Solving The Equation 2 Ln E^(ln 2x) - Ln E^(ln 10x) = Ln 30

In the realm of mathematical equations, finding the true solution often requires a blend of algebraic manipulation, logarithmic understanding, and a keen eye for detail. Today, we embark on a journey to decipher the equation 2 ln e^(ln 2x) - ln e^(ln 10x) = ln 30, a problem that appears complex at first glance but yields to a systematic approach. This exploration will not only reveal the solution but also illuminate the fundamental principles governing logarithmic functions and their interplay with exponential expressions. Let's delve into the intricacies of this equation, unraveling its layers step by step, and ultimately, arriving at the definitive answer. The beauty of mathematics lies in its precision and logical structure, and this equation serves as a perfect canvas to showcase these qualities. Through careful application of logarithmic properties and algebraic techniques, we can transform this seemingly daunting problem into a manageable and ultimately solvable puzzle. So, let's begin this mathematical adventure, armed with our knowledge of logarithms and a determination to uncover the true solution that lies within.

Demystifying Logarithmic and Exponential Relationships

Before we dive into the specifics of the equation, let's solidify our understanding of the fundamental relationship between logarithms and exponentials. The logarithm, often denoted as "ln" for the natural logarithm (base e), is the inverse operation of exponentiation. In simpler terms, if we have an equation like e^y = x, then the natural logarithm of x is y, written as ln(x) = y. This inverse relationship is crucial for simplifying expressions and solving equations involving both logarithms and exponentials. A key property to remember is that e^(ln(x)) = x and ln(e^x) = x. These identities allow us to seamlessly transition between exponential and logarithmic forms, making complex equations more manageable. For instance, in our given equation, we encounter terms like ln e^(ln 2x) and ln e^(ln 10x). Recognizing the inverse relationship, we can immediately simplify these terms, paving the way for further algebraic manipulation. Furthermore, understanding the properties of logarithms, such as the power rule (ln(x^n) = n ln(x)), the product rule (ln(xy) = ln(x) + ln(y)), and the quotient rule (ln(x/y) = ln(x) - ln(y)), is essential for effectively solving logarithmic equations. These rules provide the tools to combine, separate, and simplify logarithmic expressions, ultimately leading us closer to the true solution. In the context of our equation, we will leverage these properties to condense and rearrange terms, transforming the equation into a more solvable form. Therefore, a strong grasp of logarithmic and exponential relationships is not just helpful, but absolutely necessary for unraveling the mysteries of this equation.

Step-by-Step Solution: Unraveling the Equation

Now, let's embark on the journey of solving the equation 2 ln e^(ln 2x) - ln e^(ln 10x) = ln 30 step by step. This methodical approach will ensure clarity and precision, leading us to the true solution with confidence. Our initial focus should be on simplifying the logarithmic and exponential terms. As we discussed earlier, the inverse relationship between logarithms and exponentials allows us to simplify ln e^(ln 2x) to ln 2x and ln e^(ln 10x) to ln 10x. Substituting these simplifications into the equation, we get: 2 ln 2x - ln 10x = ln 30. The next step involves applying the power rule of logarithms, which states that ln(x^n) = n ln(x). Using this rule, we can rewrite 2 ln 2x as ln (2x)^2, which simplifies to ln 4x^2. Now, our equation transforms to: ln 4x^2 - ln 10x = ln 30. To further simplify, we utilize the quotient rule of logarithms, which states that ln(x/y) = ln(x) - ln(y). Applying this rule, we can combine the left-hand side of the equation into a single logarithm: ln (4x^2 / 10x) = ln 30. Simplifying the fraction inside the logarithm, we get ln (2x / 5) = ln 30. At this point, we have successfully condensed the equation into a form where we can directly compare the arguments of the logarithms. Since the logarithms on both sides are equal, their arguments must also be equal. This leads us to the equation: 2x / 5 = 30. Solving for x involves multiplying both sides by 5, giving us 2x = 150, and then dividing both sides by 2, resulting in x = 75. Therefore, the solution to the equation is x = 75. This step-by-step process demonstrates how a seemingly complex equation can be systematically solved by applying the fundamental principles of logarithms and algebraic manipulation.

