Solving Logarithmic Equations Log₄(10-z) + Log₄(z) = 2

Hey guys! Today, we're diving into a fun little math problem involving logarithms. We've got the equation log₄(10-z) + log₄(z) = 2, and our mission, should we choose to accept it, is to find the value(s) of 'z' that make this equation true. We'll break it down step-by-step, making sure everyone's on board, even if logs aren't your everyday jam. So, buckle up, and let's get started!

Understanding Logarithms

Before we jump into solving, let's do a quick refresh on what logarithms actually are. Think of a logarithm as the inverse operation of exponentiation. Basically, if we have an equation like b^x = y, then the logarithm (base b) of y is x. We write this as log_b(y) = x. So, the logarithm answers the question: "To what power must we raise the base (b) to get y?"

In our case, we're dealing with logarithms base 4. This means we're asking, "To what power must we raise 4 to get a certain number?" For example, log₄(16) = 2 because 4² = 16. Understanding this relationship between exponents and logarithms is crucial for solving logarithmic equations.

Now, let’s talk about some key properties of logarithms that we’ll be using. The most important one for this problem is the product rule of logarithms. This rule states that the logarithm of the product of two numbers is equal to the sum of the logarithms of those numbers (provided they have the same base). Mathematically, it looks like this: log_b(m * n) = log_b(m) + log_b(n). This rule is going to be our best friend in simplifying the left side of our equation.

Another thing to keep in mind is the definition of a logarithm. For log_b(x) to be defined, x must be greater than 0. We can't take the logarithm of a negative number or zero. This is super important because it means we'll need to check our solutions at the end to make sure they don't result in taking the logarithm of a non-positive number.

Finally, remember that the logarithmic function is only defined for positive arguments. This means that when we solve for z, we need to make sure that both 10 - z > 0 and z > 0. This will give us a range of possible values for z that we need to consider. This is a crucial step in solving logarithmic equations, so don't skip it!

Applying Logarithmic Properties

Okay, now that we've got our logarithm fundamentals down, let's get back to our equation: log₄(10-z) + log₄(z) = 2. The first thing we're going to do is use that handy product rule of logarithms we just talked about. Remember, log_b(m) + log_b(n) = log_b(m * n). So, we can combine the two logarithms on the left side of our equation into a single logarithm:

log₄((10-z) * z) = 2

This simplifies our equation quite a bit. We've gone from having two separate logarithms to just one. Now, let's distribute that 'z' inside the parentheses:

log₄(10z - z²) = 2

We're getting closer! Now we have a single logarithm equal to a number. To get rid of the logarithm, we need to use the definition of a logarithm to rewrite the equation in exponential form. Remember, log_b(y) = x is the same as b^x = y. So, in our case, log₄(10z - z²) = 2 is the same as:

4² = 10z - z²

See how we've transformed the logarithmic equation into a quadratic equation? This is a key step in solving these types of problems. By using the properties of logarithms, we've managed to get rid of the logs and end up with something we know how to solve.

So, now we have 4² = 10z - z², which simplifies to 16 = 10z - z². We're in familiar territory now. We've turned our logarithmic equation into a quadratic equation, which we can solve using our trusty algebra skills.

Solving the Quadratic Equation

Alright, we've arrived at the quadratic equation: 16 = 10z - z². To solve this, we first want to get everything on one side and set the equation equal to zero. This is the standard form for a quadratic equation, and it makes it easier to solve by factoring, completing the square, or using the quadratic formula.

So, let's add z² and subtract 10z from both sides of the equation. This gives us:

z² - 10z + 16 = 0

Now we have a quadratic equation in the form ax² + bx + c = 0, where a = 1, b = -10, and c = 16. The next step is to try and factor this quadratic. We're looking for two numbers that multiply to 16 and add up to -10. Can you think of any?

Well, -2 and -8 fit the bill perfectly! (-2) * (-8) = 16, and (-2) + (-8) = -10. So, we can factor our quadratic equation as follows:

(z - 2)(z - 8) = 0

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if (z - 2)(z - 8) = 0, then either z - 2 = 0 or z - 8 = 0.

Solving these two simple equations gives us our potential solutions for z:

z - 2 = 0 => z = 2 z - 8 = 0 => z = 8

So, we have two possible solutions: z = 2 and z = 8. But remember, we're not done yet! We need to check these solutions to make sure they're valid in our original logarithmic equation.

Checking for Extraneous Solutions

Okay, we've got our potential solutions, z = 2 and z = 8, but we need to be super careful here. Remember how we talked about the domain of logarithms? We can only take the logarithm of positive numbers. This means we need to check if our solutions make the arguments of the logarithms in the original equation positive.

Our original equation was log₄(10-z) + log₄(z) = 2. Let's plug in our solutions and see what happens.

First, let's check z = 2:

log₄(10-2) + log₄(2) = log₄(8) + log₄(2)

Both 8 and 2 are positive, so z = 2 looks promising.

Now, let's check z = 8:

log₄(10-8) + log₄(8) = log₄(2) + log₄(8)

Again, both 2 and 8 are positive, so z = 8 also seems like a valid solution.

Since both of our potential solutions satisfy the domain requirements of the logarithms, we can confidently say that they are both valid solutions to the equation. This is fantastic news! We've done all the hard work, and we're just about ready to wrap things up.

Final Answer

Phew! We've taken quite the journey, haven't we? We started with a logarithmic equation, used the properties of logarithms to simplify it, transformed it into a quadratic equation, solved the quadratic, and then checked our solutions for validity. That's a whole lot of math!

So, after all that, what's our final answer? We found two solutions that satisfy the original equation: z = 2 and z = 8. Therefore, the solution set for the equation log₄(10-z) + log₄(z) = 2 is {2, 8}.

And that's it! We've successfully solved the logarithmic equation. Give yourselves a pat on the back, guys! Logarithmic equations can seem intimidating at first, but with a solid understanding of the properties of logarithms and a systematic approach, you can conquer them. Remember to always check your solutions, and you'll be golden. Keep practicing, and you'll become a logarithm pro in no time!