Solve $4^{\log _4(x+8)}=4^2$ Easily

Hey guys! Today, we're diving into a fun little mathematical problem that involves solving an exponential equation. Don't worry, it's not as intimidating as it sounds! We're going to break it down step-by-step so everyone can follow along. The equation we're tackling is $4^{\log _4(x+8)}=4^2$. Our mission is to find the value of x that makes this equation true. We have some answer options lined up: A. $x=-8$, B. $x=-4$, C. $x=4$, and D. $x=8$. So, let's put on our thinking caps and get started!

Understanding the Basics: Exponents and Logarithms

Before we jump into solving, let's quickly refresh our understanding of exponents and logarithms, as these are the key players in our equation. Think of exponents as a shorthand way of showing repeated multiplication. For example, $4^2$ means 4 multiplied by itself, which equals 16. The number 4 is the base, and 2 is the exponent, indicating how many times the base is multiplied.

Now, logarithms are like the inverse operation of exponentiation. They answer the question: "What exponent do I need to raise the base to, in order to get a certain number?" So, in the expression $\log _4(x+8)$, the base is 4, and we're asking, "What power of 4 gives us (x+8)?" Understanding this relationship between exponents and logarithms is crucial for solving our equation. They're like two sides of the same coin, and knowing how they interact will make the problem much clearer.

The Inverse Relationship

The most important thing to remember is that logarithms and exponentiation with the same base are inverse operations. This means they "undo" each other. Mathematically, this is represented as: $a^{\log _a(x)} = x$. This property is exactly what we need to simplify our equation. When we have a base raised to a logarithm with the same base, they cancel each other out, leaving us with just the argument of the logarithm. This is a fundamental concept, so make sure you've got it down! It's the key to unlocking the solution to this problem and many others like it.

Solving the Equation Step-by-Step

Okay, now that we've got the basics covered, let's get back to our equation: $4^{\log _4(x+8)}=4^2$. The first thing we notice is that we have the same base (4) on both sides of the equation. This is great news because it allows us to use the inverse relationship between exponents and logarithms to simplify things.

Step 1: Applying the Inverse Property

Look at the left side of the equation: $4^{\log _4(x+8)}$. We have a base of 4 raised to a logarithm with the same base. Remember the property we talked about? $a^{\log _a(x)} = x$. Applying this property, the base 4 and the logarithm with base 4 effectively cancel each other out, leaving us with just $(x+8)$. So, the left side simplifies to $x+8$.

Step 2: Simplifying the Right Side

Now let's look at the right side of the equation: $4^2$. This is a simple exponential term that we can easily evaluate. $4^2$ means 4 multiplied by itself, which is 16. So, the right side of the equation is equal to 16.

Step 3: Forming the Simplified Equation

After simplifying both sides, our original equation $4^{\log _4(x+8)}=4^2$ has now transformed into a much simpler equation: $x+8=16$. See how much easier that is to handle? We've used the properties of logarithms and exponents to strip away the complexity and reveal a straightforward algebraic equation.

Step 4: Solving for x

To solve for x, we need to isolate it on one side of the equation. We can do this by subtracting 8 from both sides of the equation. This gives us: $x+8-8=16-8$. Simplifying this, we get $x=8$. So, we've found our solution! The value of x that satisfies the original equation is 8.

Checking the Solution

It's always a good idea to check our solution to make sure it's correct. To do this, we'll substitute $x=8$ back into the original equation: $4^{\log _4(x+8)}=4^2$.

Substituting x = 8

Replacing x with 8, we get: $4^{\log _4(8+8)}=4^2$. This simplifies to $4^{\log _4(16)}=4^2$. Now, we need to evaluate the logarithm. Remember, $\log _4(16)$ asks, "What power of 4 gives us 16?" Since $4^2 = 16$, $\log _4(16) = 2$. So, our equation becomes $4^2 = 4^2$.

Verifying the Equality

Now we have $4^2 = 4^2$, which simplifies to $16 = 16$. This is a true statement, which means our solution $x=8$ is indeed correct. Checking our solution is a crucial step in the problem-solving process. It ensures that we haven't made any mistakes along the way and that our answer is valid.

The Final Answer

So, after carefully solving the equation and verifying our solution, we've arrived at the final answer. The solution to the equation $4^{\log _4(x+8)}=4^2$ is $x=8$. Looking back at our answer options, this corresponds to option D.

Why Other Options Are Incorrect

It's also helpful to understand why the other answer options are incorrect. Let's briefly examine them:

• A. x = -8: If we substitute $x=-8$ into the original equation, we get $4^{\log _4(-8+8)}=4^2$, which simplifies to $4^{\log _4(0)}=4^2$. The logarithm of 0 is undefined, so this option is not valid.
• B. x = -4: Substituting $x=-4$, we get $4^{\log _4(-4+8)}=4^2$, which simplifies to $4^{\log _4(4)}=4^2$. This gives us $4^1=4^2$, or $4=16$, which is false.
• C. x = 4: Substituting $x=4$, we get $4^{\log _4(4+8)}=4^2$, which simplifies to $4^{\log _4(12)}=4^2$. This doesn't simplify to a true statement, so this option is also incorrect.

By checking these options, we reinforce our understanding of why $x=8$ is the only correct solution.

Conclusion

Awesome job, guys! We've successfully solved the exponential equation $4^{\log _4(x+8)}=4^2$. We started by understanding the fundamentals of exponents and logarithms, then applied the inverse relationship property to simplify the equation. We systematically solved for x, checked our solution, and confirmed that $x=8$ is the correct answer. Remember, the key to tackling these types of problems is to break them down into smaller, manageable steps and to thoroughly understand the underlying concepts. Keep practicing, and you'll become a math whiz in no time!

This article discusses the following keywords:

• Exponential equation: An equation in which the variable appears in the exponent.
• Logarithm: The inverse operation to exponentiation. It answers the question: