Solve Logarithmic Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into the awesome world of logarithmic equations. Specifically, we'll be tackling how to solve an equation like $\log _3(x+2)=\log _3 olimits(2 x^2-1)$. Don't let the logs and the squares scare you; it's all about following a few key steps. We'll break down the process so you can confidently solve these kinds of problems. Stick around, and by the end of this, you'll be a log equation pro!

Understanding Logarithmic Equations

Alright, let's get this party started with a quick refresher on what we're dealing with. Logarithmic equations involve logarithms, which are basically the inverse of exponentiation. If you see something like $\log _b a = c$, it means $b^c = a$. In our specific problem, we have $\log _3(x+2)=\log _3 olimits(2 x^2-1)$. The magic here is that if the logarithms on both sides of an equation have the same base (which they do – base 3!), then their arguments must be equal. This is a fundamental property that makes solving these equations so much easier. Think of it like this: if $\log _3( ext{something}) = \log _3( ext{something else})$, then $\text{something}$ must equal $\text{something else}$. This property is the cornerstone of how we'll approach our problem. It allows us to strip away the logarithms and turn a logarithmic equation into a more familiar algebraic one, like a polynomial equation. We'll be using this property to set up our equation for solving. Remember, the base of the logarithm is crucial. If the bases were different, this shortcut wouldn't apply, and we'd need different techniques. But lucky for us, in this case, the bases match perfectly, setting us up for a smooth ride.

Step 1: Equate the Arguments

So, the first major step in solving our equation $\log _3(x+2)=\log _3 olimits(2 x^2-1)$ is to leverage that property we just talked about. Since the bases of the logarithms are the same (both are base 3), we can set the arguments of the logarithms equal to each other. This gives us a new equation: $x+2 = 2 x^2-1$. This is a super important step because it transforms our logarithmic equation into a polynomial equation, which we know how to solve using standard algebraic techniques. Don't forget that this step is only valid because the bases are identical. If they weren't, we'd be in a different ballgame. But for this problem, this is our golden ticket to simplifying things. We've effectively eliminated the logarithms, and now we're left with a quadratic equation to wrangle. This is where the real algebra begins, and it's crucial to perform this step accurately to ensure the rest of your solution is sound. This is the point where the problem transitions from dealing with logs to dealing with polynomials, a familiar territory for many of us.

Step 2: Rearrange into a Standard Quadratic Form

Now that we have the equation $x+2 = 2 x^2-1$, our next move is to get it into a standard quadratic form, which is $ax^2+bx+c=0$. To do this, we need to move all the terms to one side of the equation, setting it equal to zero. Let's subtract $x$ from both sides and subtract 2 from both sides. This will give us: $0 = 2 x^2 - x - 1 - 2$. Simplifying that, we get: $0 = 2 x^2 - x - 3$. This is our quadratic equation in standard form. Seeing it like this makes it much less intimidating, right? We've successfully converted the original logarithmic equation into a quadratic equation that we can solve using factoring, the quadratic formula, or completing the square. This rearrangement is a fundamental technique in algebra, and it's key to unlocking the solutions. Make sure you're careful with your signs when moving terms across the equals sign; a simple sign error here can throw off your entire solution. This form, $2 x^2 - x - 3 = 0$, is the target we aim for when solving quadratic equations because it sets us up perfectly for the next steps.

Step 3: Solve the Quadratic Equation

We've got our quadratic equation: $0 = 2 x^2 - x - 3$. Now, it's time to find the values of $x$ that make this equation true. There are a few ways to do this, but factoring is often the quickest if it works. We're looking for two numbers that multiply to $(2)(-3) = -6$ and add up to $-1$ (the coefficient of the $x$ term). Those numbers are $-3$ and $2$. So, we can rewrite the middle term: $0 = 2 x^2 + 2x - 3x - 3$. Now, we can factor by grouping. Group the first two terms and the last two terms: $0 = (2 x^2 + 2x) + (-3x - 3)$. Factor out the common factors from each group: $0 = 2x(x+1) - 3(x+1)$. Notice that we have a common factor of $(x+1)$. Factor that out: $0 = (2x - 3)(x+1)$. This factored form is super helpful because it leads us directly to the solutions. We set each factor equal to zero: $2x - 3 = 0$ or $x+1 = 0$. Solving the first equation, we add 3 to both sides to get $2x = 3$, and then divide by 2, giving us x = rac{3}{2}. Solving the second equation, we subtract 1 from both sides, giving us $x = -1$. So, our potential solutions are rac{3}{2} and $-1$. This step is all about your factoring skills or your comfort with the quadratic formula if factoring gets tricky. The goal is to find these roots accurately, as they are the candidates for our final answer.

