Solving Exponential Equations $9^{x+2}$ Equals Cube Root Of $3^{x+2}$

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Hey guys! Today, we're diving deep into the exciting world of exponential equations, and we're tackling a fascinating problem: 9x+2=3x+239^{x+2}=\sqrt[3]{3^{x+2}}. This equation might look a bit intimidating at first glance, but don't worry! We're going to break it down step by step, so you'll not only understand how to solve it but also grasp the underlying principles of exponential equations. So, grab your thinking caps, and let's get started!

Understanding Exponential Equations

Before we jump into the nitty-gritty of our specific equation, let's take a moment to understand what exponential equations are all about. Exponential equations are equations where the variable appears in the exponent. These types of equations pop up in various real-world scenarios, from calculating compound interest to modeling population growth and radioactive decay. The key to solving exponential equations lies in manipulating them so that we can compare either the bases or the exponents. Often, this involves using the properties of exponents and logarithms. In our case, we'll primarily focus on manipulating the bases to make them the same. This is a common and effective strategy when dealing with exponential equations where the bases can be expressed as powers of the same number. Remember, the goal is to simplify the equation into a form where we can directly equate the exponents. This simplification often involves rewriting numbers as powers of a common base, a technique that not only helps in solving equations but also deepens our understanding of numerical relationships. By mastering these techniques, you'll be well-equipped to tackle a wide range of exponential equation problems.

Furthermore, it’s crucial to understand the properties of exponents thoroughly. These properties are the fundamental tools we’ll use to transform and simplify our equation. For instance, the property (am)n=amn(a^m)^n = a^{mn} allows us to deal with powers raised to powers, and the property am=ana^{m} = a^{n} implies m=nm = n (if a>0a > 0 and aβ‰ 1a \neq 1) is the cornerstone of our solving strategy. Keep these properties in mind as we move forward, and you'll see how they play a crucial role in each step of our solution.

Breaking Down the Equation 9x+2=3x+239^{x+2}=\sqrt[3]{3^{x+2}}

Okay, let's zero in on our specific equation: 9x+2=3x+239^{x+2}=\sqrt[3]{3^{x+2}}. The first thing we need to do, guys, is to make the bases the same on both sides of the equation. Notice that 9 can be written as 323^2. This is our key insight. By expressing 9 as a power of 3, we pave the way for a much simpler equation. So, let's rewrite the left side of the equation:

9x+2=(32)x+29^{x+2} = (3^2)^{x+2}

Using the power of a power rule, which states that (am)n=amn(a^m)^n = a^{mn}, we can simplify this further:

(32)x+2=32(x+2)=32x+4(3^2)^{x+2} = 3^{2(x+2)} = 3^{2x+4}

Now, let’s tackle the right side of the equation, which involves a cube root. Remember that a radical expression like amn\sqrt[n]{a^m} can be rewritten as amna^{\frac{m}{n}}. In our case, we have a cube root, so n=3n = 3. Applying this to our equation, we get:

3x+23=3x+23\sqrt[3]{3^{x+2}} = 3^{\frac{x+2}{3}}

So, our original equation 9x+2=3x+239^{x+2}=\sqrt[3]{3^{x+2}} now transforms into a much more manageable form:

32x+4=3x+233^{2x+4} = 3^{\frac{x+2}{3}}

This transformation is a crucial step. By expressing both sides of the equation with the same base, we've set the stage for directly comparing the exponents. This technique is a cornerstone of solving exponential equations, and it's a skill that will serve you well in more complex problems. Always be on the lookout for opportunities to rewrite numbers as powers of a common base. It's often the key to unlocking the solution.

Equating the Exponents

We've reached a pivotal moment! Now that we have the same base (which is 3) on both sides of the equation, we can equate the exponents. This is a fundamental property of exponential equations: if am=ana^m = a^n, then m=nm = n (provided that aa is a positive number not equal to 1). This principle allows us to transform our exponential equation into a simple algebraic equation. So, let’s take the exponents from both sides of our transformed equation, 32x+4=3x+233^{2x+4} = 3^{\frac{x+2}{3}}, and set them equal to each other:

2x+4=x+232x + 4 = \frac{x+2}{3}

Voila! We've successfully converted our exponential equation into a linear equation. This is a significant breakthrough because we now have a familiar type of equation that we can solve using standard algebraic techniques. The transition from exponential to linear equations is a common strategy in solving these types of problems, and it highlights the power of mathematical transformations. By understanding and applying these transformations, you can simplify complex problems into manageable steps.

