Solving $\log _3(x+2)=\log _3(2 X^2-1)$ A Step-by-Step Guide

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Are you struggling with logarithmic equations? Do you find it challenging to navigate the steps required to find the solutions? This comprehensive guide will walk you through the process of solving the equation $\log _3(x+2)=\log _3(2 x^2-1)$, providing a clear and structured approach. Mastering logarithmic equations is crucial for various fields, including mathematics, physics, and engineering. By understanding the underlying principles and following a systematic method, you can confidently tackle even the most complex problems.

Understanding Logarithmic Equations

Before diving into the solution, let's briefly discuss logarithmic equations. A logarithmic equation is an equation that involves a logarithm of an expression containing a variable. The key to solving these equations lies in understanding the properties of logarithms and how they relate to exponential functions. Logarithms are essentially the inverse operation of exponentiation, allowing us to solve for exponents in equations. This inverse relationship is fundamental to the solution process.

The equation we're tackling, $\log _3(x+2)=\log _3(2 x^2-1)$, presents a classic example of a logarithmic equation where the variable x appears within the logarithmic arguments. The goal is to isolate x and determine the values that satisfy the equation. However, it's crucial to remember that not all solutions obtained algebraically are valid. We must check for extraneous solutions, which are values that arise during the solving process but do not satisfy the original equation. These extraneous solutions often occur because the logarithm function is only defined for positive arguments.

Step-by-Step Solution

Let's break down the solution process into manageable steps. Each step builds upon the previous one, leading us closer to the final solution. By understanding the rationale behind each step, you'll gain a deeper understanding of how to solve logarithmic equations in general.

Step 1 Equate the Arguments:

The cornerstone of solving this logarithmic equation lies in the fundamental property that if $\log _b(m) = \log _b(n)$, then m = n, provided that b is a positive number not equal to 1, and both m and n are positive. This property stems directly from the definition of logarithms as the inverse of exponentiation. If two logarithms with the same base are equal, their arguments must also be equal. Applying this property to our equation, $\log _3(x+2)=\log _3(2 x^2-1)$, we can confidently equate the arguments, setting the stage for further algebraic manipulation. This step effectively eliminates the logarithms, transforming the equation into a more familiar algebraic form.

By equating the arguments, we arrive at the equation x + 2 = 2x² - 1. This equation is a quadratic equation, a type of polynomial equation that can be solved using various techniques, such as factoring, completing the square, or using the quadratic formula. The transformation from a logarithmic equation to a quadratic equation is a critical step in the solution process, allowing us to leverage our knowledge of algebraic techniques to find potential solutions for x. However, it's important to remember that these potential solutions must be checked against the original logarithmic equation to ensure they are valid and not extraneous.

Step 2 Rearrange the Equation:

Now that we've equated the arguments, the next crucial step involves rearranging the equation to set it equal to zero. This manipulation is a standard procedure when dealing with quadratic equations, as it allows us to utilize methods like factoring or the quadratic formula to find the solutions. The goal is to bring all terms to one side of the equation, leaving zero on the other side. This rearrangement transforms the equation into its standard quadratic form, which is essential for applying the aforementioned solving techniques. Rearranging the equation not only prepares it for solving but also provides a clearer view of the coefficients and constants involved, which can be helpful in choosing the most efficient solution method.

Starting from the equation x + 2 = 2x² - 1, we need to manipulate it so that one side is equal to zero. This involves subtracting x and 2 from both sides of the equation. By performing these operations, we systematically move all terms to one side, resulting in the equation 0 = 2x² - x - 3. This rearranged equation is now in the standard quadratic form of ax² + bx + c = 0, where a = 2, b = -1, and c = -3. This standard form is crucial for applying the quadratic formula or attempting to factor the quadratic expression. With the equation in this form, we are well-positioned to proceed with solving for x.

Step 3: Factor the Quadratic:

With the quadratic equation neatly arranged in the form 0 = 2x² - x - 3, the next strategic step is to factor the quadratic expression. Factoring, when possible, provides a direct and often efficient method for finding the roots, or solutions, of the equation. The process involves decomposing the quadratic expression into the product of two binomials. This decomposition relies on identifying two numbers that, when multiplied, yield the constant term (-3) and, when added, yield the coefficient of the linear term (-1), taking into account the coefficient of the quadratic term (2). Successfully factoring the quadratic equation simplifies the problem, transforming it into two simpler linear equations that can be solved independently.

To factor the quadratic 2x² - x - 3, we seek two binomials of the form (Ax + B)(Cx + D) such that their product equals the original quadratic. This involves finding two numbers that multiply to 2 * -3 = -6 and add up to -1. These numbers are -3 and 2. We can then rewrite the middle term, -x, as -3x + 2x, allowing us to factor by grouping. This process leads to the factorization: 2x² - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3) = (2x - 3)(x + 1). Therefore, the factored form of the equation is 0 = (2x - 3)(x + 1). This factored form is a significant milestone, as it directly reveals the potential solutions for x by setting each factor equal to zero.