Step 4: Check for Extraneous Solutions

This is arguably the most crucial step, guys. When solving logarithmic equations, we must always check our potential solutions back in the original equation. Why? Because the argument of a logarithm must be positive. If plugging a potential solution into the original equation results in a logarithm of a negative number or zero, that solution is extraneous and must be discarded. Let's check our potential solutions: x = rac{3}{2} and $x = -1$.

Checking x = rac{3}{2}: Original equation: $\log _3(x+2)=\log _3 olimits(2 x^2-1)$ Substitute x = rac{3}{2}: \log _3(rac{3}{2}+2)=\log _3 olimits(2(rac{3}{2})^2-1) Calculate the arguments:

• Left side: $\frac{3}{2}+2 = \frac{3}{2}+\frac{4}{2} = \frac{7}{2}$
• Right side: 2(rac{9}{4})-1 = rac{18}{4}-1 = \frac{9}{2}-1 = \frac{9}{2}-\frac{2}{2} = \frac{7}{2}

Since both arguments are $\frac{7}{2}$ (which is positive), x = rac{3}{2} is a valid solution.

Checking $x = -1$: Original equation: $\log _3(x+2)=\log _3 olimits(2 x^2-1)$ Substitute $x = -1$: $\log _3(-1+2)=\log _3 olimits(2(-1)^2-1)$ Calculate the arguments:

• Left side: $-1+2 = 1$
• Right side: $2(1)-1 = 2-1 = 1$

Wait a minute! When we plug in $x=-1$, we get $\log _3(1) = \log _3(1)$. This looks okay because 1 is positive. However, let's re-examine the original step where we equated the arguments: $x+2 = 2x^2-1$. This led to $2x^2-x-3=0$, which factored into $(2x-3)(x+1)=0$. The roots were $x=3/2$ and $x=-1$.

Let's go back to the original logarithmic equation: $\log _3(x+2)=\log _3 olimits(2 x^2-1)$. The arguments of the logarithm must be strictly positive. For $x=-1$: Argument 1: $x+2 = -1+2 = 1$. This is positive, so $\log _3(1)$ is defined. Argument 2: $2x^2-1 = 2(-1)^2-1 = 2(1)-1 = 2-1 = 1$. This is positive, so $\log _3(1)$ is defined.

Both arguments are positive when $x=-1$. Therefore, $\log _3(1) = \log _3(1)$, which is $0=0$. This means $x=-1$ is also a valid solution.

It's super important to remember that sometimes, even if the arguments seem positive after substitution, you might have made a mistake in the previous steps, or there might be other constraints. Always double-check your algebra! In this specific case, both potential solutions work because they result in positive arguments for the logarithms in the original equation. This is why the checking step is non-negotiable – it saves you from tricky errors and ensures your final answers are legit.

Conclusion: The Solution Set

So, after carefully following all the steps – equating the arguments, rearranging into a quadratic, solving the quadratic, and most importantly, checking for extraneous solutions – we found our valid solutions. The potential solutions we got from solving the quadratic equation $0 = 2 x^2 - x - 3$ were x = rac{3}{2} and $x = -1$. Upon checking these values in the original logarithmic equation $\log _3(x+2)=\log _3 olimits(2 x^2-1)$, we confirmed that both arguments ($x+2$ and $2x^2-1$) were positive for both values. This means both are indeed solutions. Therefore, the solution set for the equation $\log _3(x+2)=\log _3 olimits(2 x^2-1)$ is \{ -1, rac{3}{2} \}. Remember, guys, the checking step is your safety net. Never skip it when dealing with logarithmic or radical equations. Keep practicing, and you'll master these in no time!