Solving Logarithmic Equations A Step-by-Step Guide

Hey guys! Logarithmic equations can seem intimidating, but don't worry, we're going to break down how to solve them step by step. In this guide, we'll tackle the equation 2 logx - log (x - 16) = 2 and explore the concepts behind it so you can confidently solve similar problems. So, grab your calculators, and let's dive in!

Understanding Logarithms: The Foundation

Before we jump into solving the equation, let's quickly refresh our understanding of logarithms. Logarithms are essentially the inverse operation of exponentiation. Think of it this way: if we have an exponential equation like b^y = x, the logarithmic form of this is log_b(x) = y. Here, 'b' is the base, 'y' is the exponent, and 'x' is the result. When we see "log" without a base specified, it usually means we're dealing with the common logarithm, which has a base of 10. So, log(x) is the same as log_10(x). This understanding is crucial because when we work with logarithmic equations, we often need to switch between logarithmic and exponential forms to isolate the variable.

Key properties of logarithms are our best friends when solving these equations. Remember these, guys, they're super important! First, the product rule: log_b(mn) = log_b(m) + log_b(n). This tells us that the logarithm of a product is the sum of the logarithms. Next, the quotient rule: log_b(m/n) = log_b(m) - log_b(n), meaning the logarithm of a quotient is the difference of the logarithms. And finally, the power rule: log_b(m^p) = p * log_b(m), which allows us to bring exponents down as coefficients. These properties are the tools we'll use to manipulate and simplify logarithmic equations. Getting comfortable with these rules is half the battle, trust me! We'll be using them extensively in solving our example equation, so keep them in mind as we proceed. By mastering these properties, you'll find that logarithmic equations become much less daunting and even, dare I say, fun to solve!

Step-by-Step Solution: 2 logx - log (x - 16) = 2

Okay, let's get our hands dirty with the equation 2 logx - log (x - 16) = 2. We're going to tackle this step-by-step, so you can see exactly how it's done.

Step 1: Apply the Power Rule

Remember the power rule? It says log_b(m^p) = p * log_b(m). We can use this to simplify the first term in our equation. We have 2 logx, which can be rewritten as log(x^2). So, our equation now looks like this: log(x^2) - log(x - 16) = 2. See how we're already making progress? This is all about breaking the problem down into smaller, manageable pieces.

Step 2: Apply the Quotient Rule

Next up, we're going to use the quotient rule: log_b(m/n) = log_b(m) - log_b(n). We have log(x^2) - log(x - 16), which fits this pattern perfectly. We can combine these two logarithms into a single logarithm: log(x^2 / (x - 16)) = 2. We've just condensed our equation, making it much simpler to deal with. Guys, these logarithmic properties are lifesavers, aren't they?

Step 3: Convert to Exponential Form

Now, let's get rid of the logarithm altogether. Remember that the common logarithm has a base of 10. So, when we have log_10(y) = z, we can rewrite it in exponential form as 10^z = y. Applying this to our equation, log(x^2 / (x - 16)) = 2, we get 10^2 = x^2 / (x - 16). This simplifies to 100 = x^2 / (x - 16). We've transformed our logarithmic equation into a regular algebraic equation, which is something we can easily handle.

Step 4: Solve the Quadratic Equation

Alright, let's solve for x. We have 100 = x^2 / (x - 16). First, multiply both sides by (x - 16) to get rid of the fraction: 100(x - 16) = x^2. Expanding this, we get 100x - 1600 = x^2. Now, let's rearrange this into a standard quadratic equation form: x^2 - 100x + 1600 = 0. To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, the equation factors nicely: (x - 20)(x - 80) = 0. This gives us two potential solutions: x = 20 and x = 80. But hold on, we're not done yet!

Step 5: Check for Extraneous Solutions

This is a crucial step, guys! When solving logarithmic equations, we need to check our solutions to make sure they're valid. Logarithms are only defined for positive arguments. So, we need to plug our potential solutions back into the original equation and make sure we're not taking the logarithm of a negative number or zero.

Let's check x = 20: 2 log(20) - log(20 - 16) = 2 log(20) - log(4). Both 20 and 4 are positive, so this solution looks promising.

Now let's check x = 80: 2 log(80) - log(80 - 16) = 2 log(80) - log(64). Again, both 80 and 64 are positive, so this solution is also looking good.

Step 6: Verify the Solutions

Even though both solutions seem valid, let's plug them back into the original equation to make sure they satisfy it.

For x = 20: 2 log(20) - log(20 - 16) = 2 log(20) - log(4) ≈ 2(1.301) - 0.602 ≈ 2.602 - 0.602 = 2. So, x = 20 is a valid solution.

For x = 80: 2 log(80) - log(80 - 16) = 2 log(80) - log(64) ≈ 2(1.903) - 1.806 ≈ 3.806 - 1.806 = 2. So, x = 80 is also a valid solution.

Final Answer

Therefore, the solutions to the equation 2 logx - log (x - 16) = 2 are x = 20 and x = 80. We did it!

Common Mistakes to Avoid

Guys, let's talk about some common pitfalls people often encounter when solving logarithmic equations. Avoiding these mistakes will save you a lot of headaches!

Forgetting to Check for Extraneous Solutions

I can't stress this enough: always check your solutions! As we saw, logarithmic functions are only defined for positive arguments. If you plug a solution back into the original equation and end up taking the logarithm of a negative number or zero, that solution is extraneous and must be discarded. It’s like a false lead in a detective movie – it looks promising, but it doesn't hold up under scrutiny. Make it a habit to check, and you'll avoid a lot of errors.

Incorrectly Applying Logarithmic Properties

The logarithmic properties are powerful tools, but they need to be used correctly. Make sure you're applying the product, quotient, and power rules accurately. A common mistake is to try and apply these rules in situations where they don't fit. For instance, log(m + n) is not equal to log(m) + log(n). Memorize the properties and practice using them in different contexts to become comfortable with them.

Errors in Algebraic Manipulation

Solving logarithmic equations often involves algebraic manipulation, such as simplifying expressions, factoring quadratic equations, and solving for x. Careless mistakes in these steps can lead to incorrect solutions. Double-check your work, especially when dealing with signs and fractions. Writing out each step clearly can help you catch errors before they snowball into bigger problems. Remember, neatness counts!

Ignoring the Domain of Logarithmic Functions

The domain of a logarithmic function is the set of all positive real numbers. This means that the argument of a logarithm (the expression inside the log) must be greater than zero. When solving logarithmic equations, keep this in mind and ensure that your solutions satisfy this condition. This is closely related to checking for extraneous solutions, but it's worth considering the domain explicitly as you work through the problem. Always think about what values of x are actually permissible in the original equation.

By being aware of these common mistakes, you can improve your accuracy and confidence in solving logarithmic equations. Keep practicing, and you'll become a pro in no time!

Practice Problems

To really master solving logarithmic equations, practice is key! Here are a few problems for you guys to try out:

1. log₂(x + 3) + log₂(x - 3) = 4
2. log(5x + 1) = log(2x + 3)
3. 2 logx = log(2x + 3)

Work through these problems step-by-step, applying the techniques we've discussed. Remember to check your solutions for extraneous roots! The more you practice, the more comfortable and confident you'll become.

Conclusion

So there you have it! We've walked through a step-by-step guide to solving the logarithmic equation 2 logx - log (x - 16) = 2. We've covered the key logarithmic properties, the importance of checking for extraneous solutions, and common mistakes to avoid. With practice and a solid understanding of the fundamentals, you guys can conquer any logarithmic equation that comes your way. Keep practicing, and happy solving!