Solving 2 Ln E^(ln 5x) = 2 Ln 15 A Step By Step Guide

Introduction: Delving into the Depths of Logarithmic Equations

In the fascinating realm of mathematics, equations often present themselves as intricate puzzles, challenging us to decipher their hidden solutions. Among these mathematical enigmas, logarithmic equations hold a special allure, demanding a keen understanding of logarithmic properties and algebraic manipulations. This article embarks on a journey to unravel the true solution to the equation 2 ln e^(ln 5x) = 2 ln 15, providing a step-by-step guide that illuminates the underlying concepts and techniques. This comprehensive exploration will not only equip you with the ability to solve this particular equation but also empower you to tackle a wide array of logarithmic challenges.

Logarithmic equations are fundamental in various scientific and engineering disciplines, appearing in models of exponential growth and decay, chemical reactions, and electrical circuits. Mastering the art of solving these equations is therefore a crucial skill for anyone venturing into these fields. Our focus on 2 ln e^(ln 5x) = 2 ln 15 will serve as a springboard for understanding more complex logarithmic problems.

This equation, at first glance, may seem daunting, but with a systematic approach and a clear grasp of logarithmic principles, we can conquer it with ease. We will begin by simplifying the equation using the inherent properties of logarithms, gradually transforming it into a more manageable form. Along the way, we will emphasize the importance of each step, ensuring that the logic behind the solution is crystal clear. The goal is not just to arrive at the correct answer but also to foster a deeper appreciation for the elegance and power of logarithmic functions. This exploration is designed to make logarithmic equations less intimidating and more accessible to learners of all levels.

Deconstructing the Equation: A Step-by-Step Solution

To solve the equation 2 ln e^(ln 5x) = 2 ln 15, we'll embark on a methodical, step-by-step journey, leveraging the fundamental properties of logarithms. This approach will not only lead us to the solution but also illuminate the underlying principles, ensuring a solid understanding of the process.

Step 1: Simplifying the Equation Using Logarithmic Properties

The cornerstone of solving logarithmic equations lies in the adept application of logarithmic properties. Our initial focus is on simplifying the left-hand side of the equation, 2 ln e^(ln 5x). Here, we encounter a crucial property: ln e^x = x. This property states that the natural logarithm of e raised to any power x is simply x. Applying this property to our equation, we can simplify ln e^(ln 5x) to ln 5x. Thus, the left-hand side of the equation transforms from 2 ln e^(ln 5x) to 2 ln 5x. This simplification is a pivotal first step, reducing the complexity of the equation and making it more amenable to further manipulation. Without this step, the equation would remain significantly more challenging to solve. This illustrates the power and necessity of understanding and applying logarithmic properties.

Step 2: Further Simplification by Dividing Both Sides

With the equation now simplified to 2 ln 5x = 2 ln 15, our next step involves reducing the coefficients. We observe that both sides of the equation share a common factor of 2. Dividing both sides by 2, we arrive at a simpler form: ln 5x = ln 15. This division is a crucial algebraic manipulation that maintains the equality of the equation while making it easier to work with. It eliminates the extraneous factor of 2, allowing us to focus on the core logarithmic relationship. This step highlights the importance of recognizing and leveraging common factors in equation solving. By dividing both sides, we effectively strip away an unnecessary layer of complexity, paving the way for a more direct solution.

Step 3: Eliminating the Natural Logarithms

Having simplified the equation to ln 5x = ln 15, we now confront the task of eliminating the natural logarithms. The fundamental principle we employ here is that if the natural logarithms of two quantities are equal, then the quantities themselves must be equal. In other words, if ln a = ln b, then a = b. Applying this principle to our equation, we can confidently deduce that 5x = 15. This step is a direct consequence of the one-to-one nature of the logarithmic function. It allows us to transition from a logarithmic equation to a simple algebraic equation, which is far easier to solve. By eliminating the logarithms, we effectively isolate the variable x, bringing us closer to the final solution.

Step 4: Solving for x

We've now arrived at the straightforward algebraic equation 5x = 15. To isolate x, we perform the basic algebraic operation of dividing both sides of the equation by 5. This yields the solution x = 3. This final step is the culmination of our step-by-step simplification process. It demonstrates the power of algebraic manipulation in solving equations. By dividing both sides by 5, we definitively determine the value of x that satisfies the original logarithmic equation. This solution, x = 3, is the key that unlocks the puzzle, representing the value that makes the equation true. The clarity and simplicity of this final step underscore the effectiveness of the methodical approach we've employed throughout the solution process.

Verifying the Solution: Ensuring Accuracy and Validity

In the realm of mathematics, arriving at a solution is only half the battle. The crucial next step is verification: ensuring that the obtained solution is not only arithmetically correct but also logically valid within the context of the original equation. This process is particularly vital when dealing with logarithmic equations, as logarithmic functions have domain restrictions that must be considered. For the equation 2 ln e^(ln 5x) = 2 ln 15, we've arrived at the solution x = 3. To verify this solution, we must substitute it back into the original equation and check if the equation holds true. This act of substitution is not merely a formality; it's a critical safeguard against extraneous solutions that may arise from algebraic manipulations. By meticulously checking our solution, we affirm its correctness and gain confidence in our problem-solving process.

The Verification Process

Let's substitute x = 3 back into the original equation: 2 ln e^(ln 5x) = 2 ln 15.

1. Substitution: Replace x with 3: 2 ln e^(ln (5 * 3)) = 2 ln 15.
2. Simplify: This becomes 2 ln e^(ln 15) = 2 ln 15.
3. Apply Logarithmic Property: Using the property ln e^x = x, we simplify the left side to 2 ln 15 = 2 ln 15.

The equation holds true! This confirms that x = 3 is indeed a valid solution.

The Importance of Checking for Extraneous Solutions

The verification step serves a critical purpose: it helps us identify and eliminate extraneous solutions. Extraneous solutions are values that emerge during the solving process due to algebraic manipulations but do not satisfy the original equation. In the context of logarithmic equations, the most common cause of extraneous solutions is the domain restriction of logarithmic functions. Logarithms are only defined for positive arguments. Therefore, any solution that results in a non-positive argument for a logarithm in the original equation must be discarded.

In our case, the argument of the inner logarithm is 5x. If we had obtained a negative value for x, substituting it back into the original equation would have resulted in taking the logarithm of a negative number, which is undefined. This would have signaled the presence of an extraneous solution. By verifying our solution, we ensure that it not only satisfies the algebraic manipulations but also adheres to the fundamental constraints of logarithmic functions.

The Correct Answer: x = 3

Through our step-by-step solution and rigorous verification, we have definitively established that the true solution to the equation 2 ln e^(ln 5x) = 2 ln 15 is x = 3. This solution aligns with option B in the original problem statement. Our journey through this equation has highlighted the importance of understanding logarithmic properties, applying algebraic manipulations with precision, and verifying solutions to ensure accuracy and validity. The ability to solve logarithmic equations is a valuable asset in various scientific and engineering disciplines, and the techniques we've explored here will serve as a solid foundation for tackling more complex mathematical challenges. The process of finding the solution, verifying it, and understanding the underlying principles is as crucial as the answer itself. It's this holistic approach that fosters true mathematical proficiency.

Common Mistakes to Avoid When Solving Logarithmic Equations

Solving logarithmic equations can be a rewarding intellectual exercise, but it's also a domain where subtle errors can lead to incorrect solutions. To navigate this landscape effectively, it's essential to be aware of common pitfalls and develop strategies to avoid them. This section will shed light on frequent mistakes encountered when solving logarithmic equations, providing you with the knowledge and insights to steer clear of these traps and enhance your problem-solving accuracy. By understanding these common errors, you can develop a more robust and reliable approach to tackling logarithmic challenges.

Neglecting the Domain of Logarithmic Functions

One of the most prevalent errors in solving logarithmic equations is overlooking the domain restriction of logarithmic functions. Logarithms are only defined for positive arguments. This means that the expression inside the logarithm must always be greater than zero. Failing to consider this restriction can lead to the acceptance of extraneous solutions – values that satisfy the transformed equation but not the original logarithmic equation.

Example: Consider an equation where a step in the solution leads to ln(x - 2) = ln(-x + 5). If we solve this by equating the arguments, we get x - 2 = -x + 5, which yields x = 3.5. However, we must check if this solution is valid in the original logarithmic expressions. Substituting x = 3.5 into x - 2 gives 1.5, which is positive. But substituting it into -x + 5 gives 1.5 as well, which is also positive. Therefore, x = 3.5 is a valid solution. However, if solving a different equation led to a potential solution that made the argument of a logarithm negative or zero, that solution would be extraneous and must be discarded.

