Solving Logarithmic Equations: A Deep Dive

Hey math whizzes! Today, we're diving deep into the intriguing world of logarithmic equations, specifically tackling the beast: $2 \log x - \log 3 = \log 3$. You know, sometimes these problems throw a curveball, and we end up with solutions that look legit but are actually bogus. We call these extraneous solutions, and figuring out which ones are the real deal and which ones are just faking it is super important. So, grab your calculators, maybe a comfy seat, and let's unravel this mystery together. We'll explore potential solutions and determine which statement is true about them. Get ready to flex those math muscles, guys!

Understanding Logarithmic Equations and Extraneous Solutions

Alright, let's get down to brass tacks. What exactly are we dealing with when we talk about solving $2 \log x - \log 3 = \log 3$? This is a logarithmic equation, meaning it's an equation where the variable we're trying to find, our elusive 'x', is tucked away inside a logarithm. Logarithms, you'll remember, are the inverse of exponentiation. So, if $y = b^x$, then $\log_b y = x$. Pretty neat, right? The 'log' usually implies a base of 10 if no base is specified, but sometimes it could be the natural logarithm (base 'e'). For this problem, we'll assume base 10 unless told otherwise. The core issue with logarithmic equations, and why we have to be extra careful, is the domain of the logarithm function. Remember, you cannot take the logarithm of a negative number or zero. That is, $\log a$ is only defined for $a > 0$. This little restriction is the birthplace of those sneaky extraneous solutions. When we manipulate the equation using logarithm properties, we might create a situation where a solution we find works algebraically but violates the original domain requirement. It's like finding a key that fits the lock but doesn't actually open the door – frustrating, but it happens! So, our mission, should we choose to accept it, is to not only find potential solutions but also to rigorously check if they are valid within the original equation's constraints. This means plugging them back into the original equation and seeing if everything holds up. If we get a logarithm of a non-positive number, BAM! That's an extraneous solution, and we gotta toss it out.

Step-by-Step Solution: Unmasking the Extraneous Solutions

Now, let's roll up our sleeves and solve $2 \log x - \log 3 = \log 3$. The first thing we want to do is simplify this bad boy using our logarithm properties. Remember the power rule: $n \log a = \log (a^n)$? We can apply that to the first term: $2 \log x$ becomes $\log (x^2)$. So our equation transforms into: $\log (x^2) - \log 3 = \log 3$. Next up is the quotient rule: $\log a - \log b = \log (a/b)$. Applying this to the left side gives us: $\log (x^2 / 3) = \log 3$. Now, we have a logarithm on both sides of the equation. If $\log A = \log B$, then it must be that $A = B$. This is because the logarithm function is one-to-one. So, we can set the arguments of the logarithms equal to each other: $x^2 / 3 = 3$. Now, this is a much simpler algebraic equation to solve! Multiply both sides by 3: $x^2 = 9$. To find x, we take the square root of both sides: $x = \pm \sqrt{9}$. This gives us two potential solutions: $x = 3$ and $x = -3$. Okay, so we've found our candidates, but remember our earlier chat about domains? This is where the crucial check comes in. We need to plug these values back into the original equation: $2 \log x - \log 3 = \log 3$. Let's test $x=3$ first. The original equation involves $\log x$. Is $\log 3$ defined? Yes, because 3 is positive. So, $2 \log 3 - \log 3 = \log 3$. This simplifies to $\log 3 = \log 3$, which is true. So, $x=3$ is a valid solution. Now, let's test $x=-3$. The original equation involves $\log x$. Is $\log (-3)$ defined? NO! The logarithm of a negative number is undefined in the realm of real numbers. Therefore, $x = -3$ is an extraneous solution. It popped up during our algebraic manipulation, but it doesn't satisfy the original conditions of the logarithmic function.

Analyzing the Potential Solutions and Conclusion

So, after all that hard work, we've discovered that our equation $2 \log x - \log 3 = \log 3$ yielded two potential solutions: $x = 3$ and $x = -3$. However, our rigorous check revealed a critical detail. The domain of the logarithmic function dictates that its argument must be strictly positive. When we tested $x = 3$ in the original equation, we found that $\log 3$ is perfectly valid, and the equation holds true. This means $x=3$ is a genuine solution. On the other hand, when we attempted to plug $x = -3$ into the original equation, we encountered $\log (-3)$. Since the logarithm of a negative number is undefined, $x = -3$ cannot be a solution. It's an extraneous solution – a phantom solution that arose from the algebraic process but doesn't actually satisfy the initial constraints of the problem. This is why it's so important to always check your solutions in the original equation when dealing with logarithmic or radical equations, guys. The mathematical properties we use to simplify equations sometimes broaden the set of possible solutions, and we need to prune those that don't fit the original context. Therefore, out of the two potential solutions, only $x = 3$ is valid, and $x = -3$ is extraneous. This leads us directly to the correct statement about the potential solutions. We can now confidently say that only -3 is an extraneous solution. The other potential solution, 3, is a perfectly legitimate, non-extraneous solution. It's a common pitfall, so understanding this process is key to mastering these types of problems. Keep practicing, and you'll become a pro at spotting those tricky extraneous solutions!