Solve Logarithmic Equations: Find All X Values

Hey math whizzes and number crunchers! Ever stared at a logarithmic equation and felt like you were deciphering an ancient scroll? You know, those pesky things with logs, exponents, and variables all jumbled up? Well, fear not, because today we're diving deep into one such puzzle: log₅(x² - 6) - log₅(4x + 6) = 0. We've also got some proposed solutions to check out: x = -2 and x = 6. Let's break this down, get to the bottom of it, and make sure we find all the correct values of 'x' that make this equation sing. Ready to level up your math game? Let's go!

Understanding the Core Concepts of Logarithms

Before we even think about tackling our specific equation, let's get a solid grip on what logarithms are all about. At its heart, a logarithm is the inverse operation to exponentiation. So, if you have an equation like $5^y = x$, the logarithmic form is $log_5(x) = y$. The base of the logarithm (in this case, 5) is the same base as the exponent. The log tells you what power you need to raise the base to in order to get the other number. Pretty neat, right? Understanding this fundamental relationship is crucial for manipulating and solving logarithmic equations.

Now, when we're dealing with equations involving logarithms, we often need to use some handy properties. These properties are like secret codes that help us simplify complex expressions. Here are a few of the biggies:

• Product Rule: $log_b(M) + log_b(N) = log_b(M * N)$. This means if you're adding two logs with the same base, you can combine them into a single log by multiplying their arguments.
• Quotient Rule: $log_b(M) - log_b(N) = log_b(M / N)$. This is the one we'll be using extensively today, guys! If you're subtracting two logs with the same base, you can combine them into a single log by dividing their arguments.
• Power Rule: $log_b(M^p) = p * log_b(M)$. This lets you bring an exponent down in front of the log.

Remembering and applying these properties correctly is your golden ticket to simplifying logarithmic equations. Without them, you'd be stuck trying to solve equations that look way more intimidating than they actually are. So, commit these to memory, jot them down, whatever works for you – they're your best friends in the world of logs!

Step-by-Step Solution for log₅(x² - 6) - log₅(4x + 6) = 0

Alright, team, let's get our hands dirty with our main event: log₅(x² - 6) - log₅(4x + 6) = 0. Our first move, as skilled log-wrestlers, is to use the quotient rule we just talked about. Since we're subtracting two logarithms with the same base (base 5, in this case), we can combine them into a single logarithm.

Applying the quotient rule, we get:

$log_5((x² - 6) / (4x + 6)) = 0$

Now, this looks a lot simpler, right? We've transformed a subtraction of two logs into a single log. Our next step is to get rid of the logarithm. To do this, we need to remember the definition of a logarithm: if $log_b(A) = C$, then $b^C = A$. In our equation, the base $b$ is 5, the argument $A$ is $(x² - 6) / (4x + 6)$, and the result $C$ is 0.

So, converting our logarithmic equation to its exponential form, we have:

$5^0 = (x² - 6) / (4x + 6)$

Now, we know that any non-zero number raised to the power of zero is equal to 1. So, $5^0 = 1$. This simplifies our equation to:

$1 = (x² - 6) / (4x + 6)$

To solve for 'x', we need to get rid of the fraction. We can do this by multiplying both sides of the equation by the denominator, $(4x + 6)$. But here's a crucial point, guys: we need to make sure that the denominator is not zero, because division by zero is undefined. So, we must have $4x + 6 ≠ 0$, which means $4x ≠ -6$, and thus $x ≠ -6/4$, or $x ≠ -3/2$. Keep this restriction in mind as we move forward!

