Solving Logarithmic Equations Step By Step \$\log_3(x+2) = \log_3(2x^2-1)\\$

Solving logarithmic equations can seem daunting, but by following a structured approach, you can navigate through the process with confidence. This guide will walk you through the steps to solve the equation $\log_3(x+2) = \log_3(2x^2-1)$, providing a clear and comprehensive understanding of each stage. We will break down the problem into manageable steps, ensuring you grasp the underlying principles and techniques involved.

Step-by-Step Solution

1. Equate the Arguments

The cornerstone of solving logarithmic equations lies in understanding the properties of logarithms. When we have two logarithmic expressions with the same base set equal to each other, we can equate their arguments. This is because the logarithmic function is one-to-one, meaning that if $\log_b(a) = \log_b(c)$, then $a = c$. In our equation, $\log_3(x+2) = \log_3(2x^2-1)$, both logarithms have the base 3. Therefore, we can equate the arguments:

$x + 2 = 2x^2 - 1$

This step simplifies the equation significantly, transforming it from a logarithmic equation into a quadratic equation, which we can solve using algebraic methods. This initial step is crucial as it lays the foundation for the rest of the solution. By understanding and applying this property of logarithms, you can effectively simplify complex equations and pave the way for finding the solutions.

2. Rearrange the Equation

Now that we have equated the arguments, our goal is to manipulate the resulting equation into a standard quadratic form. The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable we want to solve for. To achieve this, we need to rearrange the equation $x + 2 = 2x^2 - 1$. We can start by subtracting $x$ and 2 from both sides of the equation:

$0 = 2x^2 - 1 - x - 2$

Combining the constant terms, we get:

$0 = 2x^2 - x - 3$

This equation is now in the standard quadratic form, making it easier to identify the coefficients and apply methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. Rearranging the equation is a vital step in solving for $x$, as it sets up the equation in a familiar and manageable format.

3. Factor the Quadratic Equation

With our quadratic equation in the standard form $2x^2 - x - 3 = 0$, we can now proceed to factor it. Factoring involves expressing the quadratic expression as a product of two binomials. To factor $2x^2 - x - 3$, we look for two numbers that multiply to the product of the leading coefficient (2) and the constant term (-3), which is -6, and add up to the middle coefficient (-1). These two numbers are -3 and 2. We can rewrite the middle term using these numbers:

$2x^2 - 3x + 2x - 3 = 0$

Now, we can factor by grouping:

$x(2x - 3) + 1(2x - 3) = 0$

$(2x - 3)(x + 1) = 0$

Factoring the quadratic equation is a crucial step as it allows us to find the potential solutions for $x$. By expressing the quadratic as a product of two binomials, we can set each binomial equal to zero and solve for $x$, leading us to the possible values that satisfy the equation.

4. Set Each Factor to Zero

Having factored the quadratic equation into $(2x - 3)(x + 1) = 0$, we can now apply the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero:

$2x - 3 = 0$ or $x + 1 = 0$

This step is essential because it breaks down the quadratic equation into two simpler linear equations, which are much easier to solve. By setting each factor to zero, we create a direct path to finding the potential solutions for $x$. This application of the zero-product property is a fundamental technique in solving factored equations.

5. Solve for x

Now that we have two linear equations, $2x - 3 = 0$ and $x + 1 = 0$, we can solve each one for $x$. For the first equation, $2x - 3 = 0$, we add 3 to both sides:

$2x = 3$

Then, we divide by 2:

$x = \frac{3}{2}$

For the second equation, $x + 1 = 0$, we subtract 1 from both sides:

$x = -1$

Thus, we have found two potential solutions for $x$: $\frac{3}{2}$ and -1. Solving for $x$ in each linear equation is a straightforward algebraic process, but it is a critical step in determining the possible values that satisfy the original equation. These potential solutions must now be checked to ensure they are valid within the context of the original logarithmic equation.

6. Check for Extraneous Solutions

In the realm of logarithmic equations, it's crucial to verify that our solutions don't lead to taking the logarithm of a non-positive number. Logarithms are only defined for positive arguments, so any solution that makes the argument of a logarithm zero or negative is considered an extraneous solution and must be discarded. Our potential solutions are $x = \frac{3}{2}$ and $x = -1$. Let's substitute each value back into the original equation, $\log_3(x+2) = \log_3(2x^2-1)$, and check if the arguments of the logarithms remain positive.

For $x = \frac{3}{2}$:

$x + 2 = \frac{3}{2} + 2 = \frac{7}{2} > 0$

$2x^2 - 1 = 2(\frac{3}{2})^2 - 1 = 2(\frac{9}{4}) - 1 = \frac{9}{2} - 1 = \frac{7}{2} > 0$

Since both arguments are positive, $x = \frac{3}{2}$ is a valid solution.

For $x = -1$:

$x + 2 = -1 + 2 = 1 > 0$

$2x^2 - 1 = 2(-1)^2 - 1 = 2(1) - 1 = 1 > 0$

However, if we proceed with the solution, we need to recognize a critical detail that is not immediately apparent:

Substitute $x = -1$ into the original equation:

$\log_3(-1 + 2) = \log_3(2(-1)^2 - 1)$

$\log_3(1) = \log_3(1)$

$0 = 0$

The solution appears to hold up when substituted back into the original equation. But, in the intermediate step where we equated the arguments, the solution may behave differently.

If we substitute x = -1 into the simplified equation from step 1:

$x + 2 = 2x^2 - 1$

$-1 + 2 = 2(-1)^2 - 1$

$1 = 1$

This is valid.

If the step above the solution is the factored quadratic expression, this is where issues may be visible:

$(2x - 3)(x + 1) = 0$

Since $(x+1)$ is a factor of the above, the issue was already apparent here.

We know that factoring is a proper mathematical operation, so we do not expect errors by factoring.

Therefore, both arguments are positive, x = -1 is also a valid solution, even though there may be discussion amongst mathematicians.

Discussion Points on Extraneous Solutions

In some contexts, mathematicians might define extraneous solutions more strictly. For example, if the domain of the original logarithmic functions is considered before any simplification steps, the solution $x=-1$ might be deemed extraneous by some because substituting $x=-1$ into the original equation requires careful consideration of the domain of the logarithmic functions involved.

In the expression $\log_3(x+2)$, the argument $x+2$ must be strictly greater than 0, meaning $x > -2$. Similarly, for $\log_3(2x^2-1)$, the argument $2x^2-1$ must also be strictly greater than 0, implying $2x^2 > 1$, and $x^2 > \frac{1}{2}$. This leads to $x > \frac{1}{\sqrt{2}}$ or $x < -\frac{1}{\sqrt{2}}$.

If we apply these domain restrictions rigorously from the start, $x=-1$ technically falls into the domain of each individual logarithmic expression but it is essential to verify it within the context of the entire equation due to potential inconsistencies that can arise from the logarithmic properties used during simplification.

However, the solution does hold in the original equation.

Therefore, potential solutions are $\frac{3}{2}$ and -1.

Conclusion

By following these steps, you can effectively solve logarithmic equations. Remember to always check your solutions to ensure they are valid and not extraneous. Understanding the properties of logarithms and applying algebraic techniques are key to mastering these types of problems.