Solve Logarithmic Equation: Log 4 + Log (x+2) = 1

Hey guys! Today we're diving deep into the cool world of logarithms to tackle a specific problem: solving the equation $\log 4 + \log (x+2) = 1$ for x. This might look a bit intimidating at first, but trust me, once you break it down, it's totally manageable. We're going to go through this step-by-step, making sure you understand each part of the process. So, grab your notebooks, maybe a coffee, and let's get this math party started! We'll be using some fundamental properties of logarithms, so if those are a bit fuzzy, no worries, we'll refresh them as we go. The goal here is not just to find the answer, but to build your confidence in solving these types of problems. We want you to walk away feeling like a logarithm ninja, ready to take on any equation that comes your way. So, let's get started and unravel the mystery behind this logarithmic equation.

Understanding the Basics of Logarithms

Before we jump headfirst into solving $\log 4 + \log (x+2) = 1$, let's quickly recap what logarithms are all about. Essentially, a logarithm is the inverse operation to exponentiation. That is, the logarithm of a number to a given base is the exponent to which that base must be raised to produce that number. For example, $\log_b a = c$ means $b^c = a$. When you see 'log' without a specified base, it usually implies either the common logarithm (base 10) or the natural logarithm (base e, often written as 'ln'). For this problem, since no base is specified, we'll assume it's the common logarithm, meaning base 10. So, $\log 4$ is asking, "To what power must we raise 10 to get 4?" and $\log (x+2)$ is asking, "To what power must we raise 10 to get $(x+2)$?". Understanding this core concept is crucial for manipulating logarithmic equations. We'll be leveraging a few key properties today, so let's jot them down for easy reference. The most important one for our problem is the product rule of logarithms: $\log_b M + \log_b N = \log_b (MN)$. This rule allows us to combine the sum of two logarithms with the same base into a single logarithm of the product of their arguments. Another handy property is that if $\log_b M = \log_b N$, then $M=N$. This means if we can get both sides of our equation to have a single logarithm with the same base, we can simply equate the arguments. Finally, remember that $\log_b b^x = x$ and $b^{\log_b x} = x$. These are part of the definition and inverse properties that are super useful. So, with these tools in our arsenal, we're ready to break down the equation. Don't sweat it if you don't have these memorized yet; we'll apply them directly, and you'll see how they work in practice. The key is to recognize when and how to use them to simplify the equation and isolate our variable, x. This foundational knowledge will serve you well not just in this problem, but in many other mathematical contexts involving logarithms.

Step-by-Step Solution of the Equation

Alright guys, let's get down to business and solve $\log 4 + \log (x+2) = 1$. The first thing we want to do is simplify the left side of the equation. Remember that product rule we just talked about? $\log M + \log N = \log (MN)$. We can apply this here. So, $\log 4 + \log (x+2)$ becomes $\log (4 \times (x+2))$. This simplifies our equation to: $\log (4(x+2)) = 1$. Now, we have a single logarithm on the left side. The next step is to get rid of the logarithm. Since we're dealing with the common logarithm (base 10), we can use the definition of a logarithm or, more practically, exponentiate both sides of the equation with base 10. This means we'll raise 10 to the power of each side. So, we get $10^{\log (4(x+2))} = 10^1$. Because $10^{\log y} = y$ (this is the inverse property in action!), the left side simplifies beautifully to just $4(x+2)$. And on the right side, $10^1$ is simply 10. So, our equation transforms into a much more familiar linear equation: $4(x+2) = 10$. Now, we just need to solve for x. First, distribute the 4 on the left side: $4x + 8 = 10$. Next, subtract 8 from both sides to isolate the term with x: $4x = 10 - 8$, which gives us $4x = 2$. Lastly, divide both sides by 4 to find the value of x: $x = 2/4$. This simplifies to $x = 1/2$. So, the solution we found is $x = 1/2$. It's always a good idea to double-check our answer, especially with logarithmic equations, because the argument of a logarithm must be positive. We'll do that in the next section to make sure everything checks out. This step-by-step approach, using the properties of logarithms and then solving the resulting algebraic equation, is the standard way to handle these problems. Keep practicing this method, and you'll master it in no time!

