Solving Logarithmic Equations A Step-by-Step Guide For Ln(x-2) - Ln(2) = -3

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In the realm of mathematics, solving for variables within logarithmic equations is a fundamental skill. This article delves into the process of solving the equation ln⁡(x−2)−ln⁡2=−3\ln (x-2)-\ln 2=-3, providing a comprehensive step-by-step guide to ensure clarity and understanding. We will explore the properties of logarithms, apply them strategically, and arrive at the solution, rounding to the nearest hundredth as required. Mastering this technique will empower you to tackle a wide range of logarithmic problems with confidence.

Understanding Logarithmic Properties

Before diving into the solution, it's crucial to grasp the core properties of logarithms. Logarithms, in essence, are the inverse operations of exponentiation. The expression log⁡ba=c\log_b a = c signifies that bc=ab^c = a, where 'b' is the base, 'a' is the argument, and 'c' is the logarithm. In our given equation, we encounter the natural logarithm, denoted as 'ln', which has a base of 'e' (Euler's number, approximately 2.71828). Understanding these fundamental logarithmic relationships is paramount to manipulating and solving equations effectively. Logarithmic functions are particularly useful for modeling phenomena with exponential growth or decay, which appear in various fields such as finance, physics, and computer science. The key to successfully navigating logarithmic equations lies in the ability to translate between logarithmic and exponential forms, as well as a firm grasp of logarithmic properties such as the product rule, quotient rule, and power rule. These properties allow us to simplify complex expressions and isolate the variable we are solving for. The natural logarithm, in particular, is prevalent in calculus and other advanced mathematical disciplines, making its mastery an essential stepping stone for further mathematical exploration. In the context of problem-solving, a clear understanding of the domain of logarithmic functions is also vital; the argument of a logarithm must always be positive, which introduces constraints on the solutions we can accept. By considering these constraints alongside the algebraic manipulations, we can ensure the validity of our solutions and avoid common pitfalls. Therefore, before attempting to solve any logarithmic equation, a thorough review of these fundamental concepts and properties is highly recommended.

Applying the Quotient Rule of Logarithms

The given equation, ln⁡(x−2)−ln⁡2=−3\ln (x-2)-\ln 2=-3, presents an opportunity to apply the quotient rule of logarithms. This rule states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator. Mathematically, this is expressed as ln⁡a−ln⁡b=ln⁡(a/b)\ln a - \ln b = \ln (a/b). Applying this rule to our equation, we can combine the two logarithmic terms on the left-hand side into a single logarithm. This quotient rule is a powerful tool in simplifying logarithmic expressions, allowing us to condense multiple logarithmic terms into one, which often makes the equation more manageable. The rule stems directly from the properties of exponents, where dividing exponential terms with the same base involves subtracting the exponents. The logarithmic equivalent of this operation translates to subtracting the logarithms, hence the quotient rule. In the context of our equation, identifying the difference of two logarithms immediately signals the applicability of the quotient rule. By recognizing this pattern, we can strategically transform the equation, bringing us closer to isolating the variable x. Moreover, understanding the relationship between the quotient rule and the exponential properties reinforces a deeper understanding of logarithms as inverse operations of exponentiation. The ability to seamlessly apply this rule not only simplifies the equation at hand but also builds a stronger foundation for tackling more complex logarithmic problems in the future. Therefore, mastering the quotient rule is an indispensable skill in the arsenal of any mathematician or student delving into logarithmic equations.

By applying the quotient rule, we transform the equation ln⁡(x−2)−ln⁡2=−3\ln (x-2)-\ln 2=-3 into ln⁡x−22=−3\ln \frac{x-2}{2} = -3. This step consolidates the logarithmic terms, paving the way for further simplification.

