Solve $\log _5(3 X+2)=\log _5(-x)$

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of logarithms and tackling a specific equation that might look a little intimidating at first glance: $\log _5(3 x+2)=\log _5(-x)$. Don't worry, though! By the end of this article, you'll not only understand how to solve this problem but also grasp the underlying principles that make logarithmic equations tick. We're going to break it down step-by-step, making sure every part is crystal clear. So, grab your thinking caps, and let's get started on this mathematical adventure!

Understanding the Basics of Logarithms

Before we jump into solving our specific equation, let's refresh our memory on what logarithms actually are. In simple terms, a logarithm is the inverse operation to exponentiation. That means if we have an equation like $b^y = x$, the logarithmic form of this is $\log_b(x) = y$. Here, 'b' is the base, 'x' is the argument, and 'y' is the exponent. The crucial thing to remember about logarithms is that the argument (the number inside the log) must always be positive. This is a fundamental rule that will be key to solving our equation later on.

The One-to-One Property of Logarithms

One of the most powerful tools we have when dealing with logarithmic equations is the one-to-one property. This property states that if $\log_b(M) = \log_b(N)$, then it must be true that $M = N$. Essentially, if the logarithms on both sides of an equation have the same base, we can simply set their arguments equal to each other. This property is what allows us to eliminate the logarithms and turn our equation into a more manageable algebraic one. It's like a magic key that unlocks the equation, letting us work with simpler terms. Keep this property in mind, as it's the cornerstone of our solving strategy today.

Step-by-Step Solution

Alright, let's get down to business with our equation: $\log _5(3 x+2)=\log _5(-x)$. Our first step is to use the one-to-one property we just discussed. Since both logarithms have the same base (which is 5), we can equate their arguments:

$$3x + 2 = -x$$

See? The logarithms are gone, and we're left with a simple linear equation. Now, we just need to solve for 'x'. The goal here is to get all the 'x' terms on one side of the equation and the constant terms on the other. Let's add 'x' to both sides:

$$3x + x + 2 = -x + x$$

This simplifies to:

$$4x + 2 = 0$$

Now, let's isolate the '4x' term by subtracting 2 from both sides:

$$4x + 2 - 2 = 0 - 2$$

Which gives us:

$$4x = -2$$

Finally, to find the value of 'x', we divide both sides by 4:

$$x = \frac{-2}{4}$$

Simplifying the fraction, we get:

$$x = -\frac{1}{2}$$

So, based on our algebraic manipulation, it looks like $x = -1/2$ is our potential solution. But hold on, guys, we're not quite done yet! There's a super important final step when working with logarithms.

The Crucial Check: Domain Restrictions

Remember that fundamental rule we talked about earlier? The argument of a logarithm must be positive. This means that for our original equation, both $(3x+2)$ and $(-x)$ must be greater than zero. Let's check our potential solution, $x = -1/2$, against these conditions.

First, let's check the argument $(3x+2)$:

$$3\left(-\frac{1}{2}\right) + 2 = -\frac{3}{2} + 2 = -\frac{3}{2} + \frac{4}{2} = \frac{1}{2}$$

This is positive ($\frac{1}{2} > 0$), so that part is good!

Now, let's check the argument $(-x)$:

$$- \left(-\frac{1}{2}\right) = \frac{1}{2}$$

This is also positive ($\frac{1}{2} > 0$). Awesome!

Since both arguments are positive when $x = -1/2$, our solution is valid. If either of these had resulted in a non-positive number, then $x = -1/2$ would have been an extraneous solution, and there would be no solution to the original equation. It's always, always important to perform this check. It's the difference between getting the right answer and just a number that looks right but doesn't actually work in the original equation. Think of it as the final quality control for your math problems!

Why This Check Matters: Extraneous Solutions

It's worth spending a moment to really understand why we do this check for extraneous solutions. When we use the one-to-one property, we're essentially expanding the possible solutions because the algebraic steps might introduce values of 'x' that are valid for the simplified equation but not for the original logarithmic one. For instance, if we had an equation like $\log(x) + \log(x-3) = 1$, and we combined the logs to get $\log(x(x-3)) = 1$, we'd then solve $x(x-3) = 10$. This might give us solutions like $x=5$ and $x=-2$. However, if we plug $x=-2$ back into the original equation, we'd have $\log(-2)$ and $\log(-5)$, which are undefined in the real number system. Therefore, $x=-2$ would be an extraneous solution. Our initial equation $\log _5(3 x+2)=\log _5(-x)$ is a bit simpler because the structure allows us to directly compare the arguments, but the principle of checking the domain remains paramount. Always ensure your 'x' values make the original logarithmic expressions valid.

Conclusion

And there you have it, folks! We successfully solved the logarithmic equation $\log _5(3 x+2)=\log _5(-x)$ and found that the solution is $x = -1/2$. The key takeaways here are understanding the one-to-one property of logarithms, which allows us to equate the arguments when bases are the same, and the critical importance of checking our solutions against the domain restrictions of logarithmic functions. Never skip that final check, guys – it's your mathematical safety net!

Keep practicing these types of problems, and you'll become a logarithm master in no time. If you have any questions or want to tackle another equation, drop us a comment below. Until next time, keep those brains buzzing!