Step 4: Set Each Factor to Zero:

The factored form of the equation, 0 = (2x - 3)(x + 1), holds the key to unlocking the potential solutions for x. The principle at play here is the zero-product property, a fundamental concept in algebra. This property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. Applying this principle to our factored equation, we recognize that either (2x - 3) must equal zero, or (x + 1) must equal zero, or both. This realization transforms the single quadratic equation into two simpler linear equations, each of which can be solved independently to find a potential value for x. By setting each factor to zero, we create a pathway to isolate x and determine the possible solutions that satisfy the original equation.

Following the zero-product property, we set each factor in the equation 0 = (2x - 3)(x + 1) equal to zero, creating two separate equations: 2x - 3 = 0 and x + 1 = 0. These are linear equations, which are straightforward to solve. The equation 2x - 3 = 0 can be solved by adding 3 to both sides, resulting in 2x = 3, and then dividing both sides by 2, yielding x = 3/2. Similarly, the equation x + 1 = 0 can be solved by subtracting 1 from both sides, yielding x = -1. These two values, x = 3/2 and x = -1, are the potential solutions to the quadratic equation and, by extension, the original logarithmic equation. However, it is crucial to remember that these are only potential solutions until they are verified against the original equation to eliminate any extraneous solutions.

Step 5: Solve for x

Having set each factor to zero, the next step is to isolate x in each of the resulting linear equations. This process involves basic algebraic manipulations, such as adding or subtracting constants from both sides and dividing both sides by the coefficient of x. The goal is to obtain values for x that potentially satisfy the original logarithmic equation. These values represent the candidate solutions, but it is crucial to remember that not all solutions obtained algebraically are valid in the context of logarithmic equations. The domain of logarithmic functions is restricted to positive arguments, so we must verify that each candidate solution does not lead to the logarithm of a non-positive number in the original equation. This verification step is essential to identify and discard any extraneous solutions.

From the equation 2x - 3 = 0, we add 3 to both sides to get 2x = 3. Then, dividing both sides by 2, we find x = 3/2. This is one potential solution. From the equation x + 1 = 0, we subtract 1 from both sides to get x = -1. This is another potential solution. Thus, we have two candidate solutions: x = 3/2 and x = -1. These values represent the points where the quadratic expression equals zero. However, before declaring them as the final solutions to the logarithmic equation, we must meticulously check them against the original equation to ensure they are valid within the domain of the logarithmic functions involved. This verification process is a critical safeguard against extraneous solutions that may arise due to the nature of logarithms.

Step 6: Check for Extraneous Solutions

The final and arguably most critical step in solving logarithmic equations is checking for extraneous solutions. Extraneous solutions are values that emerge as potential solutions during the algebraic solving process but do not actually satisfy the original equation. These solutions arise because the domain of a logarithmic function is restricted to positive arguments. In other words, you can only take the logarithm of a positive number. Therefore, any value of x that results in a negative argument or zero within the logarithm is an extraneous solution and must be discarded. This verification step is paramount to ensure the accuracy and validity of the final solution set.

To check for extraneous solutions, we substitute each potential solution back into the original equation, $\log _3(x+2)=\log _3(2 x^2-1)$, and verify that the arguments of the logarithms remain positive. Let's start with x = 3/2. Substituting this value into the equation, we get $\log _3(3/2+2)=\log _3(2 (3/2)^2-1)$, which simplifies to $\log _3(7/2)=\log _3(7/2)$. Since both arguments are positive, x = 3/2 is a valid solution. Now, let's check x = -1. Substituting this value, we get $\log _3(-1+2)=\log _3(2 (-1)^2-1)$, which simplifies to $\log _3(1)=\log _3(1)$. Again, both arguments are positive, so x = -1 appears to be a valid solution. However, it is essential to remember that we are checking the arguments after simplification, and it's crucial to also check the arguments before simplification to ensure no issues arise. In this case, both solutions are valid.

The Solution

After carefully following each step, from equating the arguments to checking for extraneous solutions, we have arrived at the solution to the logarithmic equation $\log _3(x+2)=\log _3(2 x^2-1)$. The valid solutions are x = 3/2 and x = -1. These values satisfy the original equation and do not result in taking the logarithm of a non-positive number. By understanding the properties of logarithms and adhering to a systematic approach, you can confidently solve a wide range of logarithmic equations. Remember to always check for extraneous solutions to ensure the accuracy of your results.

Conclusion

Solving logarithmic equations requires a blend of algebraic manipulation and an understanding of the properties of logarithms. By following the steps outlined in this guide – equating arguments, rearranging the equation, factoring the quadratic, setting each factor to zero, solving for x, and, most importantly, checking for extraneous solutions – you can effectively tackle these types of problems. Mastering logarithmic equations is a valuable skill in mathematics and related fields, enabling you to solve a variety of real-world problems. Remember, practice makes perfect, so continue to work through examples and hone your problem-solving abilities.

Final Answer:

The final answer is 3/2,−1\boxed{3/2, -1}