Solve For X: 9^(2x+1) = 3^(x-4)

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When faced with an equation like 9^(2x+1) = 3^(x-4), the immediate goal is to solve for x. This type of problem falls under the umbrella of exponential equations, where the variable we're trying to find is in the exponent. The key to unlocking these equations often lies in recognizing that the bases of the exponents can be made the same. In this specific case, we can observe that 9 is simply 3 squared (323^2). This fundamental relationship allows us to rewrite the entire equation with a common base, which is the crucial first step in isolating and solving for x. By manipulating the equation to have identical bases, we transform a seemingly complex exponential problem into a more manageable algebraic one. This strategic approach is not only effective but also elegant, showcasing the power of understanding the properties of exponents. It's about seeing the underlying structure and using mathematical rules to simplify and conquer. This initial transformation sets the stage for the subsequent steps, where we'll use exponent rules to further simplify and eventually find the value of x. Remember, in mathematics, often the first step is the most important, and recognizing common bases is a cornerstone of solving many exponential equations.

Rewriting the Equation with a Common Base

To solve for x in the equation 9^(2x+1) = 3^(x-4), our primary strategy is to ensure both sides of the equation have the same base. As noted earlier, 9 can be expressed as 323^2. This allows us to substitute 323^2 for 9 in the equation. So, the left side of the equation, 92x+19^{2x+1}, becomes (32)2x+1(3^2)^{2x+1}. Now, we apply the power of a power rule for exponents, which states that (am)n=amimesn(a^m)^n = a^{m imes n}. Multiplying the exponents 22 and (2x+1)(2x+1), we get 2imes(2x+1)=4x+22 imes (2x+1) = 4x + 2. Therefore, the left side of the equation is now 34x+23^{4x+2}. The right side of the equation, 3xβˆ’43^{x-4}, already has a base of 3. Now, our equation looks much simpler: 34x+2=3xβˆ’43^{4x+2} = 3^{x-4}. This step is critical because when two exponential expressions with the same base are equal, their exponents must also be equal. This principle allows us to move from an exponential equation to a linear one, making the process of finding x much more straightforward. It’s a beautiful demonstration of how consistent application of exponent rules can simplify complex expressions. This common base transformation is a gateway to solving the problem, paving the way for equating the exponents and performing basic algebraic manipulations to find the unknown variable.

Equating the Exponents and Solving for x

With the equation transformed into 34x+2=3xβˆ’43^{4x+2} = 3^{x-4}, we can now leverage the property that if am=ana^m = a^n and aeq1,a>0a eq 1, a > 0, then m=nm = n. This means we can simply equate the exponents to form a linear equation: 4x+2=xβˆ’44x + 2 = x - 4. This is the pivotal moment where the exponential problem becomes a standard algebraic equation. Our goal now is to isolate x. We can begin by subtracting x from both sides of the equation to gather all the x terms on one side: 4xβˆ’x+2=xβˆ’xβˆ’44x - x + 2 = x - x - 4, which simplifies to 3x+2=βˆ’43x + 2 = -4. Next, we want to isolate the term with x, so we subtract 2 from both sides: 3x+2βˆ’2=βˆ’4βˆ’23x + 2 - 2 = -4 - 2, resulting in 3x=βˆ’63x = -6. Finally, to find the value of x, we divide both sides by 3: rac{3x}{3} = rac{-6}{3}. This gives us our solution: x=βˆ’2x = -2. This process, from recognizing common bases to equating exponents and solving the resulting linear equation, is a systematic approach that works for many exponential problems. It underscores the importance of understanding exponent rules and basic algebraic manipulation. The solution for x is indeed -2, and this value can be plugged back into the original equation to verify its correctness.

Verifying the Solution

To ensure our calculated value of x = -2 is correct for the equation 92x+1=3xβˆ’49^{2x+1} = 3^{x-4}, we must perform a verification step. This involves substituting x=βˆ’2x = -2 back into the original equation and checking if both sides yield the same result. Let's start with the left side: 92x+19^{2x+1}. Substituting x = -2, we get 92(βˆ’2)+1=9βˆ’4+1=9βˆ’39^{2(-2)+1} = 9^{-4+1} = 9^{-3}. Now, we know that 9=329 = 3^2, so 9βˆ’39^{-3} can be rewritten as (32)βˆ’3(3^2)^{-3}. Applying the power of a power rule, (am)n=amimesn(a^m)^n = a^{m imes n}, we get 32imes(βˆ’3)=3βˆ’63^{2 imes (-3)} = 3^{-6}.

