Solving Logarithmic Equations Step-by-Step

Hey guys! Let's dive into solving this logarithmic equation together. You know, logarithmic equations might seem intimidating at first, but trust me, with a bit of understanding and practice, they become quite manageable. The key here is to remember the fundamental properties of logarithms and how they relate to exponential functions. When you see an equation like the one we have, $\log (x+3) - \log (x+1) = 2$, the first thing that should pop into your head is the properties of logarithms, specifically the quotient rule. This rule is our best friend in simplifying expressions where we have the difference of two logarithms with the same base. Remember, the quotient rule states that $\log_b(M) - \log_b(N) = \log_b(\frac{M}{N})$. Applying this rule allows us to condense the left side of our equation into a single logarithmic term, making it much easier to handle. Once we've combined the logarithms, we can then think about converting the logarithmic equation into its equivalent exponential form. This is a crucial step because it allows us to get rid of the logarithm altogether and work with a simple algebraic equation. In this specific problem, the base of the logarithm is not explicitly written, which means it's a common logarithm, base 10. So, when we convert to exponential form, we're essentially rewriting the equation in terms of powers of 10. Now, after converting to exponential form, we will typically end up with a rational equation or polynomial equation that we can solve using standard algebraic techniques. For example, we might need to cross-multiply, expand terms, collect like terms, and so on. It's like piecing together a puzzle, where each step gets us closer to the final solution. Remember to always keep an eye out for potential extraneous solutions. These are solutions that we obtain algebraically but don't actually satisfy the original equation. This often happens with logarithmic equations because the logarithm function is only defined for positive arguments. So, after you've found your potential solutions, it's super important to plug them back into the original equation to check if they make sense. If a solution results in taking the logarithm of a negative number or zero, it's an extraneous solution and we have to discard it. Solving logarithmic equations is like a cool mathematical journey that requires a mix of logarithmic properties, algebraic skills, and careful checking. So let's get started and see how it works step by step!

Step-by-Step Solution

Okay, let's break down this equation $\log (x+3) - \log (x+1) = 2$ step by step, making it super clear and easy to follow. First up, we're going to use the quotient rule of logarithms, which, as we discussed, is our secret weapon for combining those logs. Remember, this rule tells us that when we subtract logs with the same base, we can rewrite it as a single log of the quotient. So, in our case, $\log (x+3) - \log (x+1)$ transforms into $\log \left( \frac{x+3}{x+1} \right)$. This is a huge step because we've gone from two separate logarithmic terms to just one, making the equation much cleaner and less scary. Now our equation looks like this: $\log \left( \frac{x+3}{x+1} \right) = 2$. Next, we need to get rid of that logarithm altogether. To do this, we're going to switch gears and rewrite the equation in exponential form. Since the base of the logarithm isn't explicitly written, we know it's a common logarithm, which means it has a base of 10. So, we're going to rewrite our equation in the form $10^{\text{exponent}} = \text{result}$. In our case, this means $10^2 = \frac{x+3}{x+1}$. See how we've transformed the logarithmic equation into a good old algebraic equation? Now we have $100 = \frac{x+3}{x+1}$. Time to roll up our sleeves and solve for $x$. To do this, we'll first get rid of the fraction by multiplying both sides of the equation by $(x+1)$. This gives us $100(x+1) = x+3$. Next up, we distribute the 100 on the left side, which turns our equation into $100x + 100 = x + 3$. Now it's just a matter of moving things around to isolate $x$. Let's subtract $x$ from both sides to get $99x + 100 = 3$. Then, we subtract 100 from both sides, leaving us with $99x = -97$. Finally, we divide both sides by 99 to solve for $x$, which gives us $x = \frac{-97}{99}$. So, we've found a potential solution, but we're not done yet. Remember, we need to check for those sneaky extraneous solutions. We have to plug $x = \frac{-97}{99}$ back into the original equation to make sure it makes sense. This is super important because logarithms are only defined for positive arguments. If plugging in our solution results in taking the logarithm of a negative number or zero, we know it's an extraneous solution and we have to throw it out. Let's plug $x = \frac{-97}{99}$ into the original equation: $\log (x+3) - \log (x+1) = 2$. We need to check if both $(x+3)$ and $(x+1)$ are positive when $x = \frac{-97}{99}$. First, let's look at $x+3$. We have $\frac{-97}{99} + 3$. To add these, we need a common denominator, so we rewrite 3 as $\frac{297}{99}$. This gives us $\frac{-97}{99} + \frac{297}{99} = \frac{200}{99}$, which is positive. Great! Now let's check $x+1$. We have $\frac{-97}{99} + 1$. Similarly, we rewrite 1 as $\frac{99}{99}$, which gives us $\frac{-97}{99} + \frac{99}{99} = \frac{2}{99}$, which is also positive. Since both $(x+3)$ and $(x+1)$ are positive when $x = \frac{-97}{99}$, we know that this solution is valid. So, we've officially solved for $x$! The exact value of $x$ is $\frac{-97}{99}$. If we need to round this to four decimal places, we get approximately -0.9798. So there you have it, guys! We've conquered this logarithmic equation step by step, from applying the quotient rule to checking for extraneous solutions. Keep practicing, and you'll become a pro at solving these types of problems in no time!