Verification: Ensuring the Accuracy of Our Solution

With the solution x = 75 in hand, it's crucial to verify its accuracy. This step ensures that our algebraic manipulations and logarithmic operations were performed correctly and that the solution satisfies the original equation. To verify, we substitute x = 75 back into the original equation: 2 ln e^(ln 2x) - ln e^(ln 10x) = ln 30. Plugging in x = 75, we get: 2 ln e^(ln (275))* - ln e^(ln (1075))* = ln 30. Simplifying the expressions inside the logarithms, we have: 2 ln e^(ln 150) - ln e^(ln 750) = ln 30. Applying the inverse relationship between logarithms and exponentials, we simplify further: 2 ln 150 - ln 750 = ln 30. Now, we use the properties of logarithms to condense the left-hand side. First, we apply the power rule: ln 150^2 - ln 750 = ln 30. This simplifies to: ln 22500 - ln 750 = ln 30. Next, we apply the quotient rule: ln (22500 / 750) = ln 30. Simplifying the fraction, we get: ln 30 = ln 30. This final equality confirms that our solution, x = 75, is indeed the true solution to the equation. The verification process not only validates our answer but also reinforces our understanding of the logarithmic and exponential relationships we utilized throughout the solving process. It highlights the importance of rigor and precision in mathematics, ensuring that each step is logically sound and that the final answer is accurate.

Common Pitfalls and How to Avoid Them

Solving logarithmic equations can be tricky, and it's easy to fall into common pitfalls. Recognizing these pitfalls and understanding how to avoid them is crucial for achieving accurate solutions. One common mistake is misapplying the properties of logarithms. For example, incorrectly applying the power rule, product rule, or quotient rule can lead to significant errors. To avoid this, always double-check the rules and ensure they are being applied correctly in the specific context of the equation. Another pitfall is neglecting the domain of logarithmic functions. Logarithms are only defined for positive arguments, so it's essential to check that the solution obtained does not lead to taking the logarithm of a negative number or zero. In our equation, we found x = 75, and substituting this back into the original equation does not result in any undefined logarithms, so it is a valid solution. However, in other equations, this check might reveal extraneous solutions that need to be discarded. A further mistake is failing to simplify expressions before applying logarithmic properties. Simplifying terms as much as possible before manipulating logarithms can make the equation more manageable and reduce the chances of errors. In our step-by-step solution, we emphasized simplifying exponential and logarithmic terms early on, which streamlined the solving process. Lastly, overlooking the inverse relationship between logarithms and exponentials can hinder the solution process. Recognizing that e^(ln(x)) = x and ln(e^x) = x allows for significant simplifications, as we demonstrated in our solution. By being mindful of these common pitfalls and consistently applying the principles of logarithmic and exponential functions, we can navigate these challenges and confidently arrive at the true solution.

Conclusion: The Power of Logarithmic Understanding

In conclusion, the journey to solve the equation 2 ln e^(ln 2x) - ln e^(ln 10x) = ln 30 has been a testament to the power of logarithmic understanding and algebraic manipulation. Through a methodical approach, we successfully navigated the complexities of the equation, simplified its terms, and arrived at the true solution, x = 75. This solution was not merely a number, but a culmination of careful application of logarithmic properties, including the inverse relationship between logarithms and exponentials, the power rule, and the quotient rule. The verification process further solidified our confidence in the accuracy of the solution, highlighting the importance of rigor in mathematical problem-solving. Moreover, we explored common pitfalls that can arise when solving logarithmic equations, such as misapplying logarithmic properties, neglecting the domain of logarithmic functions, and overlooking the inverse relationship between logarithms and exponentials. By recognizing these pitfalls and understanding how to avoid them, we can enhance our problem-solving skills and approach future challenges with greater confidence. This exploration underscores the fundamental role that logarithms play in various branches of mathematics and their applications in diverse fields such as physics, engineering, and computer science. A strong grasp of logarithmic principles not only empowers us to solve equations but also provides a deeper understanding of the mathematical world around us. Therefore, the true solution to this equation is not just a number; it's a symbol of our ability to unravel mathematical complexities through knowledge, persistence, and a systematic approach.