Solving Exponential Equations Finding The Value Of 3^(x1+x2)

Introduction

In the realm of algebra, exponential equations present a unique challenge and fascination. These equations, where the variable appears in the exponent, often require clever manipulation and a deep understanding of exponential properties to solve. One such intriguing problem involves finding the value of an expression involving the roots of a given exponential equation. In this article, we will delve into the equation 3 ⋅ 2^(x+1) = 48 + 6^x - 8 ⋅ 3^x, explore its roots, and ultimately determine the value of 3^(x1+x2), where x1 and x2 are the roots of the equation. This journey will not only enhance our problem-solving skills but also deepen our appreciation for the elegance and power of exponential functions.

The core of this problem lies in manipulating the given exponential equation into a more manageable form. The presence of different bases (2, 3, and 6) suggests that we might need to rewrite the equation using a common base or employ substitution techniques to simplify it. We'll also need to recall the fundamental properties of exponents, such as the product rule (a^(m+n) = a^m ⋅ a^n) and the power of a product rule ((ab)^n = a^n ⋅ b^n). These properties will be instrumental in transforming the equation and isolating the variable x. Furthermore, understanding the relationship between the roots of a quadratic equation and its coefficients will prove crucial in the final step of finding 3^(x1+x2). This article will provide a step-by-step guide to tackling this problem, ensuring clarity and understanding at each stage.

Solving the Exponential Equation

To solve the exponential equation 3 ⋅ 2^(x+1) = 48 + 6^x - 8 ⋅ 3^x, our initial step involves simplifying the equation and attempting to rewrite it in a more manageable form. We can begin by applying the product rule of exponents to the term 2^(x+1), which allows us to express it as 2^x ⋅ 2^1 or simply 2 ⋅ 2^x. Substituting this back into the original equation, we get 3 ⋅ 2 ⋅ 2^x = 48 + 6^x - 8 ⋅ 3^x, which simplifies to 6 ⋅ 2^x = 48 + 6^x - 8 ⋅ 3^x. This transformation helps to isolate the exponential terms and paves the way for further manipulation.

Next, we recognize that 6^x can be expressed as (2 ⋅ 3)^x, which, using the power of a product rule, is equivalent to 2^x ⋅ 3^x. Substituting this into our equation, we now have 6 ⋅ 2^x = 48 + 2^x ⋅ 3^x - 8 ⋅ 3^x. This step is crucial as it introduces terms with both 2^x and 3^x, suggesting that we might be able to rearrange the equation into a form that resembles a quadratic equation. To make this resemblance clearer, we can rearrange the terms to get 2^x ⋅ 3^x - 8 ⋅ 3^x - 6 ⋅ 2^x + 48 = 0. This rearrangement is a strategic move, allowing us to group terms and potentially factor the expression.

Now, we can apply a factoring technique to the left-hand side of the equation. By grouping the first two terms and the last two terms, we can factor out common factors. From the first two terms, 2^x ⋅ 3^x - 8 ⋅ 3^x, we can factor out 3^x, leaving us with 3^x (2^x - 8). From the last two terms, -6 ⋅ 2^x + 48, we can factor out -6, which gives us -6(2^x - 8). Now, our equation looks like 3^x (2^x - 8) - 6(2^x - 8) = 0. Notice that we now have a common factor of (2^x - 8) in both terms. Factoring this out, we get (2^x - 8)(3^x - 6) = 0. This factorization is a significant milestone, as it transforms the original exponential equation into a product of two factors, each of which can be easily solved.

Setting each factor equal to zero gives us two separate equations: 2^x - 8 = 0 and 3^x - 6 = 0. Solving the first equation, 2^x - 8 = 0, we add 8 to both sides to get 2^x = 8. Since 8 can be expressed as 2^3, we have 2^x = 2^3, which implies that x = 3. Solving the second equation, 3^x - 6 = 0, we add 6 to both sides to get 3^x = 6. To solve for x in this case, we can take the logarithm of both sides. Taking the logarithm base 3, we get log3(3^x) = log3(6), which simplifies to x = log3(6). Therefore, the two roots of the original exponential equation are x1 = 3 and x2 = log3(6). These roots are the foundation for our next step, where we will calculate the value of 3^(x1+x2).

Finding the Value of 3^(x1+x2)

Now that we have determined the roots of the equation, x1 = 3 and x2 = log3(6), we can proceed to calculate the value of 3^(x1+x2). This involves substituting the values of x1 and x2 into the expression and simplifying using the properties of exponents and logarithms. The first step is to find the sum of the roots, x1 + x2, which is 3 + log3(6). This sum will be the exponent of 3 in our final calculation.

Substituting this sum into the expression 3^(x1+x2), we get 3^(3 + log3(6)). To simplify this, we can use the product rule of exponents, which states that a^(m+n) = a^m ⋅ a^n. Applying this rule, we can rewrite our expression as 3^3 ⋅ 3^(log3(6)). This separation of the exponent into two parts allows us to simplify each part individually.

The first part, 3^3, is straightforward to calculate: 3^3 = 3 ⋅ 3 ⋅ 3 = 27. The second part, 3^(log3(6)), involves an exponent that is a logarithm. Here, we can use the fundamental property of logarithms that states a^(loga(x)) = x. In our case, this means that 3^(log3(6)) = 6. This property is a direct consequence of the definition of logarithms and is a powerful tool in simplifying expressions.

Now, we can substitute these simplified values back into our expression. We have 3^(x1+x2) = 3^3 ⋅ 3^(log3(6)) = 27 ⋅ 6. Multiplying 27 by 6 gives us 162. Therefore, the value of 3^(x1+x2) is 162. This result is the culmination of our efforts in solving the exponential equation and applying the properties of exponents and logarithms.

Conclusion

In this exploration of exponential equations and their roots, we have successfully navigated the complexities of the equation 3 ⋅ 2^(x+1) = 48 + 6^x - 8 ⋅ 3^x. By applying strategic algebraic manipulations, factoring techniques, and the fundamental properties of exponents and logarithms, we were able to determine the roots of the equation and ultimately find the value of 3^(x1+x2). This journey underscores the importance of a solid understanding of exponential functions and their properties in solving mathematical problems.

The process began with simplifying the original equation by rewriting terms and applying exponent rules. We then strategically rearranged the equation to facilitate factoring, a crucial step in isolating the roots. The factorization led us to two simpler equations, each of which could be solved using basic algebraic techniques and logarithmic properties. Once the roots were found, we applied the product rule of exponents and the fundamental property of logarithms to calculate the final answer.

The result, 3^(x1+x2) = 162, highlights the interconnectedness of various mathematical concepts. This problem not only tested our knowledge of exponential equations but also our understanding of factoring, logarithms, and the properties that govern these mathematical entities. By mastering these concepts, we can approach a wide range of mathematical challenges with confidence and skill. This exercise serves as a valuable reminder of the power and beauty of mathematical problem-solving.

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