Solving Logarithmic Equations A Step-by-Step Guide For Log₃(x) + Log₃(x² + 2) = 1 + 2log₃(x)

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Introduction

In this comprehensive article, we will delve into the step-by-step process of solving the logarithmic equation log₃(x) + log₃(x² + 2) = 1 + 2log₃(x). Logarithmic equations often present a challenge, but with a systematic approach and a clear understanding of logarithmic properties, we can effectively find the solutions. This exploration will not only provide the solutions but also enhance your problem-solving skills in the realm of mathematics. We will cover the fundamental concepts, the step-by-step solution, and finally, the verification of the solutions. This article aims to serve as a valuable resource for students, educators, and anyone interested in mastering logarithmic equations.

Understanding Logarithmic Equations

Before we dive into solving the specific equation, it's essential to grasp the basics of logarithms. A logarithm is essentially the inverse operation to exponentiation. The expression logₐ(b) = c can be rewritten in exponential form as aᶜ = b, where 'a' is the base, 'b' is the argument, and 'c' is the exponent. Understanding this relationship is crucial for manipulating and solving logarithmic equations. Key logarithmic properties such as the product rule (logₐ(mn) = logₐ(m) + logₐ(n)), quotient rule (logₐ(m/n) = logₐ(m) - logₐ(n)), and power rule (logₐ(mᵖ) = p*logₐ(m)) are indispensable tools in our arsenal. These properties allow us to simplify complex logarithmic expressions and transform equations into a more manageable form. Furthermore, it's crucial to remember that the argument of a logarithm must be positive, as the logarithm of a non-positive number is undefined. This constraint often leads to restrictions on the possible solutions of logarithmic equations, which we will need to consider when verifying our answers.

Step-by-Step Solution

1. Combine Logarithmic Terms:

Our initial goal is to consolidate the logarithmic terms on one side of the equation. We start with the equation log₃(x) + log₃(x² + 2) = 1 + 2log₃(x). To achieve this, we subtract 2log₃(x) from both sides, resulting in log₃(x) + log₃(x² + 2) - 2log₃(x) = 1. This step brings all the logarithmic terms to the left side, setting the stage for further simplification. The next crucial step involves applying the power rule of logarithms to rewrite 2log₃(x) as log₃(x²). This transformation is essential as it allows us to combine the logarithmic terms using the product rule.

2. Apply Logarithmic Properties:

Now, we apply the power rule of logarithms, which states that p*logₐ(m) = logₐ(mᵖ). Applying this rule, we rewrite 2log₃(x) as log₃(x²). Our equation now looks like log₃(x) + log₃(x² + 2) - log₃(x²) = 1. The next step is to use the product rule, which states that logₐ(m) + logₐ(n) = logₐ(mn), to combine the first two terms. This gives us log₃(x(x² + 2)) - log₃(x²) = 1. Subsequently, we use the quotient rule, which states that logₐ(m) - logₐ(n) = logₐ(m/n), to combine the remaining logarithmic terms. This simplifies the equation to log₃((x(x² + 2))/x²) = 1. By applying these logarithmic properties, we've significantly simplified the equation, paving the way for the next step in solving for x.

3. Simplify and Convert to Exponential Form

Following the application of logarithmic properties, our equation stands as log₃((x(x² + 2))/x²) = 1. The next step involves simplifying the argument of the logarithm. We can simplify the expression inside the logarithm by canceling out an 'x' from the numerator and denominator, resulting in log₃((x² + 2)/x) = 1. Now, we convert the logarithmic equation into its equivalent exponential form. Recall that logₐ(b) = c is equivalent to aᶜ = b. Applying this principle, we rewrite our equation as 3¹ = (x² + 2)/x. This transformation is crucial because it eliminates the logarithm, allowing us to work with a simpler algebraic equation. The equation 3 = (x² + 2)/x is now a rational equation that we can solve using standard algebraic techniques.

4. Solve the Quadratic Equation

After converting the logarithmic equation to exponential form, we have the equation 3 = (x² + 2)/x. To solve for x, we first multiply both sides by x to eliminate the fraction, resulting in 3x = x² + 2. Next, we rearrange the equation to form a standard quadratic equation by subtracting 3x from both sides, giving us x² - 3x + 2 = 0. Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is the most straightforward approach. We look for two numbers that multiply to 2 and add up to -3. These numbers are -1 and -2. Thus, we can factor the quadratic equation as (x - 1)(x - 2) = 0. Setting each factor equal to zero gives us two possible solutions: x - 1 = 0 and x - 2 = 0. Solving these linear equations, we find x = 1 and x = 2. These are the potential solutions to our original logarithmic equation. However, it's crucial to remember that we must verify these solutions to ensure they are valid.

Verification of Solutions

1. Substitute Solutions into the Original Equation

Verification is a critical step in solving logarithmic equations because logarithmic functions have domain restrictions. Specifically, the argument of a logarithm must be positive. We found two potential solutions: x = 1 and x = 2. To verify these solutions, we substitute each value back into the original equation, log₃(x) + log₃(x² + 2) = 1 + 2log₃(x), and check if the equation holds true. First, let's substitute x = 1: log₃(1) + log₃(1² + 2) = 1 + 2log₃(1). Simplifying, we get 0 + log₃(3) = 1 + 2(0), which further simplifies to 1 = 1. This confirms that x = 1 is a valid solution. Next, we substitute x = 2: log₃(2) + log₃(2² + 2) = 1 + 2log₃(2). Simplifying, we get log₃(2) + log₃(6) = 1 + 2log₃(2). Using the product rule, we can rewrite log₃(2) + log₃(6) as log₃(2 * 6) = log₃(12). So, the equation becomes log₃(12) = 1 + 2log₃(2). Rewriting 1 as log₃(3), we have log₃(12) = log₃(3) + log₃(2²), which simplifies to log₃(12) = log₃(3) + log₃(4). Using the product rule again, we get log₃(12) = log₃(3 * 4) = log₃(12). This confirms that x = 2 is also a valid solution.

2. Check for Extraneous Solutions

After substituting the potential solutions back into the original equation, we verified that both x = 1 and x = 2 satisfy the equation. However, it's essential to explicitly check for extraneous solutions, which are solutions that arise during the solving process but do not satisfy the original equation. Extraneous solutions often occur in logarithmic equations due to the domain restrictions of logarithmic functions. In our case, we need to ensure that the arguments of all logarithms in the original equation are positive for both x = 1 and x = 2. For x = 1, the arguments are x = 1 and x² + 2 = 3, both of which are positive. For x = 2, the arguments are x = 2 and x² + 2 = 6, which are also positive. Since both potential solutions satisfy the domain restrictions and the original equation, we can confidently conclude that they are valid solutions and not extraneous. This thorough verification process ensures the accuracy of our solutions.

Conclusion

In conclusion, we have successfully solved the logarithmic equation log₃(x) + log₃(x² + 2) = 1 + 2log₃(x). By systematically applying logarithmic properties, converting the equation to exponential form, solving the resulting quadratic equation, and verifying the solutions, we found that the solutions are x = 1 and x = 2. This exercise highlights the importance of understanding logarithmic properties and the necessity of verifying solutions to avoid extraneous results. Mastering these techniques will undoubtedly enhance your ability to tackle a wide range of logarithmic equations and related mathematical problems. Remember to always consider the domain restrictions of logarithmic functions when solving equations, and verification should always be the final step in the problem-solving process. This article serves as a comprehensive guide to solving logarithmic equations, providing a clear and methodical approach for students, educators, and anyone interested in deepening their mathematical knowledge.