Solving Logarithmic Equations Step-by-Step Guide

Hey guys! Today, we're diving into the world of logarithmic equations, and we're going to tackle a specific one: log(-11z + 27) = 2. Don't worry if this looks intimidating; we'll break it down step by step. By the end of this guide, you'll not only know how to solve this particular equation but also have a solid understanding of how to approach similar problems. So, grab your calculators, and let's get started!

Understanding Logarithms

Before we jump into solving the equation, let's make sure we're all on the same page about what logarithms actually are. At its core, a logarithm is just the inverse operation of exponentiation. Think of it this way: if 10^2 = 100, then the logarithm base 10 of 100 is 2. We write this as log₁₀(100) = 2. In general, if b^y = x, then log_b(x) = y. Here, 'b' is the base of the logarithm, 'x' is the argument, and 'y' is the exponent. When you see a logarithm written without a base, like our equation log(-11z + 27) = 2, it's usually implied that the base is 10. This is called the common logarithm.

Key Concepts to Remember

• The logarithm is the inverse of exponentiation.
• If b^y = x, then log_b(x) = y.
• The common logarithm has a base of 10 (log(x) is the same as log₁₀(x)).
• The argument of a logarithm (the value inside the parentheses) must be greater than zero.

Understanding these basics is crucial because the properties of logarithms dictate how we manipulate and solve logarithmic equations. For instance, the fact that the argument must be positive will play a significant role in determining the validity of our solution later on.

Converting the Logarithmic Equation to Exponential Form

The first key step in solving our equation, log(-11z + 27) = 2, is to convert it from logarithmic form to exponential form. This is where the definition of a logarithm comes into play. Remember, the logarithm tells us what exponent we need to raise the base to in order to get a certain number. In our case, the base is 10 (since it's a common logarithm), the exponent is 2, and the result is the argument of the logarithm, which is (-11z + 27).

Using the relationship between logarithms and exponentials, we can rewrite the equation log(-11z + 27) = 2 as 10^2 = -11z + 27. This transformation is the heart of the solution process, as it allows us to get rid of the logarithm and work with a more familiar algebraic equation. Now, instead of dealing with logarithms, we have a linear equation in terms of 'z', which is much easier to solve. This step highlights the power of understanding the fundamental relationship between logarithmic and exponential functions.

Solving the Linear Equation

Now that we've converted our logarithmic equation into the linear equation 10^2 = -11z + 27, the next step is to solve for 'z'. This involves a series of algebraic manipulations to isolate 'z' on one side of the equation. First, let's simplify 10^2, which is simply 100. So, our equation becomes 100 = -11z + 27. The goal here is to get 'z' by itself, and we'll do this by performing inverse operations.

To isolate the term with 'z', we subtract 27 from both sides of the equation: 100 - 27 = -11z + 27 - 27, which simplifies to 73 = -11z. Now, to completely isolate 'z', we divide both sides of the equation by -11: 73 / -11 = -11z / -11. This gives us z = -73/11. So, we've found a potential solution for 'z'! However, we're not quite done yet. We need to verify this solution to make sure it's valid in the original logarithmic equation. This is a crucial step in solving logarithmic equations, as we'll see in the next section.

Verifying the Solution

Okay, we've found a potential solution for 'z', which is z = -73/11. But here's the thing about logarithmic equations: not every solution we find algebraically is actually a valid solution. This is because the argument of a logarithm must be greater than zero. In other words, we can only take the logarithm of positive numbers. So, we need to plug our solution back into the original equation to make sure the argument of the logarithm is positive.

Our original equation is log(-11z + 27) = 2. Let's substitute z = -73/11 into the argument: -11*(-73/11) + 27. This simplifies to 73 + 27, which equals 100. Since 100 is a positive number, our solution z = -73/11 is valid! This step is super important because if we had gotten a negative number or zero for the argument, we would have had to discard that solution. Think of it like a detective double-checking their evidence – we need to make sure our solution fits the crime scene (or, in this case, the equation!).

Expressing the Solution

Great news, guys! We've successfully solved the logarithmic equation log(-11z + 27) = 2. We found that z = -73/11 is a valid solution. Now, the question asks us to express the solution as an integer, fraction, or decimal rounded to five decimal places. We already have the solution as a fraction, -73/11, but let's convert it to a decimal to satisfy all the requirements.

Dividing -73 by 11, we get approximately -6.636363636... The question specifies rounding to five decimal places, so we round -6.636363636... to -6.63636. Therefore, our solution, expressed as a decimal rounded to five decimal places, is -6.63636. We've covered all bases here – we've found the solution, verified it, and expressed it in the requested formats. It's like getting a gold star on our math homework!

Conclusion

So there you have it! We've successfully tackled the logarithmic equation log(-11z + 27) = 2. We walked through each step, from understanding the basics of logarithms to converting the equation to exponential form, solving the linear equation, verifying the solution, and expressing it in the required formats. Remember, the key to solving logarithmic equations is understanding the relationship between logarithms and exponentials, and always verifying your solutions.

Key Takeaways

• Convert logarithmic equations to exponential form using the definition of a logarithm.
• Solve the resulting algebraic equation for the variable.
• Always verify your solution by plugging it back into the original equation to ensure the argument of the logarithm is positive.
• Express the solution in the format requested (integer, fraction, or decimal).

I hope this guide has been helpful! Logarithmic equations might seem tricky at first, but with practice and a solid understanding of the fundamentals, you can conquer them. Keep practicing, and you'll become a logarithm-solving pro in no time!