Solve: Ln(x) + Ln(x - 4) = Ln(4x)

Hey guys! Let's dive into this math problem together and figure out the value of 'x' in the equation ln(x) + ln(x - 4) = ln(4x). This looks like a fun one, so grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into solving, let's break down what we're dealing with. We have a logarithmic equation here, which means we're working with logarithms. Specifically, we're using the natural logarithm, denoted as 'ln'. Remember, the natural logarithm is just the logarithm to the base 'e', where 'e' is Euler's number (approximately 2.71828). The equation involves three logarithmic terms: ln(x), ln(x - 4), and ln(4x). Our goal is to find the value(s) of 'x' that make this equation true. Now, when dealing with logarithms, it's super important to keep in mind that we can only take the logarithm of positive numbers. This means that x, (x - 4), and (4x) must all be greater than zero. This is a crucial point that will help us determine the valid solutions later on. So, to summarize, we need to find the 'x' that satisfies the equation while also ensuring that the arguments of the logarithms are positive. This involves using the properties of logarithms to simplify the equation and then solving for 'x'. Let's dive into the solution steps now!

Solving the Equation

Alright, let's get our hands dirty and solve this equation! The first thing we're going to do is use a property of logarithms that will help us simplify things. Remember the product rule for logarithms? It states that ln(a) + ln(b) = ln(a * b). This rule is going to be our best friend here. Applying the product rule to the left side of our equation, ln(x) + ln(x - 4) = ln(4x), we can combine the two logarithmic terms into a single term: ln(x(x - 4)) = ln(4x). Awesome! Now our equation looks a lot simpler. We have a single natural logarithm on each side of the equation. So, we have ln(x(x - 4)) = ln(4x). The next step is to get rid of those logarithms. Since we have the same logarithm (natural logarithm) on both sides, we can simply equate the arguments. In other words, if ln(a) = ln(b), then a = b. Applying this to our equation, we get x(x - 4) = 4x. We've successfully eliminated the logarithms and now we're left with a good old algebraic equation. Now, let's expand the left side: x^2 - 4x = 4x. To solve for 'x', we need to rearrange this into a quadratic equation. Let's subtract 4x from both sides: x^2 - 4x - 4x = 0, which simplifies to x^2 - 8x = 0. We're almost there! Now we have a quadratic equation in the standard form. Let's see how to crack this next.

Finding the Possible Solutions

Okay, we've arrived at the quadratic equation x^2 - 8x = 0. There are a couple of ways we can solve this. One way is to use the quadratic formula, but in this case, there's a simpler method: factoring. Notice that both terms in the equation have 'x' in them. We can factor out an 'x' from the equation: x(x - 8) = 0. Now we have a product of two factors that equals zero. This means that at least one of the factors must be zero. So, either x = 0 or (x - 8) = 0. Let's consider each case separately. If x = 0, that's one possible solution. If (x - 8) = 0, then adding 8 to both sides gives us x = 8. So, we have two possible solutions: x = 0 and x = 8. But hold on! We're not done yet. Remember when we talked about the domain of logarithms? We need to check if these solutions are valid in the original equation. Logarithms can only accept positive arguments, so we have to make sure that x, (x - 4), and (4x) are all greater than zero. This is a crucial step in solving logarithmic equations, so let's not skip it. Next, we'll verify our potential solutions.

Verifying the Solutions

Alright, we've got two potential solutions: x = 0 and x = 8. But before we declare victory, we need to make sure these solutions actually work in the original equation and, more importantly, that they don't violate the domain of the logarithms. Remember, logarithms are only defined for positive arguments. So, let's go back to our original equation, ln(x) + ln(x - 4) = ln(4x), and check each solution. First, let's consider x = 0. If we plug x = 0 into the equation, we get ln(0) + ln(0 - 4) = ln(4 * 0). This simplifies to ln(0) + ln(-4) = ln(0). Uh oh! We've got a problem. Logarithms of zero and negative numbers are undefined. So, x = 0 is not a valid solution. It's an extraneous solution, which means it's a solution that we obtained algebraically but doesn't actually satisfy the original equation. Now, let's check x = 8. Plugging x = 8 into the equation, we get ln(8) + ln(8 - 4) = ln(4 * 8). This simplifies to ln(8) + ln(4) = ln(32). Now, let's use the product rule of logarithms again: ln(8 * 4) = ln(32), which simplifies to ln(32) = ln(32). This is a true statement! Also, if we check the arguments of the original logarithms, we have x = 8 which is positive, (x - 4) = (8 - 4) = 4 which is also positive, and (4x) = (4 * 8) = 32 which is positive as well. So, x = 8 is a valid solution. Hooray! We've found our solution, and we've verified it. So, the only valid solution to the equation ln(x) + ln(x - 4) = ln(4x) is x = 8.

Final Answer

So there you have it, guys! After working through the problem step by step, applying the properties of logarithms, solving the resulting quadratic equation, and, most importantly, verifying our solutions, we've found that the only valid solution to the equation ln(x) + ln(x - 4) = ln(4x) is x = 8. Remember, when dealing with logarithmic equations, always check your solutions to make sure they don't violate the domain of the logarithms. This extra step can save you from extraneous solutions and ensure you get the correct answer. I hope this explanation was clear and helpful! Keep practicing, and you'll become a pro at solving logarithmic equations in no time. Happy math-ing!