Now, let’s move on to solving this linear equation for x. We’ll start by clearing the fraction, which will make the equation easier to handle. Remember, our goal is to isolate x on one side of the equation, and we'll use algebraic manipulations to achieve this. So, stay with me as we unravel the solution step by step.

Solving the Linear Equation

Alright, let's dive into solving the linear equation we obtained: 2x+4=x+232x + 4 = \frac{x+2}{3}. The first step is to get rid of the fraction. We can do this by multiplying both sides of the equation by 3. This ensures that we maintain the equality while eliminating the denominator:

3(2x+4)=3(x+23)3(2x + 4) = 3(\frac{x+2}{3})

Distributing the 3 on the left side gives us:

6x+12=x+26x + 12 = x + 2

Now, we need to isolate x. Let's start by subtracting x from both sides of the equation:

6xβˆ’x+12=xβˆ’x+26x - x + 12 = x - x + 2

This simplifies to:

5x+12=25x + 12 = 2

Next, we subtract 12 from both sides to get the term with x by itself:

5x+12βˆ’12=2βˆ’125x + 12 - 12 = 2 - 12

Which simplifies to:

5x=βˆ’105x = -10

Finally, we divide both sides by 5 to solve for x:

5x5=βˆ’105\frac{5x}{5} = \frac{-10}{5}

So, we get:

x=βˆ’2x = -2

And there we have it! We've found the solution to our linear equation. But remember, we're not done yet. We need to make sure that this solution is also the solution to our original exponential equation. This is a crucial step in solving any equation, as it helps us avoid extraneous solutions.

Verifying the Solution

Fantastic! We've arrived at a potential solution, x=βˆ’2x = -2. But before we celebrate, it's absolutely crucial to verify our solution. Plugging our answer back into the original equation is like the ultimate fact-check for our mathematical journey. It ensures that our solution not only works in the simplified steps but also holds true for the initial problem. This step is particularly important in dealing with exponential equations, as they can sometimes lead to extraneous solutions – values that satisfy the transformed equation but not the original one.

So, let's take x=βˆ’2x = -2 and substitute it back into our original equation:

9x+2=3x+239^{x+2} = \sqrt[3]{3^{x+2}}

Plugging in x=βˆ’2x = -2, we get:

9βˆ’2+2=3βˆ’2+239^{-2+2} = \sqrt[3]{3^{-2+2}}

Simplifying the exponents:

90=3039^0 = \sqrt[3]{3^0}

Remember that any non-zero number raised to the power of 0 is 1. So, we have:

1=131 = \sqrt[3]{1}

The cube root of 1 is indeed 1, so our equation holds true:

1=11 = 1

This confirms that x=βˆ’2x = -2 is indeed a valid solution to our original equation. Yay! We've successfully navigated through the problem, found a solution, and verified it. This process not only gives us the correct answer but also reinforces our understanding of the underlying mathematical principles. Verifying solutions is a hallmark of a thorough problem-solver, and it’s a habit that will serve you well in all your mathematical endeavors.

Final Answer

Alright guys, after a thorough journey through the realm of exponential equations, we've arrived at our final destination! We successfully solved the equation 9x+2=3x+239^{x+2}=\sqrt[3]{3^{x+2}} and verified our solution. So, the final answer is:

x=βˆ’2x = -2

This wasn't just about getting the right answer; it was about understanding the process. We started by recognizing the structure of the equation, then strategically manipulated it using the properties of exponents. We transformed the equation into a more manageable form, solved it, and crucially, verified our solution. This holistic approach is what truly cements our understanding and equips us to tackle similar problems in the future.

Remember, mathematics is not just about memorizing formulas; it's about developing a way of thinking. By breaking down complex problems into smaller, manageable steps, we can conquer any mathematical challenge. This problem served as a fantastic example of how we can use these strategies to unravel exponential equations. Keep practicing, keep exploring, and most importantly, keep questioning! The world of mathematics is vast and fascinating, and each problem we solve is a step further on our journey of discovery. So, until next time, keep those mathematical gears turning!