x)=2$. To solve this equation, we'll follow a series of steps, emphasizing the importance of checking for extraneous solutions along the way. This methodical approach ensures we arrive at the correct solution and avoid the pitfall of including invalid answers. Our initial step involves using the properties of logarithms to simplify the equation. Specifically, we'll apply the quotient rule of logarithms, which states that the logarithm of a quotient is equal to the difference of the logarithms. This rule is a cornerstone in simplifying logarithmic expressions and is crucial for solving logarithmic equations. By applying this rule, we can condense the two logarithmic terms into a single, more manageable term. This simplification not only makes the equation easier to work with but also brings us closer to isolating the variable and finding its value.

Applying the quotient rule of logarithms, we can rewrite the given equation as: $\log _7\left(\frac{3 x^3+x}{x}\right)=2$. This step is significant because it combines the two logarithmic terms into one, which simplifies the equation and allows us to proceed with further algebraic manipulations. The quotient rule is a fundamental property of logarithms, and its correct application is essential for solving logarithmic equations efficiently. Now that we have a single logarithmic term, our next goal is to eliminate the logarithm altogether. To achieve this, we'll convert the logarithmic equation into its equivalent exponential form. This conversion is a standard technique for solving logarithmic equations, and it relies on the fundamental relationship between logarithms and exponentials. By expressing the equation in exponential form, we effectively undo the logarithm, making it easier to isolate the variable and solve for its value.

To convert the logarithmic equation to exponential form, we recall the definition of a logarithm: $\log_b(a) = c$ is equivalent to $b^c = a$. Applying this definition to our equation, we get: $\frac{3 x^3+x}{x}=72$. This transformation is a crucial step in solving the equation because it eliminates the logarithm, allowing us to work with a more familiar algebraic expression. Now we have a rational equation, which we can simplify further. The next step involves simplifying the equation by canceling out common factors and performing algebraic manipulations. This simplification is essential for making the equation easier to solve and for isolating the variable. By reducing the complexity of the equation, we can more readily identify the steps needed to find the solution. This process often involves algebraic techniques such as factoring, canceling common terms, and combining like terms.

Simplifying the fraction on the left side, we can factor out an 'x' from the numerator: $\frac{x(3 x^2+1)}{x}=49$. Now, we can cancel the 'x' terms in the numerator and denominator, provided that $x \neq 0$. This condition is important because we must remember that dividing by zero is undefined, and we need to account for this restriction when considering potential solutions. After canceling the 'x' terms, we are left with a simpler equation: $3 x^2+1=49$. This equation is a quadratic equation, which we can solve using standard algebraic techniques. The process of simplifying the equation is crucial because it reduces the complexity and allows us to work with a more manageable expression. In the next step, we will solve this quadratic equation to find the potential values of 'x'.

Solving the Quadratic Equation and Identifying Potential Solutions

Now that we have the simplified equation $3 x^2+1=49$, we can proceed to solve for 'x'. This involves isolating the $x^2$ term and then taking the square root of both sides. This process will yield two potential solutions, one positive and one negative. Solving quadratic equations is a fundamental skill in algebra, and it's essential to be comfortable with the techniques involved. These techniques often include rearranging terms, factoring, completing the square, or using the quadratic formula. In this case, we'll use a straightforward approach to isolate the variable and find its potential values. It's important to remember that quadratic equations can have up to two distinct solutions, and we need to find both of them to ensure we haven't missed any possibilities.

First, we subtract 1 from both sides of the equation: $3 x^2=48$. Next, we divide both sides by 3: $x^2=16$. Now, we take the square root of both sides: $x=\pm 4$. This gives us two potential solutions: $x = 4$ and $x = -4$. It's crucial to recognize that taking the square root of a number yields both a positive and a negative solution because both values, when squared, will result in the same positive number. These potential solutions are the values of 'x' that satisfy the simplified equation. However, we must remember that these are just potential solutions. We still need to check them against the original equation to ensure they are valid and not extraneous. The process of checking for extraneous solutions is particularly important in logarithmic equations, as the domain of logarithmic functions is restricted to positive values. In the next section, we will substitute these potential solutions back into the original equation to determine whether they are valid or extraneous.

Verifying Solutions and Identifying Extraneous Roots

Having obtained the potential solutions $x = 4$ and $x = -4$, the next critical step is to verify these solutions in the original logarithmic equation: $\log _7(3 x^3+x)-\log _7(x)=2$. This verification process is paramount because it helps us identify and eliminate any extraneous solutions that may have arisen during the algebraic manipulation process. Extraneous solutions are values that satisfy the transformed equation but not the original equation, often due to restrictions in the domain of the functions involved. In the case of logarithmic equations, the domain is restricted to positive arguments, meaning that the expressions inside the logarithms must be greater than zero. To verify our solutions, we will substitute each potential solution back into the original equation and check if the equation holds true. If a solution leads to the logarithm of a non-positive number, it is an extraneous solution and must be discarded. This careful checking process ensures that we only accept valid solutions that satisfy the original equation.

Let's start by checking $x = 4$. Substituting this value into the original equation, we get: $\log _7(3 (4)^3+4)-\log _7(4)=2$ $\log _7(3 \cdot 64+4)-\log _7(4)=2$ $\log _7(192+4)-\log _7(4)=2$ $\log _7(196)-\log _7(4)=2$. Now, we can use the quotient rule of logarithms to simplify the left side: $\log _7\left(\frac{196}{4}\right)=2$ $\log _7(49)=2$. Since $7^2 = 49$, the equation holds true, and $x = 4$ is a valid solution. Next, we need to check the other potential solution, $x = -4$. Substituting this value into the original equation, we get: $\log _7(3 (-4)^3+(-4))-\log _7(-4)=2$ $\log _7(3 \cdot (-64)-4)-\log _7(-4)=2$ $\log _7(-192-4)-\log _7(-4)=2$ $\log _7(-196)-\log _7(-4)=2$. Here, we encounter a problem. The arguments of both logarithms are negative. Since the logarithm of a negative number is undefined, $x = -4$ is an extraneous solution. This means that while $x = -4$ might have satisfied the simplified equation, it does not satisfy the original equation due to the domain restrictions of logarithms. Therefore, we must discard this solution. This example clearly demonstrates the importance of verifying solutions in the original equation to eliminate any extraneous roots. By performing this check, we ensure that our final answer is mathematically sound and consistent with the original problem. In conclusion, $x = 4$ is the only valid solution to the given logarithmic equation, and $x = -4$ is an extraneous solution.

Based on our step-by-step solution and verification, we can confidently identify the extraneous solution. The extraneous solution is the value that satisfies the simplified equation but not the original logarithmic equation. In this case, we found that $x = -4$ is the extraneous solution because it leads to the logarithm of a negative number, which is undefined. Therefore, the correct answer is B. This detailed analysis highlights the importance of not only solving equations but also verifying the solutions to ensure their validity. Extraneous solutions are a common pitfall in logarithmic equations, and a thorough verification process is essential to avoid them. By understanding the domain restrictions of logarithmic functions and carefully checking each potential solution, we can confidently solve these types of equations and arrive at the correct answer.

Final Answer

Therefore, the extraneous solution to the logarithmic equation $\log _7(3 x^3+x)-\log _7(x)=2$ is: B. $x=-4$ This comprehensive guide has walked you through the process of solving logarithmic equations and identifying extraneous solutions. By understanding the properties of logarithms, applying algebraic techniques, and meticulously verifying solutions, you can confidently tackle these types of problems. Remember, the key to success in mathematics is not just finding potential solutions but also ensuring their validity within the context of the original equation.