Now let's evaluate the right side of the original equation: 3xβˆ’43^{x-4}. Substituting x = -2, we get 3βˆ’2βˆ’4=3βˆ’63^{-2-4} = 3^{-6}.

Since the left side evaluates to 3βˆ’63^{-6} and the right side also evaluates to 3βˆ’63^{-6}, we can confidently conclude that 3βˆ’6=3βˆ’63^{-6} = 3^{-6}. This confirms that our solution x = -2 is correct. Verification is a vital part of the problem-solving process, especially in algebra and higher mathematics, as it helps catch errors and builds confidence in the results obtained. It’s a good habit to always double-check your answers, especially when dealing with exponents and variables.

Understanding the Concepts Behind the Solution

The ability to solve for x in equations like 92x+1=3xβˆ’49^{2x+1} = 3^{x-4} hinges on a few fundamental mathematical concepts. The most critical of these is the understanding of exponent rules. Specifically, two rules are paramount here: the power of a power rule, (am)n=amimesn(a^m)^n = a^{m imes n}, and the property that if am=ana^m = a^n (where a>0a > 0 and aeq1a eq 1), then m=nm = n. The first rule allows us to simplify expressions where an exponent is raised to another exponent, which is exactly what happens when we rewrite the base 9 as 323^2. The expression (32)2x+1(3^2)^{2x+1} neatly transforms into 32(2x+1)3^{2(2x+1)} using this rule. The second property is the cornerstone of solving exponential equations; it allows us to equate the exponents once the bases are the same, thereby converting the exponential equation into a linear one that is much easier to solve using standard algebraic techniques.

Furthermore, this problem implicitly tests the understanding of number bases. Recognizing that 9 is a power of 3 (9=329 = 3^2) is the key insight that enables the simplification. This ability to express numbers in terms of a common base is a recurring theme in mathematics, from simplifying radicals to working with logarithms. The process also involves basic algebraic manipulation, such as adding, subtracting, and dividing terms to isolate the variable x. These are the foundational skills that are built upon throughout one's mathematical journey. Mastering these concepts not only helps in solving this specific problem but also equips individuals with the tools needed to tackle a wide range of more complex mathematical challenges. It's a clear illustration of how different branches of mathematics interconnect and support each other. The elegance of the solution lies in its reliance on these core principles.

Conclusion: Mastering Exponential Equations

In conclusion, solving the exponential equation 92x+1=3xβˆ’49^{2x+1} = 3^{x-4} for x involves a systematic approach that relies on fundamental principles of mathematics. We began by recognizing that the bases, 9 and 3, could be unified by expressing 9 as a power of 3 (9=329 = 3^2). This crucial step allowed us to rewrite the equation as 34x+2=3xβˆ’43^{4x+2} = 3^{x-4}. Subsequently, we utilized the property that equal exponents imply equal bases (when the base is the same and not equal to 1 or 0) to equate the exponents: 4x+2=xβˆ’44x + 2 = x - 4. This transformation led us to a simple linear equation, which we then solved using basic algebraic operations to find x=βˆ’2x = -2. The verification step confirmed the accuracy of our solution, demonstrating the importance of checking one's work. The concepts employedβ€”understanding exponent rules, manipulating number bases, and applying algebraic techniquesβ€”are vital for proficiency in mathematics. By mastering these techniques, you gain the ability to tackle a wide array of problems, not just in algebra but across various scientific and technical fields. The journey of solving such equations is not just about finding a number; it's about honing critical thinking and problem-solving skills. These foundational mathematical abilities are invaluable assets.

For further exploration into exponential functions and equations, you can refer to resources from reputable educational institutions. A great place to start is the Khan Academy website, which offers comprehensive lessons and practice exercises on a wide range of mathematical topics, including algebra and precalculus. Their explanations are clear, concise, and designed to build a strong understanding of complex concepts. You can also find valuable information on the Brilliant.org platform, which provides interactive learning experiences that make mathematics engaging and accessible.