Solve Logarithmic Equations: A Step-by-Step Guide

by Henrik Larsen 50 views

Hey guys! Let's dive into solving this logarithmic equation together. Logarithmic equations might seem tricky at first, but with a systematic approach, they become much easier to handle. In this article, we'll break down the process step-by-step, ensuring you understand each part clearly. We'll focus on the equation $\log _2(x-2)+\log _2 x=3$, and by the end, you'll not only know the correct answer but also the why behind it. So, grab your thinking caps, and let's get started!

Understanding Logarithmic Equations

Before we jump into solving the specific equation, it's important to understand what logarithmic equations are and the key properties that govern them. Logarithmic equations involve logarithms of variables, and solving them often requires converting them into exponential form. Remember, a logarithm is essentially the inverse operation of exponentiation. For example, if we have $\log_b a = c$, this means that $b^c = a$. This fundamental relationship is crucial in solving logarithmic equations.

Key Properties of Logarithms: There are a few key properties we need to keep in mind:

  1. Product Rule: $\log_b(mn) = \log_b m + \log_b n$ This rule states that the logarithm of a product is the sum of the logarithms.
  2. Quotient Rule: $\log_b(m/n) = \log_b m - \log_b n$ The logarithm of a quotient is the difference of the logarithms.
  3. Power Rule: $\log_b(m^p) = p \log_b m$ The logarithm of a number raised to a power is the product of the power and the logarithm of the number.
  4. Logarithm of the Base: $\log_b b = 1$ The logarithm of the base to itself is always 1.
  5. Logarithm of 1: $\log_b 1 = 0$ The logarithm of 1 to any base is always 0.

These properties are our toolkit when it comes to simplifying and solving logarithmic equations. Understanding these rules thoroughly will make solving equations like $\log _2(x-2)+\log _2 x=3$ much more manageable.

Now that we have refreshed our understanding of logarithmic properties, we can move on to the specific steps for solving our equation. The first critical step involves using the product rule to combine the logarithmic terms. This will simplify the equation and bring us closer to isolating the variable x. Remember, the key is to methodically apply these properties and keep track of each step to avoid common mistakes. Let's proceed by combining the logarithms, setting the stage for converting the equation into a more solvable form. So, with our toolkit ready, we're well-equipped to tackle the equation head-on!

Step-by-Step Solution of the Equation

Let's tackle the equation $\log _2(x-2)+\log _2 x=3$ step by step. The first thing we want to do is to simplify the left side of the equation. We have a sum of two logarithms with the same base (base 2), which is a perfect setup for applying the product rule of logarithms. Remember, the product rule states that $\log_b m + \log_b n = \log_b(mn)$. So, let's apply this rule to our equation:

log⁡2(x−2)+log2x=log2((x−2)x)\\\\\log _2(x-2)+\\\\log _2 x = \\\\log _2((x-2)x)

Now our equation looks like this:

log2((x−2)x)=3\\\\log _2((x-2)x) = 3

This is much simpler! The next step is to get rid of the logarithm altogether. To do this, we'll convert the logarithmic equation into its equivalent exponential form. Recall that $\log_b a = c$ is the same as $b^c = a$. Applying this to our equation, we get:

23=(x−2)x2^3 = (x-2)x

Since $2^3 = 8$, our equation now becomes:

8=(x−2)x8 = (x-2)x

Now we have a simple quadratic equation to solve. Let's expand the right side and rearrange the equation to get it into the standard quadratic form ( $ax^2 + bx + c = 0$ ):

8=x2−2x8 = x^2 - 2x

Subtract 8 from both sides to set the equation to zero:

x2−2x−8=0x^2 - 2x - 8 = 0

Now we need to factor this quadratic equation. We're looking for two numbers that multiply to -8 and add to -2. Those numbers are -4 and 2. So, we can factor the equation as follows:

(x−4)(x+2)=0(x - 4)(x + 2) = 0

To find the solutions for x, we set each factor equal to zero:

x−4=0textorx+2=0x - 4 = 0 \\text{ or } x + 2 = 0

Solving these gives us two possible solutions:

x=4textorx=−2x = 4 \\text{ or } x = -2

But hold on! We're not done yet. We need to check these solutions in the original equation to make sure they are valid. Logarithms are only defined for positive arguments, so we need to make sure that the values inside the logarithms ( $x-2$ and $x$ ) are positive. This is a crucial step to avoid extraneous solutions. Let's move on to the next section to check our solutions.

Checking for Extraneous Solutions

Alright, guys, we've found two potential solutions for x: 4 and -2. But before we celebrate, we need to make sure these solutions actually work in our original equation: $\log _2(x-2)+\log _2 x=3$. This is a super important step because logarithms have certain restrictions. Specifically, the argument of a logarithm (the thing inside the parentheses) must be positive. If we plug in a value for x that makes the argument of any logarithm negative or zero, that solution is called an extraneous solution, and we have to discard it.

Let's start by checking $x = 4$:

Plugging $x = 4$ into the original equation, we get:

log2(4−2)+log24=log2(2)+log24\\\\log _2(4-2)+\\\\log _2 4 = \\\\log _2(2) + \\\\log _2 4

We know that $\\log _2 2 = 1$ and $\\log _2 4 = 2$, so:

1+2=31 + 2 = 3

This checks out! So, $x = 4$ is a valid solution.

Now, let's check $x = -2$:

Plugging $x = -2$ into the original equation, we get:

log2(−2−2)+log2(−2)=log2(−4)+log2(−2)\\\\log _2(-2-2)+\\\\log _2 (-2) = \\\\log _2(-4) + \\\\log _2 (-2)

Here's the problem: we have logarithms of negative numbers. Logarithms of negative numbers are not defined in the real number system. Therefore, $x = -2$ is an extraneous solution, and we must reject it.

So, after checking our solutions, we find that only one of them is valid. This highlights the importance of checking solutions in logarithmic equations. We can't just assume that every solution we find algebraically is correct. We need to make sure it makes sense in the context of the original equation. This step often saves us from choosing the wrong answer on a test or in real-world applications.

Now that we've checked our solutions, we can confidently state the solution set for the equation. We've gone through the process of combining logarithms, converting to exponential form, solving the quadratic equation, and, most importantly, checking for extraneous solutions. So, let's state our final answer!

The Correct Answer

After walking through the steps of solving the equation $\log _2(x-2)+\log _2 x=3$, we've arrived at the correct answer. We combined the logarithms using the product rule, converted the equation to exponential form, solved the resulting quadratic equation, and crucially, we checked for extraneous solutions.

We found two potential solutions: $x = 4$ and $x = -2$. However, when we plugged these values back into the original equation, we discovered that $x = -2$ resulted in taking the logarithm of a negative number, which is undefined in the real number system. Therefore, $x = -2$ is an extraneous solution.

On the other hand, $x = 4$ worked perfectly in the original equation. Plugging it in, we got $\log _2(4-2)+\log _2 4 = \log _2(2) + \log _2 4 = 1 + 2 = 3$, which confirms that $x = 4$ is a valid solution.

Therefore, the solution set to the equation $\log _2(x-2)+\log _2 x=3$ is:

B. $x=4$

So, there you have it! We've successfully solved the logarithmic equation. Remember, the key to solving these types of problems is to follow a systematic approach: use logarithmic properties to simplify, convert to exponential form, solve the resulting equation, and always, always check for extraneous solutions. This careful approach will help you tackle even the trickiest logarithmic equations with confidence. Keep practicing, and you'll become a pro at solving these in no time! Great job, guys!