Solving (5/2)^x = 7^(1-x) A Step-by-Step Guide With Exact And Approximate Solutions

When dealing with exponential equations, the primary goal is to isolate the variable, which often resides in the exponent. In this article, we delve into the intricacies of solving the equation (5/2)^x = 7^(1-x), showcasing a step-by-step approach to arrive at both the exact and approximate solutions. Exponential equations like this one, where the variable appears in the exponent, are fundamental in various fields, including mathematics, physics, engineering, and finance. Understanding how to solve them is crucial for tackling real-world problems such as population growth, radioactive decay, compound interest, and many more. This guide provides a comprehensive approach, ensuring clarity and understanding at each step.

Before we dive into solving the equation (5/2)^x = 7^(1-x), let's establish a firm understanding of what exponential equations are and the techniques commonly used to solve them. Exponential equations are equations in which the variable appears in the exponent. The general form of an exponential equation is a^x = b, where a and b are constants and x is the variable. Solving such equations often involves using logarithms, which are the inverse operations of exponentiation. The logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. For instance, if 2^3 = 8, then the logarithm of 8 to the base 2 is 3, written as log₂8 = 3. Logarithms are essential tools for solving exponential equations because they allow us to bring the exponent down as a coefficient, making the equation easier to handle.

There are several properties of logarithms that are particularly useful in solving exponential equations. One of the most important is the power rule, which states that log_b(a^x) = x * log_b(a). This rule allows us to move the exponent x from the exponent position to a coefficient, which is crucial for isolating the variable. Another useful property is the change of base formula, which allows us to convert logarithms from one base to another. The formula is log_b(a) = log_c(a) / log_c(b), where c is a new base. This is particularly helpful when dealing with calculators that only compute logarithms to base 10 or base e (natural logarithms). Additionally, understanding the properties of exponents, such as a^(m+n) = a^m * a^n and a^(m-n) = a^m / a^n, is essential for manipulating exponential expressions. These properties allow us to simplify and rewrite equations in a form that is more amenable to logarithmic techniques. In the following sections, we will apply these principles to solve the given exponential equation, providing a detailed and clear explanation of each step.

To solve the equation (5/2)^x = 7^(1-x), we will follow a series of steps that involve applying logarithms to both sides of the equation. This approach allows us to bring the exponents down and isolate the variable x. First, we take the natural logarithm (ln) of both sides of the equation. The natural logarithm is the logarithm to the base e, where e is an irrational number approximately equal to 2.71828. Taking the natural logarithm of both sides is a common technique because it simplifies the equation by allowing us to use logarithmic properties.

Applying the natural logarithm to both sides of the equation (5/2)^x = 7^(1-x) gives us ln((5/2)^x) = ln(7^(1-x)). Next, we use the power rule of logarithms, which states that ln(a^b) = b * ln(a). Applying this rule to both sides of the equation, we get x * ln(5/2) = (1-x) * ln(7). This step is crucial because it moves the exponents x and (1-x) from the exponent positions to coefficients, making the equation linear in terms of x. Now, we need to expand the right side of the equation by distributing ln(7) across (1-x). This gives us x * ln(5/2) = ln(7) - x * ln(7). The next step is to gather all terms containing x on one side of the equation and constant terms on the other side. To do this, we add x * ln(7) to both sides of the equation, which results in x * ln(5/2) + x * ln(7) = ln(7). Now, we can factor out x from the left side of the equation, yielding x * (ln(5/2) + ln(7)) = ln(7). To isolate x, we divide both sides of the equation by (ln(5/2) + ln(7)). This gives us the exact solution for x, which is x = ln(7) / (ln(5/2) + ln(7)). In the following sections, we will simplify this expression further and find its decimal approximation.

The exact form solution of the equation (5/2)^x = 7^(1-x) is given by x = ln(7) / (ln(5/2) + ln(7)). This expression represents the precise value of x that satisfies the equation. However, for practical purposes and better understanding, it is often useful to simplify this expression using logarithmic properties and to obtain a decimal approximation. To simplify the exact form, we can use the logarithm product rule, which states that ln(a) + ln(b) = ln(a * b). Applying this rule to the denominator of the solution, we have ln(5/2) + ln(7) = ln((5/2) * 7) = ln(35/2). Therefore, the exact form solution can be rewritten as x = ln(7) / ln(35/2). This form is slightly more concise and easier to understand compared to the original expression.

However, it is still in an exact form, meaning it contains logarithms and does not provide a numerical value. While this form is mathematically accurate, it is often necessary to obtain a decimal approximation for practical applications. The exact form solution is valuable because it avoids any rounding errors that may occur during the calculation of the decimal approximation. It is the most accurate representation of the solution and is preferred in theoretical contexts where precision is paramount. In summary, the exact form solution x = ln(7) / ln(35/2) provides a precise representation of the value of x that satisfies the equation (5/2)^x = 7^(1-x). It is derived using logarithmic properties and is essential for maintaining mathematical accuracy. In the next section, we will calculate the decimal approximation of this solution to provide a more intuitive understanding of its value.

