Solving Logarithmic Equations: Find X In Log₂(3x²+9x-20)=2

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Solving Logarithmic Equations: Find x in log₂(3x²+9x-20)=2

Hey guys! Today, we're diving into a fun math problem: solving a logarithmic equation. Specifically, we're tackling the equation log₂(3x² + 9x - 20) = 2. Logarithmic equations might seem intimidating at first, but with a step-by-step approach, you'll see they're totally manageable. Our main goal here is to find all possible values of x that satisfy this equation. So, let’s roll up our sleeves and get started!

Understanding Logarithmic Equations

Before we jump into solving our specific problem, let's make sure we're all on the same page about what logarithms actually are. At its heart, a logarithm is just the inverse operation of exponentiation. Think of it this way: if we have an exponential equation like b**y = x, we can rewrite it in logarithmic form as logb(x) = y. Here, b is the base, y is the exponent, and x is the result of raising b to the power of y. This simple conversion is the key to unlocking many logarithmic equations.

Now, let's break down the components of our equation, log₂x(3x² + 9x - 20) = 2. In this equation:

  • The base of the logarithm is 2x. This is a crucial part because the base significantly impacts the solution.
  • The argument of the logarithm is 3x² + 9x - 20. This expression must be positive for the logarithm to be defined, which is a critical condition we'll need to check later.
  • The result of the logarithm is 2. This means that 2x raised to the power of 2 should equal the argument, 3x² + 9x - 20.

Key Properties to Remember

To effectively solve logarithmic equations, we need to keep some key properties in mind:

  1. The Definition of a Logarithm: As we discussed, logb(x) = y is equivalent to b**y = x. This is the foundational principle we'll use to convert our logarithmic equation into a more manageable form.
  2. The Argument Must Be Positive: The argument of a logarithm (the expression inside the log) must be greater than zero. We can't take the logarithm of a negative number or zero. This gives us an important restriction on the possible values of x.
  3. The Base Must Be Positive and Not Equal to 1: The base of a logarithm must be a positive number and cannot be equal to 1. If the base were 1, the logarithm would be undefined, and negative or zero bases lead to complex issues.

Keeping these properties in mind, we can confidently tackle the equation at hand. Remember, logarithmic equations are all about carefully unraveling the relationship between the base, the exponent, and the argument. With these basics down, let’s move on to the next step and start solving!

Step-by-Step Solution

Alright, let's get our hands dirty and solve the equation log₂x(3x² + 9x - 20) = 2 step by step. We're going to break it down so it's super clear and easy to follow. Remember, the key here is to use the properties of logarithms to transform the equation into something we can actually work with.

Step 1: Convert the Logarithmic Equation to Exponential Form

The first thing we need to do is get rid of that logarithm. To do this, we'll use the fundamental relationship between logarithms and exponents. Recall that logb(x) = y is equivalent to b**y = x. Applying this to our equation, where the base is 2x, the exponent is 2, and the result is 3x² + 9x - 20, we get:

(2x)² = 3x² + 9x - 20

See? We've successfully transformed our logarithmic equation into a quadratic equation, which is something we know how to handle. This is a huge step forward. Now, let's simplify this equation and get it into a standard form.

Step 2: Simplify and Rearrange the Equation

Next up, we need to simplify the left side of the equation and then rearrange everything to get a standard quadratic form. Squaring 2x gives us 4x², so our equation becomes:

4x² = 3x² + 9x - 20

Now, let’s move all the terms to one side to set the equation equal to zero. Subtracting 3x², 9x, and adding 20 to both sides, we have:

4x² - 3x² - 9x + 20 = 0

This simplifies to:

x² - 9x + 20 = 0

Awesome! We now have a standard quadratic equation. This is a form we can easily solve using factoring, completing the square, or the quadratic formula. In this case, factoring looks like the easiest route.

Step 3: Solve the Quadratic Equation

Okay, let's solve the quadratic equation x² - 9x + 20 = 0. We need to find two numbers that multiply to 20 and add up to -9. Those numbers are -4 and -5. So, we can factor the quadratic equation as follows:

(x - 4)(x - 5) = 0

Now, we set each factor equal to zero and solve for x:

  • x - 4 = 0 => x = 4
  • x - 5 = 0 => x = 5

So, we have two potential solutions: x = 4 and x = 5. But hold on! We're not done yet. We need to check these solutions against the original logarithmic equation to make sure they're valid.

Checking for Extraneous Solutions

Alright, we've got two potential solutions for x: 4 and 5. But before we high-five each other, we need to do a crucial step: check for extraneous solutions. Extraneous solutions are values that we get when solving an equation, but they don't actually work in the original equation. This often happens with logarithmic and radical equations because of the restrictions on the domain.

Why We Need to Check

Remember those key properties of logarithms we talked about earlier? The base of the logarithm must be positive and not equal to 1, and the argument of the logarithm must be positive. If our potential solutions don't meet these conditions, they're extraneous and we have to throw them out.

Checking x = 4

Let's start with x = 4. Plug it back into the original equation: log₂x(3x² + 9x - 20) = 2.

  • Base: 2x = 2(4) = 8. This is positive and not equal to 1, so it's good.
  • Argument: 3x² + 9x - 20 = 3(4)² + 9(4) - 20 = 3(16) + 36 - 20 = 48 + 36 - 20 = 64. This is positive, so it's also good.

Since both the base and the argument are valid, x = 4 is a valid solution.

Checking x = 5

Now let's check x = 5. Again, we plug it back into the original equation:

  • Base: 2x = 2(5) = 10. This is positive and not equal to 1, so we're good.
  • Argument: 3x² + 9x - 20 = 3(5)² + 9(5) - 20 = 3(25) + 45 - 20 = 75 + 45 - 20 = 100. This is positive, so it's also good.

Since the base and argument are valid for x = 5 as well, this is also a valid solution.

The Verdict

After carefully checking both potential solutions, we can confidently say that both x = 4 and x = 5 are valid solutions to the equation log₂x(3x² + 9x - 20) = 2. Woohoo! We did it!

Final Answer

So, guys, we’ve journeyed through the world of logarithms, tackled a tricky equation, and emerged victorious! To recap, we wanted to solve for all values of x in the equation log₂x(3x² + 9x - 20) = 2. We broke it down step-by-step:

  1. We converted the logarithmic equation to exponential form.
  2. We simplified and rearranged the equation into a standard quadratic form.
  3. We solved the quadratic equation by factoring.
  4. And most importantly, we checked our potential solutions for extraneous values.

After all that, we found that the valid solutions are:

x = 4 and x = 5

These are the values of x that make the original equation true. It’s crucial to remember that last step of checking for extraneous solutions. It’s like the final boss in a video game – you can’t skip it if you want to win!

Tips for Tackling Logarithmic Equations

Before we wrap up, here are a few extra tips to keep in mind when you’re dealing with logarithmic equations:

  • Always Remember the Definition: The relationship between logarithms and exponents is your best friend. Master it, and you'll be able to convert equations with ease.
  • Watch Out for Extraneous Solutions: This is so important it’s worth repeating! Make it a habit to check your answers, especially in exams.
  • Know Your Properties: Familiarize yourself with the key properties of logarithms, like the product rule, quotient rule, and power rule. They can simplify complex equations.
  • Practice Makes Perfect: The more you practice, the more comfortable you'll become with these types of problems. Seek out different examples and challenge yourself.

Solving logarithmic equations can feel like a puzzle, but with a systematic approach and a good understanding of the underlying principles, you'll be able to crack them every time. Keep practicing, stay curious, and you’ll be a math whiz in no time! Until next time, happy solving!