Unraveling Logarithmic Equations: A Step-by-Step Guide

Hey guys! Ready to dive into the world of logarithms? Today, we're tackling a logarithmic equation, and I'll walk you through it step by step. We'll be solving the equation $\log_{12}(x+3) + \log_{12}(x-1) = \log_{12}(x-2)$. Don't worry if it looks intimidating at first; we'll break it down into manageable chunks. Our goal is to find the value(s) of x that satisfy this equation. We'll express our answer either as an exact expression or as a decimal approximation rounded to four decimal places. If, by some chance, there's no solution, we'll indicate that with "No Solution (Ø)". So, grab your pencils, and let's get started!

Understanding the Basics of Logarithms

Before we jump into the equation, let's quickly recap some essential logarithm properties. These properties are the keys to unlocking and solving logarithmic equations. Remember, the logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. In our equation, the base is 12. Two fundamental properties will be super helpful for us today. First, the product rule: $\log_b(m) + \log_b(n) = \log_b(mn)$. This rule tells us that the sum of logarithms with the same base is the logarithm of the product of the arguments. Second, the power rule or the inverse of the product rule: $\log_b(m) - \log_b(n) = \log_b(m/n)$. Knowing these properties allows us to combine and simplify logarithmic expressions, which is exactly what we need to do to solve the equation. Also, it's crucial to understand that the argument of a logarithm (the part inside the parentheses) must always be greater than zero. This restriction is crucial, and we'll need to check our solutions to make sure they fit this requirement. If we get a solution that makes the argument negative, we have to discard it, because logarithms of negative numbers are undefined in the real number system. Keep these basics in mind, and you'll be well on your way to mastering logarithmic equations. Furthermore, the base of the logarithm plays a crucial role. In our case, the base is 12, which means the logarithm is asking the question, "To what power must I raise 12 to get this number?" It's a fundamental question in mathematics, and understanding it will help you solve problems with ease. The properties we mentioned above make solving equations like these manageable, but it's important to keep in mind the underlying principles. Ready? Let's get down to the business of solving our particular equation.

Applying Logarithmic Properties to Simplify the Equation

Alright, let's start solving $\log_{12}(x+3) + \log_{12}(x-1) = \log_{12}(x-2)$. The first step is to use the product rule on the left side of the equation. This simplifies the equation, making it easier to solve. According to the product rule, $\log_b(m) + \log_b(n) = \log_b(mn)$. Applying this to our equation, we get $\log_{12}((x+3)(x-1)) = \log_{12}(x-2)$. Now that both sides of the equation have a single logarithm with the same base, we can remove the logarithms. Basically, if $\log_b(m) = \log_b(n)$, then m = n. So, we're left with $(x+3)(x-1) = x-2$. We've essentially transformed our logarithmic equation into a quadratic equation, which is much easier to work with. Expanding the left side gives us $x^2 + 2x - 3 = x - 2$. This is a big step, guys, because it has reduced the problem to something we know how to deal with - simple algebra. The key takeaway from this phase is that applying logarithmic properties is crucial to simplifying the equation. Simplifying the equation makes it much easier to solve and eventually find the value of x. The goal is always to get the x out of the logarithm and to do this, we need to know the properties. The quadratic equation formed can be solved by different methods, such as factoring, completing the square, or using the quadratic formula. Let's move on to the next step and see how we solve this quadratic equation and find the value of x.

Solving the Resulting Quadratic Equation

Now that we have the quadratic equation, $x^2 + 2x - 3 = x - 2$, let's bring all the terms to one side to set the equation to zero. Subtracting x and adding 2 to both sides gives us $x^2 + x - 1 = 0$. We now have a standard quadratic equation in the form of $ax^2 + bx + c = 0$. Since this quadratic equation doesn't factor easily, we can use the quadratic formula to solve for x. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In our equation, $a = 1$, $b = 1$, and $c = -1$. Plugging these values into the quadratic formula, we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$. Simplifying further, we get $x = \frac{-1 \pm \sqrt{1 + 4}}{2}$, which means $x = \frac{-1 \pm \sqrt{5}}{2}$. Therefore, the two potential solutions are $x = \frac{-1 + \sqrt{5}}{2}$ and $x = \frac{-1 - \sqrt{5}}{2}$. But hold your horses, we aren't done yet! Remember those restrictions we talked about earlier? We need to check if these solutions are valid by substituting them back into the original logarithmic equation to ensure the arguments of the logarithms are positive. This check is crucial because it helps us identify extraneous solutions—solutions that arise during the solving process but don't actually satisfy the original equation. Let's start this check, before we are done, to know which is the correct solution.

Checking for Valid Solutions and Extraneous Solutions

Okay, guys, here comes the crucial part: checking our potential solutions to make sure they're valid. We've got two possible values for x: $x = \frac{-1 + \sqrt{5}}{2}$ and $x = \frac{-1 - \sqrt{5}}{2}$. We'll plug each of these back into the original equation, $\log_{12}(x+3) + \log_{12}(x-1) = \log_{12}(x-2)$, to see if they work. First, let's consider $x = \frac{-1 + \sqrt{5}}{2}$. Plugging this into the arguments of the logarithms, we get $(x+3)$, $(x-1)$, and $(x-2)$. If we calculate these values, we'll see that $x + 3 \approx 2.618$, $x - 1 \approx 0.618$, and $x - 2 \approx -1.382$. Notice that $x-2$ is negative. Because the argument of a logarithm cannot be negative, this means that $x = \frac{-1 + \sqrt{5}}{2}$ is an extraneous solution and is not valid. Now, let's consider the second potential solution, $x = \frac{-1 - \sqrt{5}}{2}$. This is approximately $-1.618$. Plugging this value into the original equation, we'd find that all the arguments of the logarithms become negative. Specifically, $x + 3 \approx 1.382$, $x - 1 \approx -2.618$, and $x - 2 \approx -3.618$. Since we cannot have negative arguments in a logarithm, $x = \frac{-1 - \sqrt{5}}{2}$ is also an extraneous solution. Because both of our potential solutions are extraneous, we can conclude that the original logarithmic equation has no valid solutions. Therefore, the answer is