Solve Logarithmic Equations: Find X = -6

Hey math whizzes! Today, we're diving deep into the world of logarithmic equations and tackling a super common problem: finding out which equation has a specific solution. Our star player today is x = -6. We've got four contenders (A, B, C, and D), and only one of them will give us that sweet, sweet solution of x = -6. So, grab your calculators, sharpen those pencils, and let's break down each option to see which one is the true champion!

Option A: $\log _x 36=2$

Alright guys, let's kick things off with option A: $\log _x 36=2$. This equation is asking us, "To what power do we need to raise 'x' to get 36?" And the answer, according to the equation, is 2. We can rewrite this logarithmic equation in its exponential form, which is usually the key to unlocking these problems. Remember, the general rule is that if you have $\log _b a = c$, it's the same as saying $b^c = a$. Applying this to our equation, we get $x^2 = 36$. Now, this is a pretty straightforward quadratic equation. To find 'x', we need to take the square root of both sides. So, $x = \pm\sqrt{36}$. This gives us two possible solutions: $x = 6$ and $x = -6$. However, there's a crucial rule about logarithms that we must remember: the base of a logarithm can never be negative or equal to 1. In our equation, 'x' is the base. Therefore, $x = -6$ is not a valid solution because it violates the fundamental rule of logarithms. So, even though it pops up as a mathematical possibility from $x^2 = 36$, it's disqualified from being a solution to the original logarithmic equation. This means option A, while it might seem like a contender at first glance, isn't our winning ticket. It's super important to always check the domain restrictions for logarithms, guys! Missing these can lead you down the wrong path.

Option B: $\log _3(2 x-9)=3$

Moving on to option B, we have the equation $\log _3(2 x-9)=3$. Here, the base of the logarithm is 3, which is perfectly fine (it's positive and not 1). The argument of the logarithm is $(2x-9)$, and the equation tells us that when we take the log base 3 of this argument, we get 3. Let's convert this into its exponential form. Using our rule $\log _b a = c \implies b^c = a$, we get $3^3 = 2x-9$. Now, we know that $3^3$ is $3 \times 3 \times 3$, which equals 27. So, the equation becomes $27 = 2x-9$. Our goal is to isolate 'x'. First, let's add 9 to both sides of the equation: $27 + 9 = 2x$. This simplifies to $36 = 2x$. Finally, to find 'x', we divide both sides by 2: $x = 36 / 2$. And guess what? $x = 18$. So, for option B, the solution is $x = 18$. This is definitely not the $x = -6$ we are looking for. Keep in mind, we also need to ensure that the argument of the logarithm, $(2x-9)$, is positive. If $x=18$, then $2(18) - 9 = 36 - 9 = 27$, which is indeed positive. So, $x=18$ is a valid solution for this equation, but it's not the one we need. We're still on the hunt for that $x = -6$!

Option C: $\log _5 218=x$

Let's check out option C, which is $\log _5 218=x$. This equation is already in a pretty neat form. It's essentially asking, "What is the power we need to raise 5 to in order to get 218?" and that power is 'x'. This equation directly gives us the value of 'x' in terms of a logarithm. To find the numerical value of 'x', we would need a calculator. If we plug $\log _5 218$ into a calculator, we'll get an approximate value. Using the change of base formula, $\log _5 218 = \frac{\log 218}{\log 5} \approx \frac{2.338}{0.699} \approx 3.34$. So, for option C, $x \approx 3.34$. This is nowhere near our target solution of $x = -6$. So, option C is definitely not the equation we're looking for. It's important to recognize when an equation directly gives you the variable or requires conversion. In this case, 'x' is already isolated, but its value isn't what we need.

Option D: $\log _3(-2 x-3)=2$

Finally, guys, we arrive at option D: $\log _3(-2 x-3)=2$. This looks promising! The base is 3, which is valid. The argument is $(-2x-3)$, and the result of the logarithm is 2. Let's convert this into exponential form. Using our trusty rule $\log _b a = c \implies b^c = a$, we get $3^2 = -2x-3$. We know that $3^2$ is $3 \times 3$, which equals 9. So, the equation becomes $9 = -2x-3$. Now, we need to isolate 'x'. First, let's add 3 to both sides: $9 + 3 = -2x$. This simplifies to $12 = -2x$. To find 'x', we divide both sides by -2: $x = 12 / (-2)$. And BAM! We get $x = -6$. This is exactly the solution we were searching for! But wait, before we declare victory, we must check if this solution is valid within the domain of the original logarithmic equation. The argument of a logarithm must always be positive. So, we need to check if $(-2x-3) > 0$ when $x = -6$. Let's substitute $x = -6$ into the argument: $-2(-6) - 3$. This becomes $12 - 3$, which equals 9. Since 9 is positive, our solution $x = -6$ is valid for this logarithmic equation! Therefore, option D is the correct answer.

Conclusion: Which Equation Has x = -6 as the Solution?

After diligently working through each option, we've found our winner! Option A gave us potential solutions of $x=6$ and $x=-6$, but $x=-6$ was invalid as a base. Option B yielded $x=18$, not our target. Option C gave us an approximate value of $x \approx 3.34$. It was only Option D: $\log _3(-2 x-3)=2$ that provided us with the valid solution of x = -6. Remember, folks, the key to solving these problems is understanding how to convert between logarithmic and exponential forms and always, always checking the domain restrictions for logarithms. Keep practicing, and you'll be a log equation master in no time! Happy solving!