Find X: Solving Logarithmic Equations - Easy Math

by Kenji Nakamura 50 views

Hey guys! Let's dive into the exciting world of logarithms and tackle a fun equation together. We've got a classic logarithmic problem here, and we're going to break it down step by step so you can see exactly how to solve it. Get ready to sharpen those math skills!

The Logarithmic Challenge

Our mission, should we choose to accept it, is to find the value of x that makes the following equation true:

log⁡2(6x−8)−log⁡28=1\log _2(6 x-8)-\log _2 8=1

So, how do we approach this? Don't worry, it's not as intimidating as it looks! We're going to use some key properties of logarithms to simplify the equation and isolate x. Trust me, it's going to be an awesome journey into the heart of logarithmic functions.

Understanding Logarithms

Before we dive into the solution, let's quickly recap what logarithms are all about. Think of a logarithm as the inverse operation of exponentiation. In simple terms, if we have an equation like this:

2y=x2^y = x

The logarithm (base 2) of x is y. We can write this as:

log⁡2x=y\log _2 x = y

So, the logarithm answers the question: "To what power must we raise the base (in this case, 2) to get x?"

Key Properties of Logarithms

To solve our equation, we'll need to use some important properties of logarithms. These properties are like secret weapons that help us simplify and manipulate logarithmic expressions.

  1. The Quotient Rule: This is our most important tool for this problem. The quotient rule states that the logarithm of a quotient is equal to the difference of the logarithms. Mathematically, it looks like this:

    log⁡b(MN)=log⁡bM−log⁡bN\log _b (\frac{M}{N}) = \log _b M - \log _b N

    Where b is the base of the logarithm (any positive number not equal to 1), and M and N are positive numbers.

  2. Logarithm of the Base: The logarithm of the base to itself is always 1. This makes sense because any number raised to the power of 1 is itself.

    log⁡bb=1\log _b b = 1

    For example, log⁡22=1\log _2 2 = 1

  3. Power Rule: This rule is handy when dealing with exponents inside logarithms. It states that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number.

    log⁡b(Mp)=plog⁡bM\log _b (M^p) = p \log _b M

  4. Product Rule: This rule is similar to the quotient rule but deals with products instead of quotients. The logarithm of a product is equal to the sum of the logarithms.

    log⁡b(MN)=log⁡bM+log⁡bN\log _b (MN) = \log _b M + \log _b N

Step-by-Step Solution

Okay, now that we've refreshed our understanding of logarithms and their properties, let's get back to our equation:

log⁡2(6x−8)−log⁡28=1\log _2(6 x-8)-\log _2 8=1

Step 1: Apply the Quotient Rule

Notice that we have a difference of two logarithms with the same base (base 2). This is a perfect opportunity to use the quotient rule. We can rewrite the left side of the equation as a single logarithm:

log⁡2(6x−88)=1\log _2(\frac{6x-8}{8}) = 1

Step 2: Convert to Exponential Form

Now, let's get rid of the logarithm altogether! Remember that logarithms are the inverse of exponents. We can rewrite the equation in exponential form. If log⁡ba=c\log _b a = c, then bc=ab^c = a. Applying this to our equation, we get:

21=6x−882^1 = \frac{6x-8}{8}

Step 3: Simplify and Solve for x

This looks much more manageable! Let's simplify and solve for x:

2=6x−882 = \frac{6x-8}{8}

Multiply both sides by 8:

16=6x−816 = 6x - 8

Add 8 to both sides:

24=6x24 = 6x

Divide both sides by 6:

x=4x = 4

Step 4: Check the Solution

It's crucial to check our solution to make sure it's valid. We need to plug x = 4 back into the original equation and see if it holds true. This is especially important with logarithmic equations because we can't take the logarithm of a negative number or zero.

log⁡2(6(4)−8)−log⁡28=1\log _2(6(4)-8)-\log _2 8=1

log⁡2(24−8)−log⁡28=1\log _2(24-8)-\log _2 8=1

log⁡2(16)−log⁡28=1\log _2(16)-\log _2 8=1

Now, we know that 24=162^4 = 16 and 23=82^3 = 8, so:

4−3=14 - 3 = 1

1=11 = 1

Our solution checks out! This means x = 4 is the correct answer.

Common Pitfalls to Avoid

Logarithmic equations can be tricky, so here are a few common mistakes to watch out for:

  • Forgetting to Check the Solution: Always, always, always check your solution in the original equation. Logarithmic functions have domain restrictions, so some solutions might not be valid.
  • Incorrectly Applying Logarithmic Properties: Make sure you understand the quotient, product, and power rules and apply them correctly.
  • Mixing Up Logarithmic and Exponential Forms: Practice converting between logarithmic and exponential forms to avoid errors.

Conclusion: You Did It!

Awesome job, guys! We've successfully navigated a logarithmic equation and found the value of x that satisfies it. Remember, the key is to break down the problem into smaller steps, use the properties of logarithms wisely, and always check your answer. Keep practicing, and you'll become a logarithm master in no time! In this equation, the value of x is 4. Logarithmic equations are useful for real-world applications such as calculations involving exponential growth and decay and are frequently encountered in fields like engineering, computer science, and finance. By practicing solving logarithmic equations, we strengthen our analytical skills and deepen our mathematical problem-solving abilities.

Repair Input Keyword

Let's make the question crystal clear. Instead of just saying "What value of xx satisfies this equation?", let's rephrase it to be super easy to understand:

"Determine the value of the variable xx that makes the given logarithmic equation true. Express your answer as a numeral."

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Find x: Solving Logarithmic Equations - Easy Math