Extraneous Solutions In Logarithmic Equations Explained

In the realm of mathematics, particularly when dealing with logarithmic equations, we often encounter solutions that, despite seemingly valid algebraically, don't satisfy the original equation. These are known as extraneous solutions. They arise due to the inherent restrictions on the domain of logarithmic functions. Understanding how to identify and discard these extraneous solutions is crucial for accurately solving logarithmic equations. This article delves into the concept of extraneous solutions, illustrates their emergence through a step-by-step solution of a logarithmic equation, and emphasizes the importance of verifying solutions in the context of logarithmic functions. Our focus will be on a specific example, shedding light on the process of pinpointing extraneous solutions and reinforcing the fundamental principles of logarithmic equations.

What are Extraneous Solutions?

Before diving into the specifics of the given equation, it's essential to define what extraneous solutions are and why they occur. In simple terms, an extraneous solution is a value that you obtain while solving an equation that, when plugged back into the original equation, does not hold true. This phenomenon is particularly common in equations involving radicals, rational expressions, and, as we'll explore here, logarithms. The reason for their appearance lies in the algebraic manipulations we perform during the solving process, which can sometimes introduce solutions that are not part of the original equation's domain.

Logarithmic functions, by their very nature, have a restricted domain. The argument of a logarithm (the value inside the logarithm) must always be positive. This is because logarithms are the inverse of exponential functions, and exponential functions always produce positive outputs. Consequently, when solving logarithmic equations, we must be vigilant about checking our solutions to ensure they don't lead to taking the logarithm of a negative number or zero. If a solution does, it's an extraneous solution and must be discarded. Ignoring extraneous solutions can lead to incorrect answers and a misunderstanding of the equation's true solution set. Therefore, the final step in solving any logarithmic equation should always involve verifying the solutions against the original equation's domain restrictions.

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

Let's tackle the logarithmic equation presented: $\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$. Our goal is to find the value(s) of x that satisfy this equation, while being mindful of potential extraneous solutions. We'll proceed step-by-step, explaining each step in detail.

1. Combine Logarithms on the Left Side

The first step is to utilize the logarithmic property that states: $\log _b(m) + \log _b(n) = \log _b(mn)$. Applying this property to the left side of our equation, we get:

$\log _4(x(x-3)) = \log _4(-7x+21)$

This simplifies the equation by combining the two logarithmic terms into a single one. This step is crucial as it allows us to eliminate the logarithms in the subsequent step.

2. Eliminate Logarithms

Since we have a single logarithm on each side of the equation with the same base (base 4), we can eliminate the logarithms by equating their arguments. This is based on the fundamental property of logarithms: if $\log _b(m) = \log _b(n)$, then m = n. Therefore, we can write:

$x(x-3) = -7x + 21$

This step transforms the logarithmic equation into a standard algebraic equation, which we can solve using familiar techniques.

3. Simplify and Rearrange

Now, let's simplify and rearrange the equation into a standard quadratic form. Expanding the left side, we get:

$x^2 - 3x = -7x + 21$

Adding 7x and subtracting 21 from both sides, we obtain:

$x^2 + 4x - 21 = 0$

We now have a quadratic equation that we can solve for x.

4. Solve the Quadratic Equation

To solve the quadratic equation $x^2 + 4x - 21 = 0$, we can either factor it or use the quadratic formula. In this case, the equation is easily factorable. We look for two numbers that multiply to -21 and add up to 4. These numbers are 7 and -3. So, we can factor the equation as:

$(x + 7)(x - 3) = 0$

This gives us two potential solutions for x:

$x + 7 = 0$ or $x - 3 = 0$

Which leads to:

$x = -7$ or $x = 3$

These are our candidate solutions. However, we must remember the crucial step of checking for extraneous solutions.

Identifying Extraneous Solutions: The Verification Process

Now comes the critical step: verifying whether our candidate solutions, x = -7 and x = 3, are indeed valid solutions or extraneous. We must substitute each value back into the original logarithmic equation and check if the equation holds true and, more importantly, if the arguments of the logarithms remain positive.

Checking x = -7

Let's substitute x = -7 into the original equation: $\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$

This gives us:

$\log _4(-7) + \log _4(-7-3) = \log _4(-7(-7) + 21)$

$\log _4(-7) + \log _4(-10) = \log _4(49 + 21)$

$\log _4(-7) + \log _4(-10) = \log _4(70)$

Notice that we have logarithms of negative numbers: $\log _4(-7)$ and $\log _4(-10)$. Since the logarithm of a negative number is undefined, x = -7 is an extraneous solution. It does not belong to the solution set of the original equation.

Checking x = 3

Next, let's substitute x = 3 into the original equation:

$\log _4(3) + \log _4(3-3) = \log _4(-7(3) + 21)$

$\log _4(3) + \log _4(0) = \log _4(-21 + 21)$

$\log _4(3) + \log _4(0) = \log _4(0)$

Here, we encounter $\log _4(0)$. The logarithm of zero is also undefined. Therefore, x = 3 is also an extraneous solution. This leaves us with no valid solutions for the original equation.

