Solving The Exponential Equation $3^{3 X-5} \cdot A^{x-2}=5^{x-2} \cdot A^{1-x}$

Introduction to Exponential Equations

In the realm of mathematics, exponential equations stand as a fascinating and crucial area of study. These equations, characterized by variables appearing in the exponents, challenge us to delve into the intricacies of exponents and their properties. Our focus today is on dissecting and solving an intriguing equation: 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}. This equation, while seemingly complex, unveils its secrets when we apply the fundamental principles of exponential manipulation and algebraic techniques. Understanding how to solve such equations is not just an academic exercise; it has practical applications in various fields, including finance, physics, and computer science. In the financial world, exponential equations help model compound interest and investment growth. In physics, they describe phenomena like radioactive decay and population growth. Even in the realm of computer science, algorithms and data structures often rely on exponential relationships. Therefore, mastering the art of solving exponential equations opens doors to a broader understanding of the world around us.

Deconstructing the Equation: 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}

Let's begin by carefully deconstructing the given equation: 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}. At first glance, it presents a blend of exponential terms with different bases and exponents involving the variable 'x'. The presence of the variable 'a' as a base adds another layer of complexity. Our initial strategy will involve employing the core properties of exponents to rearrange and simplify the equation. This includes leveraging rules like the product of powers (where a^m ullet a^n = a^{m+n}) and the quotient of powers (where $a^m / a^n = a^{m-n}$). We'll also explore the possibility of grouping terms with similar bases together. This step-by-step approach allows us to systematically untangle the equation, transforming it into a more manageable form. By isolating terms and applying logarithmic transformations, we aim to pave the way for solving for 'x'. The success of this endeavor hinges on a solid grasp of exponential rules and algebraic manipulation techniques. These are the tools that will empower us to navigate the intricacies of this equation and ultimately arrive at a solution.

Step-by-Step Solution: Isolating and Solving for 'x'

To solve the equation 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}, our first step is to strategically rearrange the terms. We aim to group terms with the same base together. Divide both sides of the equation by $a^{x-2}$ and by $5^{x-2}$. This gives us: $\frac{3^{3x-5}}{5^{x-2}} = \frac{a^{1-x}}{a^{x-2}}$. This manipulation allows us to consolidate the terms with the same base, making the equation more amenable to further simplification. Next, we apply the quotient of powers rule, which states that $\frac{a^m}{a^n} = a^{m-n}$. Applying this rule to the right side of the equation, we get: $a^{(1-x) - (x-2)} = a^{1-x-x+2} = a^{3-2x}$. Now, our equation looks like this: $\frac{3^{3x-5}}{5^{x-2}} = a^{3-2x}$. To further simplify the left side, we can rewrite the equation as: 3^{3x-5} = 5^{x-2} ullet a^{3-2x}. At this point, the equation presents a challenge because we have different bases and exponents. The next logical step involves taking the logarithm of both sides. This is a powerful technique that allows us to bring the exponents down as coefficients. By carefully applying logarithmic properties, we can transform the equation into a linear form, making it easier to solve for 'x'. This process will be the key to unlocking the final solution.

Logarithmic Transformation: Unveiling the Linear Equation

Continuing our quest to solve for 'x' in the equation 3^{3x-5} = 5^{x-2} ullet a^{3-2x}, we now employ the powerful tool of logarithmic transformation. Taking the natural logarithm (ln) of both sides of the equation, we get: ln(3^{3x-5}) = ln(5^{x-2} ullet a^{3-2x}). The beauty of logarithms lies in their ability to convert exponential relationships into multiplicative ones. Using the logarithmic property ln(a^b) = b ullet ln(a), we can rewrite the left side of the equation as: $(3x-5)ln(3)$. Similarly, on the right side, we use the property ln(a ullet b) = ln(a) + ln(b) to get: $ln(5^{x-2}) + ln(a^{3-2x})$. Applying the power rule of logarithms again, we have: $(x-2)ln(5) + (3-2x)ln(a)$. Now, our equation looks like this: $(3x-5)ln(3) = (x-2)ln(5) + (3-2x)ln(a)$. This transformation has brought us to a crucial juncture. We have successfully converted the exponential equation into a linear equation in terms of 'x'. The next step involves carefully expanding the terms and collecting like terms to isolate 'x'. This algebraic manipulation will pave the way for us to express 'x' in terms of known constants and logarithmic values. The precision and accuracy in this step are paramount to arriving at the correct solution.

