Solve Logarithmic Equations: 2 Log X - Log(x-2) = 2 Log 3

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Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a tricky logarithmic equation. If you're a math whiz or just looking to brush up on your skills, you've come to the right place. We're going to break down the equation $2 \log x-\log (x-2)=2 \log 3$ step-by-step, making sure you understand every single part of the process. Logarithmic equations can seem intimidating at first, but with the right approach and a solid understanding of logarithm properties, they become quite manageable. So, grab your notebooks, get comfortable, and let's get this solved! We'll cover everything from the basic properties of logarithms to the final checks you need to make to ensure your solution is valid. This isn't just about getting the answer; it's about building your confidence and problem-solving abilities. Let's make math fun and accessible, just the way we like it here at Plastik Magazine!

Understanding the Basics of Logarithms

Before we jump into solving our specific equation, $2 \log x-\log (x-2)=2 \log 3$, let's quickly recap some fundamental logarithm properties. These are the tools in our toolbox that will help us simplify and manipulate the equation. Remember, the logarithm of a number is the exponent to which a fixed base must be raised to produce that number. For example, $\log_b a = c$ means $b^c = a$. In our equation, we're dealing with what's commonly called the common logarithm, which has a base of 10 (though the base doesn't critically affect the solving process here, it's good to keep in mind). The properties we'll be using most frequently are:

• The Power Rule: $n \log_b M = \log_b (M^n)$. This rule is crucial because it allows us to move coefficients into the exponent of the logarithm. We'll use this on the $2 \log x$ term.
• The Quotient Rule: $\log_b M - \log_b N = \log_b (M/N)$. This rule helps us combine two logarithms with subtraction into a single logarithm. We'll apply this to the left side of our equation.
• The Equality Rule: If $\log_b M = \log_b N$, then $M=N$. This is the golden rule that allows us to eliminate the logarithms once both sides of the equation are in the form of a single logarithm.

It's also super important to remember the domain restrictions for logarithms. The argument of a logarithm (the part inside the parentheses) must always be positive. So, for $\log x$, we must have $x > 0$. For $\log (x-2)$, we must have $x-2 > 0$, which means $x > 2$. Combining these, our solution must satisfy $x > 2$. Any potential solution we find that doesn't meet this condition is an extraneous solution and must be discarded. Keeping these rules and restrictions in mind will make solving our equation a breeze. We're going to apply these properties systematically to simplify the equation and isolate our variable, $x$. Let's get started with the first step of manipulation!

Step-by-Step Solution

Alright, team, let's tackle this equation head-on: $2 \log x-\log (x-2)=2 \log 3$. The first thing we want to do is simplify each side of the equation by applying the logarithm properties we just discussed. Notice the coefficients in front of the logarithms? That's our cue to use the Power Rule. On the left side, we have $2 \log x$. Applying the power rule, this becomes $\log (x^2)$. Now, the left side of our equation looks like this: $\log (x^2) - \log (x-2)$.

Next, we can use the Quotient Rule to combine these two logarithms on the left side. Remember, subtraction of logarithms means division of their arguments. So, $\log (x^2) - \log (x-2)$ simplifies to $\log \left(\frac{x^2}{x-2}\right)$.

Now let's look at the right side of the original equation: $2 \log 3$. Again, we use the Power Rule. This simplifies to $\log (3^2)$, which is $\log 9$.

So, after applying these rules, our equation has transformed into a much simpler form: $\log \left(\frac{x^2}{x-2}\right) = \log 9$.

We're now at a point where we have a single logarithm on each side of the equation, and the bases are the same (implied base 10). This is where the Equality Rule comes into play. If $\log M = \log N$, then $M=N$. Therefore, we can set the arguments of the logarithms equal to each other:

$\frac{x^2}{x-2} = 9$

This has effectively removed the logarithms, leaving us with an algebraic equation to solve. This is a common strategy in solving logarithmic equations: manipulate them using logarithm properties until you can eliminate the logs entirely.

Solving the Resulting Algebraic Equation

We've successfully transformed our logarithmic equation into an algebraic one: $\frac{x^2}{x-2} = 9$. Now, our mission is to solve for $x$. To get rid of the fraction, we can multiply both sides of the equation by the denominator, $(x-2)$. Important Note: Remember our domain restrictions? We established that $x > 2$. This means $x-2$ will never be zero, so multiplying by it is a safe operation. If $x=2$ were a possibility, we'd have to be more cautious.