Conclusion: The Importance of Verification

In conclusion, solving logarithmic equations requires careful attention not only to the algebraic manipulations but also to the fundamental restrictions on the domain of logarithmic functions. We found that both candidate solutions, x = -7 and x = 3, turned out to be extraneous because they led to taking the logarithm of a negative number or zero. This highlights the crucial importance of the verification step in solving logarithmic equations. Always substitute your solutions back into the original equation to ensure they are valid and do not violate the domain restrictions of the logarithmic functions involved. By doing so, you can avoid the pitfall of extraneous solutions and arrive at the correct solution set. In this specific case, the original equation has no real solutions.

This section provides a detailed explanation of how to identify the extraneous solution for the logarithmic equation $\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$. We'll dissect the process, emphasizing the underlying principles and potential pitfalls. The primary focus will be on understanding why certain solutions obtained algebraically might not be valid in the context of the original equation. By carefully examining each step, we aim to provide a comprehensive understanding of extraneous solutions in logarithmic equations.

Step-by-Step Solution and Extraneous Solution Identification

Let's revisit the given equation and solve it step-by-step, explicitly pointing out where extraneous solutions might arise and how to identify them:

$\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$

1. Apply the Product Rule of Logarithms

As discussed earlier, the first step is to combine the logarithms on the left side using the product rule: $\log _b(m) + \log _b(n) = \log _b(mn)$. This gives us:

$\log _4(x(x-3)) = \log _4(-7x+21)$

This step is algebraically sound, but it's crucial to remember that we are now dealing with a product within the logarithm. This means that even if individual terms inside the logarithm might seem negative for certain values of x, their product could be positive. This is a potential source of extraneous solutions if we don't verify later.

2. Eliminate the Logarithms

Since the bases of the logarithms are the same on both sides, we can equate the arguments:

$x(x-3) = -7x + 21$

This step is valid as long as we remember that the arguments of the logarithms in the original equation must be positive. This is the most critical point to keep in mind when solving logarithmic equations.

3. Simplify and Rearrange into Quadratic Form

Expanding the left side and rearranging terms, we get:

$x^2 - 3x = -7x + 21$

$x^2 + 4x - 21 = 0$

This is a standard quadratic equation, and the solutions we obtain from this equation are potential solutions to the logarithmic equation, but they are not guaranteed to be valid until we check them.

4. Solve the Quadratic Equation

Factoring the quadratic equation, we have:

$(x + 7)(x - 3) = 0$

This gives us two candidate solutions:

$x = -7$ or $x = 3$

These are the values we need to scrutinize carefully. They are solutions to the quadratic equation, but they might not be solutions to the original logarithmic equation.

5. Check for Extraneous Solutions

This is the most important step in solving logarithmic equations. We must substitute each candidate solution back into the original logarithmic equation and check if the arguments of the logarithms are positive. If an argument is negative or zero, the solution is extraneous.

Checking x = -7

Substituting x = -7 into the original equation:

$\log _4(-7) + \log _4(-7-3) = \log _4(-7(-7) + 21)$

$\log _4(-7) + \log _4(-10) = \log _4(70)$

As we saw before, we have logarithms of negative numbers: $\log _4(-7)$ and $\log _4(-10)$. This immediately tells us that x = -7 is an extraneous solution.

Checking x = 3

Substituting x = 3 into the original equation:

$\log _4(3) + \log _4(3-3) = \log _4(-7(3) + 21)$

$\log _4(3) + \log _4(0) = \log _4(0)$

We encounter $\log _4(0)$, which is also undefined. Therefore, x = 3 is also an extraneous solution.

Key Takeaway: The Importance of Domain Restrictions

The key takeaway from this detailed solution is the critical role of domain restrictions in logarithmic equations. The argument of a logarithm must always be positive. This restriction is what leads to extraneous solutions. When solving logarithmic equations, we often perform algebraic manipulations that can mask these domain restrictions. Therefore, it is absolutely essential to check all candidate solutions in the original equation to ensure they are valid.

In summary, extraneous solutions are a common occurrence in logarithmic equations due to the restricted domain of logarithmic functions. The process of identifying them involves carefully solving the equation algebraically and then, crucially, verifying each candidate solution in the original equation. By substituting the solutions back into the original equation, we can determine whether they lead to taking the logarithm of a negative number or zero, which would make them extraneous. This verification step is paramount for obtaining the correct solution set. The given equation, $\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$, serves as a prime example of how both algebraically derived solutions can be extraneous, highlighting the necessity of thorough verification in solving logarithmic equations.

Therefore, the answer is that both x = -7 and x = 3 are extraneous solutions.