Solving Logarithmic Equations: Step-by-Step Guide

Hey guys! Let's break down how to solve logarithmic equations. We're going to take a look at the equation $\log(x^2 - 15) = \log(2x)$ and walk through each step, so you can tackle these problems with confidence. Logarithmic equations might seem intimidating at first, but with a systematic approach, they become much more manageable. We’ll order the steps from 1 to 5, making it super clear and easy to follow. So, grab your math hats, and let's dive in!

1. Setting Up the Equation: $\log(x^2 - 15) = \log(2x)$

Okay, so our main goal here is to solve for 'x' in the equation $\log(x^2 - 15) = \log(2x)$. The very first thing we need to recognize is that we're dealing with logarithms. Remember, logarithms are just the inverse operations of exponentiation. Think of it like this: if you have $\log_b(a) = c$, that's the same as saying $b^c = a$. This is super important for understanding how to manipulate logarithmic equations. In our specific case, we have logarithms on both sides of the equation. A key property of logarithms comes into play here: if $\log_b(A) = \log_b(B)$, then it must be true that $A = B$. In simpler terms, if the logarithms of two expressions are equal (and they have the same base, which is assumed to be 10 if no base is written), then the expressions themselves are equal.

So, applying this property to our equation, $\log(x^2 - 15) = \log(2x)$, we can equate the arguments inside the logarithms. This means we can safely say that $x^2 - 15 = 2x$. We've just turned a logarithmic equation into a regular algebraic equation, which is a huge step forward! This move simplifies things drastically and gets us closer to solving for 'x'. Make sure you understand this initial step, because it’s the foundation for everything that follows. We’ve essentially eliminated the logarithms and set up a quadratic equation that we can solve using familiar methods.

2. Transforming into a Quadratic Equation: $x^2 - 2x - 15 = 0$

Now that we've ditched the logarithms and have the equation $x^2 - 15 = 2x$, our next mission is to rearrange it into a standard quadratic form. Why? Because quadratic equations have well-known methods for solving them, like factoring, completing the square, or using the quadratic formula. The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where a, b, and c are constants. This form helps us easily identify the coefficients we need for these solving methods.

To get our equation into this form, we need to move all the terms to one side, leaving zero on the other side. Currently, we have $x^2 - 15 = 2x$. To get rid of the $2x$ on the right side, we simply subtract $2x$ from both sides of the equation. This maintains the balance and keeps the equation true. So, when we subtract $2x$ from both sides, we get: $x^2 - 15 - 2x = 2x - 2x$. Simplifying this, we have $x^2 - 2x - 15 = 0$. Voila! We've successfully transformed our equation into the standard quadratic form. We can now clearly see that $a = 1$, $b = -2$, and $c = -15$. This sets us up perfectly for the next step, which is to actually solve this quadratic equation. Remember, this step of rearranging into standard form is crucial because it allows us to apply the techniques we already know for solving quadratics.

3. Factoring the Quadratic Equation: $(x - 5)(x + 3) = 0$

Alright, we've got our quadratic equation in the standard form: $x^2 - 2x - 15 = 0$. Now comes the fun part – factoring! Factoring is a method of breaking down the quadratic expression into a product of two binomials. Think of it as the reverse of expanding brackets. The idea is to find two numbers that multiply to give the constant term (which is -15 in our case) and add up to give the coefficient of the x term (which is -2). If we can find these numbers, we can rewrite the quadratic expression in factored form.

So, let's brainstorm. We need two numbers that multiply to -15 and add up to -2. After a bit of thought, we can see that the numbers -5 and +3 fit the bill perfectly. Why? Because (-5) * (+3) = -15 and (-5) + (+3) = -2. Great! Now that we've found these numbers, we can rewrite our quadratic equation in factored form. The factored form will look like this: $(x + first\ number)(x + second\ number) = 0$. Plugging in our numbers, we get $(x - 5)(x + 3) = 0$. Awesome! We've successfully factored the quadratic equation. This is a massive step because it transforms the equation into a form where we can easily find the potential solutions for 'x'. Each factor now gives us a possible value for 'x' that makes the entire equation equal to zero. Factoring is a super useful technique, so make sure you're comfortable with it!

