What Is The True Solution To $2 \ln E^{\ln 5x} = 2 \ln 15$?A. $x = 0$ B. $ X = 3 X = 3 X = 3 [/tex] C. $x = 9$ D. $x = 15$
Introduction
Mathematical equations often seem daunting, but with the right approach, they can be solved with ease. In this article, we will delve into the world of logarithms and explore the solution to the equation $2 \ln e^{\ln 5x} = 2 \ln 15$. This equation may seem complex, but by breaking it down step by step, we can arrive at a simple and elegant solution.
Understanding the Equation
The given equation involves logarithms and exponentials. To solve it, we need to understand the properties of these mathematical functions. The equation can be rewritten as $\ln e^{\ln 5x} = \ln 15$. This is because the logarithm and exponential functions are inverses of each other, and the constant 2 can be canceled out.
Applying Logarithmic Properties
Now that we have simplified the equation, we can apply the properties of logarithms to solve it. One of the key properties of logarithms is that $\ln a^b = b \ln a$. Using this property, we can rewrite the equation as $\ln 5x = \ln 15$.
Equating the Arguments
Since the logarithm function is one-to-one, we can equate the arguments of the logarithms. This means that $5x = 15$.
Solving for x
Now that we have a simple equation, we can solve for x. Dividing both sides of the equation by 5, we get $x = 3$.
Conclusion
In this article, we have solved the equation $2 \ln e^{\ln 5x} = 2 \ln 15$ step by step. By applying the properties of logarithms and exponentials, we have arrived at a simple and elegant solution. The correct answer is $x = 3$.
Final Answer
The final answer to the equation $2 \ln e^{\ln 5x} = 2 \ln 15$ is $x = 3$. This is the only solution that satisfies the equation.
Comparison with Other Options
Let's compare our solution with the other options given in the problem.
- Option A: $x = 0$ This is not a valid solution because it does not satisfy the equation.
- Option B: $x = 3$ This is the correct solution that we have obtained.
- Option C: $x = 9$ This is not a valid solution because it does not satisfy the equation.
- Option D: $x = 15$ This is not a valid solution because it does not satisfy the equation.
Conclusion
In conclusion, the true solution to the equation $2 \ln e^{\ln 5x} = 2 \ln 15$ is $x = 3$. This is the only solution that satisfies the equation. By applying the properties of logarithms and exponentials, we have arrived at a simple and elegant solution.
Frequently Asked Questions
- Q: What is the equation $2 \ln e^{\ln 5x} = 2 \ln 15$? A: This is a mathematical equation that involves logarithms and exponentials.
- Q: How do we solve the equation $2 \ln e^{\ln 5x} = 2 \ln 15$? A: We can solve the equation by applying the properties of logarithms and exponentials.
- Q: What is the correct solution to the equation $2 \ln e^{\ln 5x} = 2 \ln 15$? A: The correct solution is $x = 3$.
References
- [1] "Logarithms and Exponentials" by Math Open Reference
- [2] "Properties of Logarithms" by Khan Academy
Related Articles
- "Solving Equations with Logarithms"
- "Properties of Exponentials"
- "Logarithmic Identities"
Introduction
In our previous article, we solved the equation $2 \ln e^{\ln 5x} = 2 \ln 15$ step by step. However, we understand that some readers may still have questions about the solution. In this article, we will address some of the most frequently asked questions about solving the equation $2 \ln e^{\ln 5x} = 2 \ln 15$.
Q: What is the equation $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: This is a mathematical equation that involves logarithms and exponentials. It is a complex equation that requires a step-by-step approach to solve.
Q: How do we solve the equation $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: We can solve the equation by applying the properties of logarithms and exponentials. Specifically, we can use the property that $\ln a^b = b \ln a$ to simplify the equation.
Q: What is the correct solution to the equation $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: The correct solution is $x = 3$. This is the only solution that satisfies the equation.
Q: Why is the solution $x = 3$ correct?
A: The solution $x = 3$ is correct because it satisfies the equation $2 \ln e^{\ln 5x} = 2 \ln 15$. When we substitute $x = 3$ into the equation, we get $2 \ln e^{\ln 15} = 2 \ln 15$, which is true.
Q: What if I get a different solution?
A: If you get a different solution, it is likely because you made a mistake in your calculations. Double-check your work and make sure you applied the properties of logarithms and exponentials correctly.
Q: Can I use a calculator to solve the equation $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: While a calculator can be a useful tool, it is not necessary to solve this equation. In fact, using a calculator can lead to errors if you are not careful. It is better to solve the equation by hand using the properties of logarithms and exponentials.
Q: What are some common mistakes to avoid when solving the equation $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: Some common mistakes to avoid include:
- Not applying the properties of logarithms and exponentials correctly
- Not simplifying the equation enough
- Not checking your work carefully
- Using a calculator without double-checking your work
Q: How can I practice solving equations like $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: You can practice solving equations like $2 \ln e^{\ln 5x} = 2 \ln 15$ by working through examples and exercises in a math textbook or online resource. You can also try solving similar equations on your own to build your skills and confidence.
Q: What are some real-world applications of solving equations like $2 \ln e^{\ln 5x} = 2 \ln 15$?
A: Solving equations like $2 \ln e^{\ln 5x} = 2 \ln 15$ has many real-world applications, including:
- Modeling population growth and decay
- Analyzing financial data and making predictions
- Solving optimization problems in business and economics
- Understanding complex systems and phenomena in science and engineering
Conclusion
In this article, we have addressed some of the most frequently asked questions about solving the equation $2 \ln e^{\ln 5x} = 2 \ln 15$. We hope that this article has been helpful in clarifying any confusion and providing additional guidance on solving this equation.