Solving Logarithmic Equations Find X In Log(x+3) - Log(x+1) = 2

Hey guys! Today, we're diving into the world of logarithmic equations and tackling a common problem: solving for x. Logarithmic equations might seem intimidating at first, but with a clear understanding of the properties of logarithms and a step-by-step approach, you'll be solving them like a pro in no time. So, let's break down the problem and get to the solution. We'll focus on the specific equation $\log (x+3) - \log (x+1) = 2$, but the techniques we'll use can be applied to a wide range of logarithmic equations. Understanding logarithms is crucial for various fields, including mathematics, physics, and computer science. Let's get started and demystify these equations together!

Understanding the Basics of Logarithms

Before we jump into solving the equation, let's quickly recap the fundamentals of logarithms. A logarithm is essentially the inverse operation of exponentiation. In simple terms, if we have an equation like $b^y = x$, the logarithm (base b) of x is y, written as $\log_b(x) = y$. This means that the logarithm tells us the exponent to which we need to raise the base b to obtain the value x. When we see “log” without a specified base, it usually implies the common logarithm, which has a base of 10 (i.e., $\log_{10}$). So, $\log(x)$ is the same as $\log_{10}(x)$. Understanding this fundamental relationship between logarithms and exponents is key to manipulating logarithmic equations. There are several key properties of logarithms that we'll use to solve equations. One of the most important is the quotient rule, which states that the logarithm of a quotient is equal to the difference of the logarithms: $\log_b(x/y) = \log_b(x) - \log_b(y)$. This property will be instrumental in simplifying our given equation. Another important property is the power rule, which states that the logarithm of a number raised to a power is equal to the power times the logarithm of the number: $\log_b(x^p) = p \log_b(x)$. While we won't directly use the power rule in this specific problem, it's a valuable tool to have in your logarithmic equation-solving arsenal. Remember, the goal when solving logarithmic equations is often to isolate the variable x. To do this, we'll need to use the properties of logarithms to combine logarithmic terms, eliminate logarithms, and eventually solve for x.

Applying Logarithmic Properties to Simplify the Equation

Okay, let's get back to our equation: $\log (x+3) - \log (x+1) = 2$. The first thing we want to do is simplify the left side of the equation. Remember that quotient rule we talked about? This is where it comes in handy! We have the difference of two logarithms, so we can combine them into a single logarithm using the quotient rule. Specifically, we can rewrite $\log (x+3) - \log (x+1)$ as $\log \frac{x+3}{x+1}$. This gives us a new, simplified equation: $\log \frac{x+3}{x+1} = 2$. Now, we have a single logarithm on the left side, which is much easier to work with. The next step is to get rid of the logarithm altogether. To do this, we need to remember the relationship between logarithms and exponents. Since we're dealing with the common logarithm (base 10), we can rewrite the equation in exponential form. The equation $\log \frac{x+3}{x+1} = 2$ is equivalent to saying that 10 raised to the power of 2 is equal to $\frac{x+3}{x+1}$. Mathematically, this is expressed as $10^2 = \frac{x+3}{x+1}$. See how we've transformed the logarithmic equation into a more manageable algebraic equation? This is a crucial step in solving for x. By applying the properties of logarithms and converting to exponential form, we've set ourselves up to isolate x and find the solution. Remember, practice makes perfect when it comes to these manipulations. The more you work with logarithmic properties, the more comfortable you'll become with applying them.

Converting to Exponential Form and Solving for x

Great! We've successfully transformed our logarithmic equation into an exponential one: $10^2 = \frac{x+3}{x+1}$. Now, let's simplify further and solve for x. First, we know that $10^2$ is simply 100, so we can rewrite the equation as $100 = \frac{x+3}{x+1}$. To get rid of the fraction, we can multiply both sides of the equation by $(x+1)$. This gives us: $100(x+1) = x+3$. Now we have a linear equation that we can easily solve. Let's distribute the 100 on the left side: $100x + 100 = x + 3$. Next, we want to get all the x terms on one side and the constant terms on the other side. Subtract x from both sides: $99x + 100 = 3$. Then, subtract 100 from both sides: $99x = -97$. Finally, to isolate x, divide both sides by 99: $x = \frac{-97}{99}$. So, we've found a potential solution for x. But, before we declare victory, there's one crucial step we need to take: checking for extraneous solutions. Extraneous solutions are values that we obtain algebraically but don't actually satisfy the original equation. This often happens with logarithmic equations because the domain of a logarithm is restricted to positive numbers. Remember, always check your solutions in the original equation to ensure they are valid.

Checking for Extraneous Solutions and the Final Answer

Alright, we've arrived at a potential solution: $x = \frac{-97}{99}$. Now, the critical step: checking for extraneous solutions. Remember, logarithms are only defined for positive arguments. This means that in our original equation, $\log (x+3) - \log (x+1) = 2$, both $(x+3)$ and $(x+1)$ must be greater than zero. Let's check if our solution satisfies these conditions. First, let's check $(x+3)$: $\frac{-97}{99} + 3 = \frac{-97 + 297}{99} = \frac{200}{99}$. This is positive, so far so good. Now, let's check $(x+1)$: $\frac{-97}{99} + 1 = \frac{-97 + 99}{99} = \frac{2}{99}$. This is also positive! Since both $(x+3)$ and $(x+1)$ are positive when $x = \frac{-97}{99}$, our solution is valid. Therefore, the solution to the equation $\log (x+3) - \log (x+1) = 2$ is $x = \frac{-97}{99}$. We've successfully solved the equation by applying the properties of logarithms, converting to exponential form, solving the resulting algebraic equation, and, most importantly, checking for extraneous solutions. Remember, the key to mastering logarithmic equations is understanding the properties, practicing the steps, and always verifying your answers. Great job, guys! You've tackled a logarithmic equation like pros!

Key Takeaways for Solving Logarithmic Equations

Before we wrap up, let's recap the key takeaways for solving logarithmic equations. These steps will help you approach any logarithmic equation with confidence. First and foremost, understand the properties of logarithms. Knowing the product rule, quotient rule, power rule, and the relationship between logarithms and exponents is crucial. Second, simplify the equation by combining logarithmic terms using the properties of logarithms. This often involves using the quotient rule to combine differences of logarithms into a single logarithm or using the product rule to combine sums of logarithms. Third, convert the equation to exponential form. This is the key step in eliminating the logarithm and transforming the equation into a more familiar algebraic form. Fourth, solve the resulting algebraic equation. This may involve solving a linear equation, a quadratic equation, or some other type of equation, depending on the specific problem. And finally, and perhaps most importantly, check for extraneous solutions. Always plug your solutions back into the original equation to ensure they are valid. Logarithms are only defined for positive arguments, so any solution that makes the argument of a logarithm negative or zero is extraneous. By following these steps, you'll be well-equipped to tackle a wide variety of logarithmic equations. Remember, solving logarithmic equations is a skill that improves with practice. So, keep practicing, and you'll become more and more comfortable with these types of problems. You've got this!