To find the decimal approximation of the solution x = ln(7) / (ln(5/2) + ln(7)), we need to use a calculator. We will compute the natural logarithm of 7, ln(5/2), and then perform the necessary arithmetic operations. First, we find that ln(7) ≈ 1.94591. Next, we calculate ln(5/2) = ln(2.5) ≈ 0.91629. Now, we can substitute these values into the expression for x: x = ln(7) / (ln(5/2) + ln(7)) x ≈ 1.94591 / (0.91629 + 1.94591) x ≈ 1.94591 / 2.86220 Now, we perform the division: x ≈ 0.67988 Rounding this to three decimal places, we get x ≈ 0.680. Therefore, the decimal approximation of the solution to the equation (5/2)^x = 7^(1-x) is approximately 0.680. This value provides a practical understanding of the solution, making it easier to use in real-world applications. It is important to note that this is an approximation, and the exact value is given by the logarithmic expression x = ln(7) / (ln(5/2) + ln(7)).

While the decimal approximation is useful for many purposes, it is crucial to remember that it is not the exact solution. The exact solution, expressed in terms of logarithms, is the most accurate representation. However, for practical applications where a numerical value is needed, the decimal approximation provides a convenient and understandable result. In summary, the decimal approximation of x ≈ 0.680 is obtained by using a calculator to evaluate the natural logarithms and performing the necessary arithmetic operations. This value is rounded to three decimal places for clarity and ease of use, providing a practical solution to the equation (5/2)^x = 7^(1-x).

While using logarithms is a standard method for solving exponential equations, there are alternative approaches that can be employed depending on the specific equation. One common method involves rewriting the equation with a common base. If it is possible to express both sides of the equation with the same base, the exponents can then be equated, leading to a simpler algebraic equation. However, this method is not always feasible, especially when the bases are not easily related, as in the equation (5/2)^x = 7^(1-x).

Another technique involves graphical methods. By plotting the functions represented by each side of the equation, the solution can be found at the point of intersection. For example, one could plot y = (5/2)^x and y = 7^(1-x) on the same graph. The x-coordinate of the point where the two graphs intersect is the solution to the equation. This method is particularly useful for visualizing the solution and for equations that are difficult to solve algebraically. Graphical methods can also provide insights into the number of solutions and their approximate values. Additionally, numerical methods, such as the Newton-Raphson method, can be used to approximate the solutions of exponential equations. These methods involve iterative calculations to refine an initial guess until a sufficiently accurate solution is obtained. Numerical methods are especially useful for equations that do not have closed-form solutions or are too complex to solve analytically.

In summary, while logarithmic techniques are the most common and versatile method for solving exponential equations, alternative approaches such as rewriting with a common base, graphical methods, and numerical methods can also be employed depending on the specific characteristics of the equation. Each method has its advantages and limitations, and the choice of method often depends on the nature of the equation and the desired level of accuracy.

In this comprehensive guide, we have thoroughly explored the process of solving the exponential equation (5/2)^x = 7^(1-x). We began by understanding the fundamental principles of exponential equations and the crucial role of logarithms in solving them. We then proceeded with a step-by-step solution, demonstrating how to apply the natural logarithm to both sides of the equation, use logarithmic properties to simplify the expression, and isolate the variable x. The exact form solution, x = ln(7) / (ln(5/2) + ln(7)), was derived using logarithmic properties, providing a precise representation of the solution. For practical applications, we also calculated the decimal approximation of the solution, rounding it to three decimal places as x ≈ 0.680.

Throughout this guide, we emphasized the importance of understanding each step and the underlying mathematical principles. We also discussed alternative methods for solving exponential equations, such as rewriting with a common base, graphical methods, and numerical methods, highlighting their advantages and limitations. By providing a detailed explanation and multiple approaches, we aimed to equip readers with a solid understanding of how to solve exponential equations effectively. The ability to solve exponential equations is essential in various fields, including mathematics, science, engineering, and finance. This guide serves as a valuable resource for students, educators, and professionals seeking to enhance their problem-solving skills in this area.

In summary, we successfully solved the exponential equation (5/2)^x = 7^(1-x) by applying logarithms. The exact solution is x = ln(7) / (ln(5/2) + ln(7)), which simplifies to x = ln(7) / ln(35/2). The decimal approximation, rounded to three decimal places, is x ≈ 0.680. This step-by-step guide provides a clear and concise method for solving exponential equations, emphasizing both the theoretical understanding and practical application of logarithmic techniques. The knowledge gained from this guide can be applied to a wide range of problems involving exponential relationships, making it a valuable tool for anyone working in quantitative fields.

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