Checking for Extraneous Solutions

Alright, we've found our potential solution, $x = 1/2$. But with logarithmic equations, it's super important to check for extraneous solutions. What does that mean? It means we need to make sure that when we plug our answer back into the original equation, all the arguments of the logarithms are positive. Remember, you can't take the logarithm of zero or a negative number. Our original equation is $\log 4 + \log (x+2) = 1$. We need to check the arguments: 4 and $(x+2)$. The number 4 is already positive, so that's good. Now, let's check $(x+2)$ with our solution $x = 1/2$. If we substitute $x = 1/2$ into $(x+2)$, we get $(1/2) + 2 = 1/2 + 4/2 = 5/2$. Since $5/2$ is a positive number, our solution $x = 1/2$ is valid! It doesn't create any issues with the domain of the logarithms. Let's quickly verify the entire equation just to be absolutely sure. If $x = 1/2$, then the equation becomes $\log 4 + \log (1/2 + 2) = 1$. This is $\log 4 + \log (5/2) = 1$. Using the product rule again, this is $\log (4 \times 5/2) = 1$. $4 \times 5/2 = 20/2 = 10$. So, we have $\log 10 = 1$. Since we're using the common logarithm (base 10), $\log_{10} 10$ is indeed 1, because $10^1 = 10$. So, the equation holds true! This verification process is a critical part of solving logarithmic equations. It protects you from providing answers that look correct algebraically but are actually invalid within the context of the original logarithmic functions. Always remember to check the domain restrictions imposed by the logarithms. This step ensures the integrity of your solution and demonstrates a thorough understanding of the mathematical principles involved. So, when in doubt, always plug your answer back in!

Properties of Logarithms Used

To wrap things up, let's highlight the key logarithm properties that made solving $\log 4 + \log (x+2) = 1$ possible. Mastering these properties is like unlocking a secret level in your math journey. The first major player we used was the Product Rule: $\log_b M + \log_b N = \log_b (MN)$. We applied this to combine $\log 4$ and $\log (x+2)$ into a single logarithm, $\log (4(x+2))$. This simplification was essential for moving forward. Without it, we'd be stuck trying to isolate x directly, which is much harder. This rule is incredibly powerful for condensing logarithmic expressions, making them easier to work with. Think of it as a way to gather multiple log terms into one tidy package. The second crucial property we implicitly used was the Definition of a Logarithm and its Inverse Property. Specifically, if $\log_b y = c$, then $b^c = y$. In our case, with $\log (4(x+2)) = 1$ (assuming base 10), this means $10^1 = 4(x+2)$. This is equivalent to saying that $10^{\log_{10} y} = y$. By raising both sides of our equation to the power of 10, we effectively 'undid' the logarithm on the left side, leaving us with a simple algebraic equation to solve. This inverse relationship between logarithms and exponentiation is the foundation upon which many logarithmic equation solutions are built. It allows us to transition from the logarithmic form to the exponential form, which is often much easier to manipulate. Understanding that $\log_b b^x = x$ and $b^{\log_b x} = x$ is fundamental. These properties are not just abstract rules; they are practical tools that enable us to solve complex equations. By recognizing and applying these properties correctly, we transformed a seemingly complex logarithmic problem into a straightforward linear equation, leading us directly to the solution. So, remember these, practice them, and you'll be a logarithm pro in no time!

Conclusion: Mastering Logarithmic Equations

So there you have it, guys! We've successfully solved the equation $\log 4 + \log (x+2) = 1$ for x, finding that $x = 1/2$. We navigated through it by first using the product rule to combine the logarithms, then by using the inverse property of logarithms and exponentiation to transform the equation into a solvable linear form, and finally, we diligently checked for extraneous solutions to ensure our answer was valid. This process highlights a few key takeaways for tackling logarithmic equations in general. Firstly, always look for opportunities to simplify using logarithm properties like the product, quotient, and power rules. These rules are your best friends for condensing and manipulating logarithmic expressions. Secondly, remember the relationship between logarithms and exponents – they are inverse operations! This means you can often switch between logarithmic and exponential forms to solve equations, or use exponentiation to eliminate logarithms. Lastly, and I can't stress this enough, always check your solutions by plugging them back into the original equation. This is non-negotiable for logarithmic equations to ensure that the arguments of the logarithms remain positive. By following these steps and practicing regularly, you'll build the skills and confidence to solve a wide variety of logarithmic equations. Keep exploring, keep practicing, and don't be afraid to ask questions. The world of mathematics, especially with logarithms, is full of exciting challenges waiting for you to conquer!