4. Finding Potential Solutions: $x - 5 = 0$ or $x + 3 = 0$

We've successfully factored our quadratic equation into $(x - 5)(x + 3) = 0$. This is fantastic because it sets us up perfectly to find the potential solutions for 'x'. The core principle we're using here is the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In simpler terms, if we have something times something else equals zero, then either the first something is zero, or the second something is zero, or both are zero.

Applying this to our factored equation, $(x - 5)(x + 3) = 0$, we can say that either $(x - 5) = 0$ or $(x + 3) = 0$. This gives us two separate linear equations to solve, which are much easier to handle than the original quadratic. Let's tackle the first one: $x - 5 = 0$. To solve for 'x', we simply add 5 to both sides of the equation. This gives us $x = 5$. Great, we've found one potential solution! Now let's look at the second equation: $x + 3 = 0$. To solve for 'x', we subtract 3 from both sides. This gives us $x = -3$. Fantastic, we've found a second potential solution! So, based on our factoring and the zero-product property, we have two potential solutions for 'x': 5 and -3. But hold on, we're not quite done yet. We need to remember that we started with a logarithmic equation, and logarithmic equations have certain restrictions on their domains. This means we need to check if these potential solutions actually work in the original equation.

5. Checking for Extraneous Solutions: $x = 5$ is Valid, $x = -3$ is Not

Okay, we've arrived at a crucial step: checking for extraneous solutions. We found two potential solutions for x: 5 and -3. But remember, we started with a logarithmic equation, $\log(x^2 - 15) = \log(2x)$. Logarithmic functions have a very important restriction: you can only take the logarithm of a positive number. You can't take the logarithm of zero or a negative number. This is because logarithms are the inverse of exponential functions, and exponential functions always produce positive results. So, any solution that makes the argument inside a logarithm zero or negative is an extraneous solution – it's a solution that we found algebraically, but it doesn't actually work in the original equation.

Let's start by checking $x = 5$. We need to plug this value back into the original equation and see if it makes sense. So we have $\log(5^2 - 15) = \log(2 * 5)$. Simplifying, we get $\log(25 - 15) = \log(10)$, which simplifies further to $\log(10) = \log(10)$. This is perfectly valid! The logarithm of 10 (which is 1) equals the logarithm of 10. So, $x = 5$ is a genuine solution to our equation. Now let's check $x = -3$. Plugging this into the original equation, we get $\log((-3)^2 - 15) = \log(2 * -3)$. Simplifying, we have $\log(9 - 15) = \log(-6)$, which becomes $\log(-6) = \log(-6)$. Uh oh! We've run into a problem. We're trying to take the logarithm of a negative number (-6), which is not allowed. This means that $x = -3$ is an extraneous solution. It doesn't work in the original logarithmic equation, even though it popped out as a solution when we solved the quadratic. So, after carefully checking our solutions, we can confidently say that the only valid solution to the equation $\log(x^2 - 15) = \log(2x)$ is $x = 5$. This step of checking for extraneous solutions is super important, guys! Don't skip it, or you might end up with the wrong answer.

Final Answer

So, to recap, here's the ordered list of steps to solve the equation $\log(x^2 - 15) = \log(2x)$:

1. $\bf{x^2 - 15 = 2x}$
2. $\bf{x^2 - 2x - 15 = 0}$
3. $\bf{(x - 5)(x + 3) = 0}$
4. $\bf{x - 5 = 0}$ or $\bf{x + 3 = 0}$
5. Potential solutions are -3 and 5, but after checking for extraneous solutions, only 5 is a valid solution.

Solving logarithmic equations can be tricky, but by following these steps, you'll be solving them like a pro in no time! Remember to always check for extraneous solutions – it's a lifesaver!