Multiplying both sides by $(4x + 6)$, we get:

$1 * (4x + 6) = x² - 6$

$4x + 6 = x² - 6$

Now, we've got a quadratic equation on our hands! To solve a quadratic equation, we typically want to set it equal to zero. So, let's move all the terms to one side. Subtracting $4x$ and $6$ from both sides gives us:

$0 = x² - 4x - 6 - 6$

$0 = x² - 4x - 12$

Or, written the standard way:

$x² - 4x - 12 = 0$

Solving the Quadratic Equation: Finding Potential 'x' Values

We've successfully transformed our tricky logarithmic equation into a standard quadratic equation: $x² - 4x - 12 = 0$. Now, we have a few methods to solve this. We can try factoring, completing the square, or using the quadratic formula. Factoring is often the quickest if it works, so let's give that a shot first.

We're looking for two numbers that multiply to -12 and add up to -4. Let's brainstorm some pairs of factors for -12:

• 1 and -12 (sum = -11)
• -1 and 12 (sum = 11)
• 2 and -6 (sum = -4)
• -2 and 6 (sum = 4)
• 3 and -4 (sum = -1)
• -3 and 4 (sum = 1)

Bingo! The pair 2 and -6 fits the bill perfectly. They multiply to -12 and add up to -4. So, we can factor our quadratic equation as:

$(x + 2)(x - 6) = 0$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for 'x':

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

2. $x - 6 = 0 x = 6$

So, our potential solutions are $x = -2$ and $x = 6$. These are the same values proposed in the problem description, which is a good sign! However, we are not done yet, guys! In logarithmic equations, we always need to check our solutions in the original equation to make sure they are valid.

The Crucial Step: Checking for Extraneous Solutions

This is perhaps the most important part of solving logarithmic equations, and it's where many people slip up. We MUST check our potential solutions in the original equation: log₅(x² - 6) - log₅(4x + 6) = 0. Why? Because the argument of a logarithm (the part inside the parentheses) must always be positive. If plugging in a value for 'x' results in a zero or negative argument for any logarithm in the original equation, that solution is called an extraneous solution and must be discarded.

Let's check our first proposed solution, x = -2:

We need to evaluate the arguments of the logarithms:

• $x² - 6 = (-2)² - 6 = 4 - 6 = -2$
• $4x + 6 = 4(-2) + 6 = -8 + 6 = -2$

Uh oh! Both arguments, $(x² - 6)$ and $(4x + 6)$, result in -2, which is a negative number. Since the argument of a logarithm cannot be negative, x = -2 is an extraneous solution and must be rejected. It doesn't work in the original equation.

Now, let's check our second proposed solution, x = 6:

Let's evaluate the arguments again:

• $x² - 6 = (6)² - 6 = 36 - 6 = 30$
• $4x + 6 = 4(6) + 6 = 24 + 6 = 30$

Both arguments, 30 and 30, are positive. This means x = 6 is a valid input for the logarithms. Let's plug it back into the original equation to see if it holds true:

$log_5(30) - log_5(30) = 0$

This is clearly true, as any number subtracted from itself equals zero. Therefore, x = 6 is a valid solution.

Final Thoughts and Key Takeaways

So, after all that hard work, we found that the only true solution to the equation log₅(x² - 6) - log₅(4x + 6) = 0 is x = 6. It's super important to remember that while solving the quadratic equation gave us two possibilities ($x = -2$ and $x = 6$), the domain restrictions of logarithms forced us to eliminate one of them. Always, always, always check your solutions in the original logarithmic equation! This step is non-negotiable.

Here’s a quick recap of our journey:

1. Understand Log Properties: Master the product, quotient, and power rules.
2. Apply Quotient Rule: Combine the logs using $log_b(M) - log_b(N) = log_b(M / N)$.
3. Convert to Exponential Form: Use $log_b(A) = C b^C = A$.
4. Solve the Resulting Equation: This often leads to a quadratic equation.
5. Check for Domain Restrictions: Ensure the arguments of all original logs are positive.
6. Verify Solutions: Plug potential solutions back into the original equation to eliminate extraneous ones.

By following these steps diligently, you guys can confidently tackle any logarithmic equation thrown your way. Keep practicing, stay curious, and happy solving!