How to Avoid: Always identify the domain restrictions imposed by the logarithmic terms in the equation before you begin solving. This typically involves setting the argument of each logarithm greater than zero and solving the resulting inequalities. Once you've obtained potential solutions, check them against these domain restrictions to eliminate any extraneous values. This proactive approach will significantly reduce the risk of accepting incorrect answers.

Incorrectly Applying Logarithmic Properties

Logarithmic properties are powerful tools for simplifying and solving equations, but they must be applied with precision. Misapplying these properties is a common source of error. For instance, the property ln(a * b) = ln(a) + ln(b) is often confused with ln(a + b), which has no direct simplification. Similarly, the power rule ln(a^b) = b * ln(a) must be applied carefully, ensuring that the exponent applies to the entire argument of the logarithm.

Example: A frequent mistake is attempting to simplify ln(x^2 + 1) as 2 ln(x) + ln(1). This is incorrect because there's no general property to simplify the logarithm of a sum. Another error is misinterpreting ln(5x) as ln(5) * ln(x) instead of the correct application of the product rule: ln(5x) = ln(5) + ln(x).

How to Avoid: Before applying any logarithmic property, take a moment to ensure you're using the correct rule and that it's applicable in the given context. Write out the property explicitly if needed to avoid misremembering it. Practice applying the properties in various scenarios to solidify your understanding. When in doubt, double-check the property in a reliable reference source.

Errors in Algebraic Manipulation

Even with a solid grasp of logarithmic properties, errors in algebraic manipulation can derail the solution process. Mistakes in arithmetic, sign errors, and incorrect factoring are common culprits. These errors can occur at any stage of the solving process, leading to a flawed final answer.

Example: Consider the equation 2 ln(x) = ln(9). Applying the power rule correctly gives ln(x^2) = ln(9). Equating the arguments leads to x^2 = 9. The correct solutions are x = 3 and x = -3. However, failing to check the domain would lead to erroneously accepting both solutions, when only x = 3 is valid since logarithms are not defined for negative numbers.

How to Avoid: Practice meticulous algebraic techniques. Double-check each step as you proceed, paying close attention to signs and arithmetic operations. If the equation involves factoring, review factoring techniques to ensure accuracy. A systematic approach, combined with careful attention to detail, will minimize the risk of algebraic errors.

Forgetting to Verify the Solution

The final, and perhaps most critical, step in solving any equation is verifying the solution. As we discussed earlier, this is particularly important for logarithmic equations due to the domain restrictions. Forgetting to substitute the solution back into the original equation can lead to the acceptance of extraneous solutions, undermining the entire solving process.

How to Avoid: Make verification an automatic part of your problem-solving routine. Once you've obtained a solution, always substitute it back into the original equation and check if it satisfies the equation. Additionally, verify that the solution does not violate any domain restrictions imposed by the logarithmic terms. This final step provides a crucial safeguard against errors and ensures the validity of your solution.

Conclusion: Mastering Logarithmic Equations

Our exploration into the solution of 2 ln e^(ln 5x) = 2 ln 15 has taken us through the essential steps of simplifying, solving, and verifying logarithmic equations. By understanding and applying logarithmic properties, performing algebraic manipulations with care, and vigilantly checking for extraneous solutions, we've not only found the correct answer (x = 3) but also gained valuable insights into the broader landscape of logarithmic problem-solving.

Mastering logarithmic equations is a journey that requires both knowledge and practice. The common mistakes we've discussed serve as important reminders of the potential pitfalls and the strategies to avoid them. Remember to always consider the domain restrictions of logarithmic functions, apply logarithmic properties accurately, pay close attention to algebraic manipulations, and never forget to verify your solutions.

The ability to confidently solve logarithmic equations is a powerful tool in mathematics and its applications. Whether you're a student grappling with algebraic concepts or a professional applying mathematical models in your field, the principles and techniques we've explored here will serve you well. Embrace the challenges that logarithmic equations present, and with persistence and a systematic approach, you'll unlock their solutions and deepen your mathematical understanding.