Solving Logarithmic Equations A Step-by-Step Guide

Hey guys! Let's dive into solving a logarithmic equation step-by-step. We're tackling the equation log₄₈(x+2) + log₄₈(x) = 1, and our goal is to find the exact value of x. Logarithmic equations might seem intimidating at first, but with a few key properties and a bit of algebraic manipulation, we can crack this nut! We'll walk through each step, making sure to explain the why behind the how. So, grab your thinking caps, and let's get started!

Understanding Logarithmic Properties

Before we jump into solving, let's quickly review the fundamental logarithmic property that we'll be using. The property states that the sum of the logarithms of two numbers with the same base is equal to the logarithm of the product of those numbers. Mathematically, this is expressed as: logₐ(m) + logₐ(n) = logₐ(mn). This property is super crucial because it allows us to combine multiple logarithmic terms into a single, more manageable term. Think of it as a way to compress information, making the equation easier to handle. We will also use the definition of logarithms, which states that logₐ(b) = c is equivalent to aᶜ = b. This definition provides a bridge between logarithmic and exponential forms, which is essential for solving logarithmic equations. Mastering these properties is like having the right tools in your toolbox – they’re essential for tackling any logarithmic problem that comes your way. Remember, practice makes perfect, so the more you use these properties, the more natural they'll become.

Combining Logarithms

The first step in solving our equation, log₄₈(x+2) + log₄₈(x) = 1, is to use the logarithmic property we just discussed to combine the two logarithmic terms on the left side. Applying the property logₐ(m) + logₐ(n) = logₐ(mn), we can rewrite the left side as a single logarithm. In our case, a is 48, m is (x+2), and n is x. So, we get: log₄₈((x+2) * x) = 1. This simplifies to log₄₈(x² + 2x) = 1. By combining the logarithms, we've transformed the equation into a simpler form where we have a single logarithmic term equal to a constant. This is a significant step forward because it allows us to use the definition of logarithms to convert the equation into an exponential form. Remember, the goal here is to isolate x, and by condensing the logarithmic terms, we're one step closer to achieving that. So, we've successfully compressed our equation, making it much easier to handle in the subsequent steps.

Converting to Exponential Form

Now that we have log₄₈(x² + 2x) = 1, it’s time to leverage the definition of logarithms to convert this equation into exponential form. As a refresher, the logarithmic equation logₐ(b) = c is equivalent to the exponential equation aᶜ = b. In our case, a is 48, b is (x² + 2x), and c is 1. Applying this conversion, we get 48¹ = x² + 2x. This simplifies to 48 = x² + 2x. This transformation is a pivotal moment in solving the equation. We've effectively eliminated the logarithm, and now we're dealing with a quadratic equation, which we know how to solve. Converting to exponential form is like switching gears in a car – it allows us to move from one type of mathematical expression to another, making the problem more accessible. So, by applying the definition of logarithms, we've successfully transitioned our equation into a familiar algebraic form, setting the stage for the next step: solving the quadratic equation.

Solving the Quadratic Equation

We've arrived at the quadratic equation x² + 2x = 48. To solve this, the first step is to rearrange the equation into the standard quadratic form, which is ax² + bx + c = 0. Subtracting 48 from both sides, we get x² + 2x - 48 = 0. Now, we need to factor this quadratic. We're looking for two numbers that multiply to -48 and add up to 2. Those numbers are 8 and -6. So, we can factor the quadratic as (x + 8)(x - 6) = 0. Setting each factor equal to zero gives us two potential solutions for x: x + 8 = 0 and x - 6 = 0. Solving these equations, we find x = -8 and x = 6. These are our candidate solutions, but we're not done yet! We need to check these solutions in the original logarithmic equation to make sure they're valid. Remember, logarithms are only defined for positive arguments, so we need to ensure that our solutions don't lead to taking the logarithm of a negative number or zero.

Checking for Extraneous Solutions

Now comes the crucial step of checking for extraneous solutions. Remember, logarithmic functions have domain restrictions – specifically, the argument of a logarithm must be positive. We need to plug our candidate solutions, x = -8 and x = 6, back into the original equation, log₄₈(x+2) + log₄₈(x) = 1, to see if they hold true and don't result in taking the logarithm of a non-positive number. Let's start with x = -8. Plugging this into the original equation, we get log₄₈(-8+2) + log₄₈(-8) = log₄₈(-6) + log₄₈(-8). Since we can't take the logarithm of a negative number, x = -8 is an extraneous solution and must be discarded. Now, let's check x = 6. Plugging this in, we get log₄₈(6+2) + log₄₈(6) = log₄₈(8) + log₄₈(6). Both 8 and 6 are positive, so this solution is potentially valid. Let's verify if it satisfies the original equation: log₄₈(8) + log₄₈(6) = log₄₈(8 * 6) = log₄₈(48) = 1. Since this holds true, x = 6 is a valid solution. Checking for extraneous solutions is like the quality control step in a manufacturing process – it ensures that our final answer is correct and doesn't violate any mathematical rules. So, we've successfully weeded out the bad solution and confirmed our valid solution.

Final Answer

After carefully checking our solutions, we've determined that the only valid solution to the equation log₄₈(x+2) + log₄₈(x) = 1 is x = 6. The solution x = -8 was extraneous because it resulted in taking the logarithm of a negative number. Therefore, our final answer, in exact form, is x = 6. We've successfully navigated through the logarithmic equation, combining logarithms, converting to exponential form, solving the quadratic, and, most importantly, checking for extraneous solutions. Remember, guys, the key to mastering logarithmic equations is understanding the properties and definitions, and always, always checking your solutions! So, there you have it – a step-by-step solution to this logarithmic puzzle!