Solving Ln(x^2 - 25) = 0: A Step-by-Step Guide
Are you grappling with the equation ln(x^2 - 25) = 0 and searching for its solutions? You've come to the right place! This guide will walk you through the process, step by step, ensuring you understand not just the how, but also the why behind each manipulation. Let's dive into the world of logarithms and algebra to find the values of x that satisfy this equation.
Understanding the Problem
Before we jump into solving, it's crucial to understand what the equation is telling us. The equation ln(x^2 - 25) = 0 involves the natural logarithm, which is the logarithm to the base e (where e is approximately 2.71828). In simpler terms, we're looking for the value(s) of x that, when plugged into the expression (x^2 - 25), will make the natural logarithm of that result equal to zero. To solve this, we need to recall the fundamental relationship between logarithms and exponentials. Remember that if ln(a) = b, then e^b = a. This is the key to unlocking our equation.
Furthermore, we need to be mindful of the domain of the natural logarithm function. The natural logarithm is only defined for positive arguments. This means that the expression inside the logarithm, (x^2 - 25), must be strictly greater than zero. This condition will be crucial later when we check our solutions to ensure they are valid. We'll delve deeper into this domain restriction as we progress.
Finally, the equation involves a quadratic expression (x^2 - 25). This suggests that we might encounter multiple solutions, potentially two distinct values for x. It's also worth noting the specific form of the quadratic: x^2 - 25 is a difference of squares, which can be factored, a technique that might prove useful in our solution process. By keeping these initial observations in mind, we'll be better equipped to navigate the steps ahead and arrive at the correct answer. We'll explore each of these concepts in more detail as we proceed, so don't worry if some of this seems a bit abstract right now. The goal is to break down the problem into manageable parts and tackle them systematically.
Step 1: Convert the Logarithmic Equation to Exponential Form
The first key step in solving ln(x^2 - 25) = 0 is to convert the logarithmic equation into its equivalent exponential form. As we discussed earlier, the fundamental relationship between logarithms and exponentials states that if ln(a) = b, then e^b = a. In our case, 'a' corresponds to (x^2 - 25) and 'b' corresponds to 0. Applying this transformation, we get:
e^0 = x^2 - 25
This simple conversion is a powerful tool because it eliminates the logarithm, transforming our equation into a more familiar algebraic form. Now, we're dealing with an equation involving an exponential term and a quadratic expression, which is something we can readily manipulate. Remember that any non-zero number raised to the power of 0 equals 1. Therefore, e^0 simplifies to 1, further simplifying our equation to:
1 = x^2 - 25
This equation is significantly easier to handle than the original logarithmic equation. We've effectively moved from the realm of logarithms to the realm of basic algebra. This step highlights the importance of understanding the fundamental relationships between different mathematical concepts. By recognizing the connection between logarithms and exponentials, we were able to perform a crucial transformation that paved the way for solving the equation. In the next step, we'll focus on isolating the x^2 term and further simplifying the equation to bring us closer to finding the solutions for x. So, with this key conversion completed, we're well on our way to unraveling the mystery of the equation ln(x^2 - 25) = 0.
Step 2: Isolate the Quadratic Term
Now that we've converted the logarithmic equation to its exponential form and simplified it to 1 = x^2 - 25, the next logical step is to isolate the quadratic term (x^2). To do this, we need to get the x^2 term by itself on one side of the equation. This involves a simple algebraic manipulation: adding 25 to both sides of the equation. This maintains the balance of the equation while moving the constant term to the left-hand side. Performing this addition, we get:
1 + 25 = x^2 - 25 + 25
This simplifies to:
26 = x^2
Now we have a much cleaner equation: x^2 equals 26. This form is significantly easier to work with and brings us closer to our goal of finding the values of x that satisfy the original equation. Isolating the variable or variable term is a common and crucial technique in solving equations of all kinds. It allows us to focus on the specific term we're trying to solve for and apply the necessary operations to unravel its value. In this case, by isolating x^2, we've set the stage for the final step: finding the square roots of both sides to solve for x. The next step will involve taking the square root, but it's essential to remember that taking the square root can yield both positive and negative solutions, a point we'll emphasize in the subsequent section. So, with x^2 neatly isolated, we're poised to uncover the solutions to our equation.
Step 3: Solve for x by Taking the Square Root
With the equation simplified to x^2 = 26, the final step in solving for x is to take the square root of both sides. It's crucial to remember that when taking the square root of both sides of an equation, we must consider both the positive and negative roots. This is because both a positive number and its negative counterpart, when squared, will result in the same positive number. Applying this principle, we get:
x = ±√26
This notation, x = ±√26, indicates that there are two potential solutions: x = √26 and x = -√26. These are the values that, when squared, will equal 26. At this point, we have found two possible solutions to our equation. However, we're not quite done yet. It's essential to remember the domain restriction we discussed earlier: the argument of the natural logarithm (x^2 - 25) must be greater than zero. This means we need to check if both of these solutions are valid within the original equation. Taking the square root is a fundamental operation in algebra, but it's vital to apply it correctly and be mindful of the implications of both positive and negative roots. This step highlights the importance of careful attention to detail and a thorough understanding of mathematical principles. In the next section, we'll perform the crucial step of verifying our solutions against the domain restriction to ensure they are indeed valid solutions to the original equation. This verification process is a critical part of problem-solving in mathematics, ensuring the accuracy and completeness of our answer.
Step 4: Verify the Solutions
We've arrived at two potential solutions: x = √26 and x = -√26. However, as we discussed earlier, the natural logarithm function, ln(x), is only defined for x > 0. This means the expression inside the logarithm in our original equation, (x^2 - 25), must be greater than zero. We need to verify whether both of our solutions satisfy this condition.
Let's first consider x = √26. Plugging this into the expression (x^2 - 25), we get:
(√26)^2 - 25 = 26 - 25 = 1
Since 1 > 0, the solution x = √26 is valid.
Now let's consider x = -√26. Plugging this into the expression (x^2 - 25), we get:
(-√26)^2 - 25 = 26 - 25 = 1
Again, we get 1 > 0, so the solution x = -√26 is also valid.
Both solutions satisfy the domain restriction of the natural logarithm. This verification step is crucial because it ensures that our solutions are not extraneous, meaning they don't arise from the algebraic manipulation but don't actually satisfy the original equation's constraints. This process highlights the importance of not just blindly applying mathematical rules but also understanding the underlying principles and limitations of those rules. In the context of logarithmic functions, the domain restriction is a critical consideration. By verifying our solutions, we've ensured the accuracy and completeness of our answer. This meticulous approach to problem-solving is a hallmark of mathematical rigor and ensures that our conclusions are sound.
Conclusion
Therefore, the potential solutions to the equation ln(x^2 - 25) = 0 are x = ±√26, which corresponds to option C. We arrived at this answer by carefully converting the logarithmic equation to exponential form, isolating the quadratic term, solving for x by taking the square root, and crucially, verifying our solutions against the domain restriction of the natural logarithm function. This step-by-step process illustrates the power of breaking down complex problems into smaller, manageable parts. By understanding the fundamental relationships between mathematical concepts, such as logarithms and exponentials, and by paying close attention to detail, we can confidently navigate and solve even seemingly challenging equations. Remember, always verify your solutions, especially when dealing with functions that have domain restrictions, to ensure the accuracy of your results. This methodical approach is key to success in mathematics and problem-solving in general.
For further exploration of logarithmic equations and their solutions, you might find helpful resources on websites like Khan Academy.