Algebraic Manipulation: Isolating 'x' and Finding the Solution

Having transformed the original exponential equation into a linear equation, we now focus on the algebraic manipulation required to isolate 'x'. Our equation stands as: $(3x-5)ln(3) = (x-2)ln(5) + (3-2x)ln(a)$. The first step is to expand the terms on both sides of the equation. This gives us: $3xln(3) - 5ln(3) = xln(5) - 2ln(5) + 3ln(a) - 2xln(a)$. Next, we want to group all the terms containing 'x' on one side of the equation and the constant terms on the other side. Rearranging the terms, we get: $3xln(3) - xln(5) + 2xln(a) = 5ln(3) - 2ln(5) + 3ln(a)$. Now, we factor out 'x' from the terms on the left side: $x(3ln(3) - ln(5) + 2ln(a)) = 5ln(3) - 2ln(5) + 3ln(a)$. Finally, to isolate 'x', we divide both sides of the equation by the coefficient of 'x': $x = \frac{5ln(3) - 2ln(5) + 3ln(a)}{3ln(3) - ln(5) + 2ln(a)}$. This expression represents the solution for 'x' in terms of the natural logarithms of 3, 5, and 'a'. It is important to note that the value of 'a' must be such that the denominator is not equal to zero. This ensures that our solution is mathematically valid. The solution we've obtained is a testament to the power of logarithmic transformation and algebraic manipulation in solving complex exponential equations.

Verification and Conditions: Ensuring a Valid Solution

With a potential solution for 'x' in hand, it's crucial to verify its validity and consider any conditions that might apply. Our solution is given by: $x = \frac{5ln(3) - 2ln(5) + 3ln(a)}{3ln(3) - ln(5) + 2ln(a)}$. Before we declare this as the definitive solution, we must address a critical condition: the denominator cannot be zero. If the denominator, $3ln(3) - ln(5) + 2ln(a)$, equals zero, our solution becomes undefined. Therefore, we need to ensure that $3ln(3) - ln(5) + 2ln(a) \neq 0$. This condition places a constraint on the possible values of 'a'. If this condition is not met, then the equation has no solution. To further verify our solution, we can substitute the expression for 'x' back into the original equation: 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}. This substitution, while algebraically intensive, provides a rigorous check on the correctness of our derived solution. If both sides of the equation are equal after the substitution, then our solution is indeed valid. This process of verification and condition checking is an indispensable part of mathematical problem-solving. It ensures that our solutions are not only mathematically sound but also practically meaningful. In the context of real-world applications, understanding these conditions is vital for interpreting the results and making informed decisions.

Conclusion: The Power of Exponential Equation Solving

In conclusion, we've successfully navigated the intricate world of exponential equations to solve 3^{3x-5} ullet a^{x-2} = 5^{x-2} ullet a^{1-x}. Our journey began with deconstructing the equation and strategically rearranging terms using the fundamental properties of exponents. We then harnessed the power of logarithmic transformation to convert the exponential equation into a linear one. This pivotal step allowed us to apply algebraic manipulation techniques, ultimately isolating 'x' and expressing it in terms of natural logarithms. However, our exploration didn't end with just finding a solution. We emphasized the crucial step of verification and condition checking, ensuring the mathematical validity and practical relevance of our result. The condition that the denominator, $3ln(3) - ln(5) + 2ln(a)$, must not be zero highlights the importance of considering constraints in mathematical problem-solving. The ability to solve exponential equations is a valuable skill that extends far beyond the realm of pure mathematics. It empowers us to model and understand phenomena in diverse fields, from finance and physics to computer science and engineering. The techniques we've employed, such as logarithmic transformation and algebraic manipulation, are versatile tools that can be applied to a wide range of mathematical challenges. As we continue to explore the world of mathematics, the lessons learned from solving exponential equations will undoubtedly serve as a solid foundation for tackling even more complex problems.