Solving The Logarithmic Equation: Find The Value Of X

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Solving the Logarithmic Equation: Find the Value of x

Hey guys! Let's dive into this interesting logarithmic equation problem together. We're going to break it down step by step, so don't worry if it looks intimidating at first. Our goal is to find the value of x that satisfies the equation:

2ln⁡eln⁡2x−ln⁡eln⁡10x=ln⁡302 \ln e^{\ln 2 x}-\ln e^{\ln 10 x}=\ln 30

We have some options to choose from:

  • A. $x=30$
  • B. $x=75$
  • C. $x=150$
  • D. $x=300$

Let’s get started and figure out the correct answer!

Understanding Logarithmic Properties

Before we jump into solving the equation, let’s quickly review some key logarithmic properties that will be super helpful. Understanding these properties is crucial for simplifying the equation and making it easier to handle.

  1. The Power Rule: This rule states that $\ln(a^b) = b \ln(a)$. In simpler terms, if you have a logarithm of something raised to a power, you can bring the power down and multiply it by the logarithm.
  2. The Inverse Property: This property is super useful when dealing with natural logarithms and exponentials. It states that $\ln(e^x) = x$ and $e^{\ln(x)} = x$. Basically, the natural logarithm and the exponential function undo each other. This is going to be a lifesaver for simplifying our equation.
  3. The Logarithm of a Product: This rule tells us that $\ln(ab) = \ln(a) + \ln(b)$. The logarithm of a product is the sum of the logarithms. While we won’t directly use this in this specific problem, it’s a good one to keep in your toolkit.
  4. The Logarithm of a Quotient: Similar to the product rule, this one states that $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$. The logarithm of a quotient is the difference of the logarithms.
  5. The Logarithm of 1: Remember that $\ln(1) = 0$. This can be useful in certain simplification scenarios.

With these properties in mind, we're well-equipped to tackle the equation. The inverse property, in particular, will be our best friend in this case. We'll use it to simplify the terms involving both natural logarithms and exponentials. By applying these rules carefully, we can transform the equation into a much simpler form that's easier to solve for x. So, let's move on and see how we can put these properties into action!

Step-by-Step Solution

Okay, let's break down the solution step-by-step. This way, you can follow along easily and see how we arrive at the correct answer. Remember our equation?

2ln⁡eln⁡2x−ln⁡eln⁡10x=ln⁡302 \ln e^{\ln 2 x}-\ln e^{\ln 10 x}=\ln 30

Step 1: Simplify using the Inverse Property

The first thing we want to do is simplify those terms that have both a natural logarithm and an exponential. Remember the inverse property? $\ln(e^x) = x$. We can apply this to our equation:

2ln⁡(eln⁡2x)=2(ln⁡2x)2 \ln (e^{\ln 2x}) = 2(\ln 2x)

ln⁡(eln⁡10x)=ln⁡10x\ln (e^{\ln 10x}) = \ln 10x

So, our equation now looks like this:

2ln⁡(2x)−ln⁡(10x)=ln⁡302 \ln (2x) - \ln (10x) = \ln 30

Step 2: Apply the Power Rule (implicitly)

Think of the 2 in front of ln(2x) as a power that was brought down. We can rewrite the equation to reflect this:

ln⁡(2x)2−ln⁡(10x)=ln⁡30\ln (2x)^2 - \ln (10x) = \ln 30

Which simplifies to:

ln⁡(4x2)−ln⁡(10x)=ln⁡30\ln (4x^2) - \ln (10x) = \ln 30

Step 3: Use the Quotient Rule

Now, we have a difference of logarithms on the left side. Let's use the quotient rule, which states that $\ln(a) - \ln(b) = \ln(\frac{a}{b})$. Applying this to our equation:

ln⁡(4x210x)=ln⁡30\ln \left( \frac{4x^2}{10x} \right) = \ln 30

We can simplify the fraction inside the logarithm:

ln⁡(2x5)=ln⁡30\ln \left( \frac{2x}{5} \right) = \ln 30

Step 4: Eliminate the Logarithms

Since we have a logarithm on both sides of the equation, we can simply drop them. This is because if $\ln(a) = \ln(b)$, then $a = b$. So, we get:

2x5=30\frac{2x}{5} = 30

Step 5: Solve for x

Now, it's just a matter of solving for x. Multiply both sides by 5:

2x=1502x = 150

Divide both sides by 2:

x=75x = 75

And there you have it! We've found the solution to the equation. So, the correct answer is:

  • B. $x=75$

Common Mistakes to Avoid

Alright, let's chat about some common pitfalls people stumble into when tackling logarithmic equations. Knowing these can save you from making errors and help you nail these problems every time!

  1. Forgetting Logarithmic Properties: This is a big one! Logarithmic properties are the bread and butter of solving these equations. If you're not solid on the power rule, quotient rule, product rule, and the inverse property, you're going to have a tough time. Make sure you review these properties and understand how they work. For instance, mixing up the quotient and product rules can lead to incorrect simplifications. Remember, $\ln(a/b) = \ln(a) - \ln(b)$, not $\ln(a) + \ln(b)$.

  2. Incorrectly Applying the Inverse Property: The inverse property, $\ln(e^x) = x$ and $e^{\ln(x)} = x$, is super handy, but it's crucial to apply it correctly. A common mistake is trying to apply it when the expression isn't in the right form. For example, you can't directly simplify $\ln(e^x + 1)$ using this property because the + 1 is outside the exponent. Always double-check that the logarithm and exponential are directly inverse operations before simplifying.

  3. Skipping Steps: It might be tempting to rush through the steps to save time, but skipping steps can easily lead to errors. Each step in solving a logarithmic equation is important, and writing them out helps you keep track of what you're doing. Plus, it makes it easier to spot any mistakes you might have made along the way. Take your time and show your work.

  4. Ignoring the Domain: Logarithms have a domain restriction: you can only take the logarithm of positive numbers. When solving logarithmic equations, you need to make sure that your solutions don't lead to taking the logarithm of a negative number or zero in the original equation. Always check your solutions by plugging them back into the original equation to ensure they're valid. For example, if you get a solution that makes the argument of a logarithm negative, you'll need to discard that solution.

  5. Misunderstanding the Order of Operations: Just like with any mathematical equation, the order of operations (PEMDAS/BODMAS) is essential. Make sure you're performing operations in the correct order. For example, if you have an expression like $2 \ln(x)$, you need to apply the logarithm before multiplying by 2. Incorrect order of operations can throw off your entire solution.

By keeping these common mistakes in mind, you'll be better equipped to solve logarithmic equations accurately and efficiently. Always double-check your work, and don't hesitate to break the problem down into smaller, manageable steps. You've got this!

Practice Problems

To really master solving logarithmic equations, practice is key! Let's try a few more problems to get you comfortable with the process. Working through these will help solidify your understanding and build your confidence. Remember to use the properties we discussed and take your time with each step.

Practice Problem 1: Solve for x:

ln⁡(x)+ln⁡(x−2)=ln⁡(3)\ln(x) + \ln(x - 2) = \ln(3)

Practice Problem 2: Solve for x:

2ln⁡(x)=ln⁡(4x−3)2 \ln(x) = \ln(4x - 3)

Practice Problem 3: Solve for x:

ln⁡(x+1)−ln⁡(x)=ln⁡(2)\ln(x + 1) - \ln(x) = \ln(2)

Solutions:

  1. x = 3
  2. x = 1, x = 3
  3. x = 1

Conclusion

So, we've successfully solved the logarithmic equation and found that x = 75. Remember, the key to tackling these types of problems is understanding the logarithmic properties and applying them step by step. Don't forget to watch out for those common mistakes! Keep practicing, and you'll become a pro at solving logarithmic equations in no time. You got this, guys!