Multiplying both sides by $(x-2)$, we get:

$x^2 = 9(x-2)$

Now, let's distribute the 9 on the right side:

$x^2 = 9x - 18$

To solve this quadratic equation, we need to set it equal to zero. We can do this by subtracting $9x$ and adding $18$ to both sides:

$x^2 - 9x + 18 = 0$

This is a standard quadratic equation in the form $ax^2 + bx + c = 0$. We can solve this by factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method if the quadratic is easily factorable. We're looking for two numbers that multiply to 18 (our 'c' term) and add up to -9 (our 'b' term). Let's think about pairs of factors for 18: (1, 18), (2, 9), (3, 6). To get a sum of -9, we need both numbers to be negative. So, let's try (-1, -18), (-2, -9), and (-3, -6). Aha! -3 and -6 add up to -9 and multiply to 18. So, we can factor our quadratic as:

$(x-3)(x-6) = 0$

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 - 3 = 0

   $x = 3$

2. $x - 6 = 0

   $x = 6$

So, we have two potential solutions: $x=3$ and $x=6$. But hold on, we're not done yet! Remember those crucial domain restrictions we talked about at the very beginning?

Checking for Extraneous Solutions

This is arguably the most important step when solving logarithmic equations, guys. We found two potential solutions, $x=3$ and $x=6$. However, we need to check if these solutions satisfy the original domain restrictions. Remember, for $\log x$, we need $x > 0$, and for $\log (x-2)$, we need $x-2 > 0$, which means $x > 2$. Therefore, our valid solutions must be greater than 2.

Let's check our first potential solution, $x=3$:

• Is $3 > 0$? Yes.
• Is $3 > 2$? Yes.

Since $x=3$ satisfies both conditions, it is a valid solution.

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

• Is $6 > 0$? Yes.
• Is $6 > 2$? Yes.

Since $x=6$ also satisfies both conditions, it is also a valid solution.

In this particular case, both of our solutions are valid. However, in many logarithmic equations, one or even both potential solutions might be extraneous. This happens when a solution makes the argument of any logarithm in the original equation zero or negative. Always plug your potential solutions back into the original equation to be absolutely sure, or at the very least, verify they meet the domain restrictions derived from the original equation.

Let's do a quick sanity check by plugging them back into the original equation $2 \log x-\log (x-2)=2 \log 3$:

For $x=3$:

$2 \log 3 - \log (3-2) = 2 \log 3$

$2 \log 3 - \log 1 = 2 \log 3$

Since $\log 1 = 0$ (because $10^0 = 1$), this becomes:

$2 \log 3 - 0 = 2 \log 3$

$2 \log 3 = 2 \log 3$

This is true! So, $x=3$ is definitely a solution.

For $x=6$:

$2 \log 6 - \log (6-2) = 2 \log 3$

$2 \log 6 - \log 4 = 2 \log 3$

Using the power rule on the left: $\log (6^2) - \log 4 = \log 36 - \log 4$

Using the quotient rule: $\log (36/4) = \log 9$

And we know that $2 \log 3 = \log (3^2) = \log 9$.

So, $\log 9 = \log 9$. This is also true! Thus, $x=6$ is also a solution.

Conclusion

And there you have it, mathletes! We've successfully solved the logarithmic equation $2 \log x-\log (x-2)=2 \log 3$. By applying the fundamental properties of logarithms – the Power Rule, Quotient Rule, and Equality Rule – we simplified the equation into a manageable algebraic form. We then solved the resulting quadratic equation, $(x-3)(x-6)=0$, to find our potential solutions, $x=3$ and $x=6$. The critical final step was checking these solutions against the domain restrictions ($x > 2$) to eliminate any extraneous solutions. In this case, both $x=3$ and $x=6$ proved to be valid solutions.

Solving logarithmic equations is a fantastic way to hone your algebraic skills and deepen your understanding of exponential and logarithmic functions. It requires careful attention to detail, especially with domain restrictions. Remember these steps for future problems:

1. Simplify Each Side: Use logarithm properties (power, product, quotient) to combine terms.
2. Isolate Logarithms: Aim to get a single logarithm on each side of the equation.
3. Use the Equality Rule: If $\log_b M = \log_b N$, then $M=N$.
4. Solve Algebraically: Solve the resulting equation (linear, quadratic, etc.).
5. Check for Extraneous Solutions: Verify that all solutions are within the domain of the original logarithmic expressions.

Keep practicing, and don't be afraid to revisit these properties whenever you need a refresher. Math is all about building blocks, and mastering these techniques will serve you well. Thanks for joining us on Plastik Magazine for this mathematical deep dive. Until next time, keep those minds sharp and keep exploring the amazing world of numbers!