Converting to Exponential Form

With the equation now in the form ln⁡x−22=−3\ln \frac{x-2}{2} = -3, the next crucial step involves converting it from logarithmic form to exponential form. Recall that the natural logarithm (ln) has a base of 'e'. Therefore, the equation ln⁡y=z\ln y = z is equivalent to ez=ye^z = y. Applying this principle to our equation, we can rewrite ln⁡x−22=−3\ln \frac{x-2}{2} = -3 as e−3=x−22e^{-3} = \frac{x-2}{2}. This transformation is a fundamental technique in solving logarithmic equations, as it allows us to eliminate the logarithm and isolate the expression containing the variable 'x'. The process of converting between logarithmic and exponential forms is rooted in the very definition of logarithms; it is the bridge that connects these two mathematical concepts. Recognizing the equivalence between these forms is essential for manipulating equations and ultimately solving for the unknown. In our specific case, converting to exponential form not only simplifies the equation but also sets the stage for the subsequent algebraic steps needed to isolate 'x'. Moreover, this step underscores the importance of understanding the inverse relationship between logarithms and exponentials, a cornerstone of mathematical literacy. Therefore, mastering the conversion between logarithmic and exponential forms is not just a procedural skill but a conceptual understanding that enhances problem-solving capabilities in a broader mathematical context.

Isolating x and Solving

Now that we have the equation in exponential form, e−3=x−22e^{-3} = \frac{x-2}{2}, the focus shifts to isolating 'x'. This involves a series of algebraic manipulations to get 'x' by itself on one side of the equation. First, we multiply both sides of the equation by 2 to eliminate the denominator: 2e−3=x−22e^{-3} = x-2. This step clears the fraction, making the equation easier to handle. Next, we add 2 to both sides to isolate 'x': x=2e−3+2x = 2e^{-3} + 2. These algebraic steps are fundamental in solving any equation, regardless of its complexity. The principle of maintaining equality by performing the same operation on both sides is the bedrock of algebraic manipulation. In our case, each step is carefully chosen to progressively isolate 'x', bringing us closer to the solution. The ability to execute these manipulations accurately and efficiently is a testament to one's algebraic proficiency. Moreover, understanding the order of operations and applying them correctly is crucial to avoid errors. As we isolate 'x', we are effectively unwinding the operations that were initially applied to it, a process that highlights the inverse nature of mathematical operations. Therefore, mastering the art of isolating variables is not just a technique for solving equations but a core skill that permeates all areas of mathematics.

Calculating the Value of x and Rounding

With the equation now solved for xx, we have x=2e−3+2x = 2e^{-3} + 2. The next step is to calculate the numerical value of this expression. Using a calculator, we find that e−3e^{-3} is approximately 0.049787. Therefore, 2e−32e^{-3} is approximately 0.099574. Adding 2 to this value gives us x≈2.099574x \approx 2.099574. The final instruction is to round the answer to the nearest hundredth. The hundredth's place is the second digit after the decimal point. Looking at the third digit after the decimal point (9), we see that it is 5 or greater, so we round up the hundredth's digit. Therefore, the value of x rounded to the nearest hundredth is 2.10. This final step underscores the importance of paying attention to the instructions and expressing the answer in the required format. Rounding is a practical skill that is used extensively in real-world applications where precise values are often impractical or unnecessary. The ability to round accurately requires understanding place value and applying the rounding rules correctly. Moreover, it is essential to be mindful of the level of precision required in a given context and to round accordingly. In our case, rounding to the nearest hundredth provides a balance between accuracy and simplicity. Therefore, this final step not only completes the solution but also reinforces the importance of numerical skills and attention to detail in mathematical problem-solving.

Therefore, x≈2.10x \approx 2.10.

Conclusion

In this comprehensive guide, we have meticulously solved the logarithmic equation ln⁡(x−2)−ln⁡2=−3\ln (x-2)-\ln 2=-3. We began by revisiting the fundamental properties of logarithms, emphasizing the quotient rule and the relationship between logarithmic and exponential forms. We then applied these principles to transform the equation, isolate 'x', and calculate its value. Finally, we rounded the answer to the nearest hundredth, adhering to the instructions provided. This step-by-step approach underscores the importance of a systematic methodology in solving mathematical problems. Each step builds upon the previous one, highlighting the logical flow of reasoning and the interconnectedness of mathematical concepts. The ability to break down a complex problem into smaller, manageable steps is a hallmark of effective problem-solving. Moreover, the process of solving this equation reinforces the value of understanding the underlying principles rather than simply memorizing formulas. By grasping the core concepts of logarithms and their properties, one can confidently tackle a wide range of logarithmic equations. This skill is not only valuable in academic pursuits but also in various fields that rely on mathematical modeling and analysis. Therefore, mastering the techniques presented in this guide is an investment in one's mathematical proficiency and problem-solving capabilities.

Final Answer: The final answer is 2.10\boxed{2.10}