Common Mistakes and How to Avoid Them

Alright, let's talk about some common pitfalls that people often stumble into when solving logarithmic equations. Recognizing these mistakes is half the battle, so you'll be well-equipped to sidestep them. One of the most frequent errors is forgetting to check for extraneous solutions. This is a biggie! Logarithmic functions are only defined for positive arguments, which means you can't take the log of a negative number or zero. So, even if you've done all the algebra perfectly, you're not in the clear until you've plugged your solutions back into the original equation to make sure they don't lead to taking the log of a non-positive number. It's like double-checking your route on a map – you might think you're on the right track, but it's always good to confirm! Another common mistake is misapplying the properties of logarithms. These properties are powerful tools, but they need to be used correctly. For instance, some people might incorrectly try to separate a logarithm of a sum or difference, like trying to say $\log(x+3)$ is equal to $\log(x) + \log(3)$. This is a no-go! The properties work for products and quotients inside the logarithm, not sums and differences. So, always double-check which property you're using and make sure it fits the situation. Also, watch out for sign errors when you're manipulating equations. It's super easy to drop a negative sign or make a mistake when distributing. A small sign error can throw off your entire solution, so it's worth taking an extra moment to double-check your work, especially when you're dealing with multiple terms and operations. Another thing to keep in mind is the base of the logarithm. If the base isn't explicitly written, it's assumed to be 10 (the common logarithm). But if you're working with a different base, like the natural logarithm (base $e$), you need to make sure you're applying the correct rules and conversions. Mixing up bases can lead to serious errors. Also, sometimes people get tripped up when converting between logarithmic and exponential forms. Remember that a logarithmic equation $\log_b(x) = y$ is equivalent to the exponential equation $b^y = x$. Make sure you're correctly identifying the base, the exponent, and the result when you make this conversion. It's like translating between two languages – you need to get the grammar right! Lastly, don't forget the basic algebraic skills needed to solve the resulting equation after you've dealt with the logarithms. You might need to factor, use the quadratic formula, or apply other algebraic techniques. So, make sure your algebra skills are sharp, too! To avoid these mistakes, the key is practice, practice, practice! The more you work with logarithmic equations, the more comfortable you'll become with the properties and the techniques involved. And remember, always double-check your work, especially for those pesky extraneous solutions. You've got this, guys!

Practice Problems

To really nail down your understanding of solving logarithmic equations, nothing beats practice. Working through a variety of problems will help you become more confident and comfortable with the different scenarios you might encounter. Let's dive into some practice problems that cover different aspects of what we've discussed. Remember, the key is to apply the properties of logarithms correctly, convert between logarithmic and exponential forms when necessary, and always, always check for extraneous solutions. For our first practice problem, let's try something similar to our example but with a slight twist: Solve for $x$: $\log_2(x+1) + \log_2(x-1) = 3$. Notice here that we have a base 2 logarithm, so we'll need to keep that in mind when we convert to exponential form. The first step here is to use the product rule of logarithms to combine the two logarithmic terms. This will simplify the equation and make it easier to work with. Once you've combined the logarithms, you can convert the equation to exponential form and solve for $x$. Don't forget to check your solutions! Next up, let's tackle an equation that involves a single logarithm: Solve for $x$: $\log(3x-2) = 1$. This one might seem simpler, but it's a great way to practice converting from logarithmic to exponential form. Remember, if the base isn't written, it's a common logarithm (base 10). So, rewrite the equation in exponential form and solve for $x$. And, of course, check for extraneous solutions. For our third problem, let's try one that involves a subtraction of logarithms, just like our original example: Solve for $x$: $\log(x+2) - \log(x-1) = 1$. This is another opportunity to use the quotient rule of logarithms. Combine the logarithms, convert to exponential form, solve for $x$, and check those solutions! Now, let's spice things up a bit with an equation that involves a natural logarithm: Solve for $x$: $\ln(x+5) = 2$. Remember that $\ln$ represents the natural logarithm, which has a base of $e$ (Euler's number, approximately 2.71828). So, when you convert to exponential form, you'll be working with powers of $e$. Solve for $x$ and check your answer. And finally, let's try a problem that might lead to a quadratic equation: Solve for $x$: $\log(x) + \log(x-3) = 1$. After you combine the logarithms and convert to exponential form, you might end up with a quadratic equation. This means you'll need to factor or use the quadratic formula to find the solutions. Remember to check both solutions in the original equation to see if they are valid. Working through these practice problems will give you a solid foundation in solving logarithmic equations. Remember to take your time, write out each step clearly, and always double-check your work. You've got this, guys! Keep practicing, and you'll become a logarithmic equation-solving master!

Alright guys, we've journeyed through the world of logarithmic equations, and hopefully, you're feeling much more confident about tackling these types of problems. We started by understanding the core concepts, like the properties of logarithms and how to convert between logarithmic and exponential forms. We then dove into a step-by-step solution of our example equation, $\log (x+3) - \log (x+1) = 2$, making sure to cover every detail so you could follow along easily. We also highlighted the importance of checking for extraneous solutions, a step that can't be skipped if you want to get the correct answer. We talked about common mistakes that people make when solving logarithmic equations, so you know what to watch out for and how to avoid those pitfalls. From misapplying logarithm properties to forgetting to check for extraneous solutions, we covered the key areas where errors often occur. And finally, we provided a set of practice problems to give you the opportunity to flex your new skills and solidify your understanding. These problems covered a range of scenarios, from different bases to equations that lead to quadratic forms, ensuring you're well-prepared for anything that comes your way. Remember, solving logarithmic equations is like any other mathematical skill – it gets easier with practice. The more you work with these types of problems, the more comfortable you'll become with the steps involved and the more confident you'll feel in your ability to solve them. So, don't be afraid to dive in and give it a try! And if you get stuck, don't worry. Go back and review the concepts, the steps we've covered, and the common mistakes to avoid. Math is a journey, and every problem you solve is a step forward. You've now got a solid toolkit for solving logarithmic equations. You know the properties of logarithms, you know how to convert between logarithmic and exponential forms, and you know the importance of checking for extraneous solutions. So go out there and conquer those logarithmic equations, guys! You've got this!