Solving 2^x + 1/(2^x) = 17/4 A Detailed Mathematical Analysis

by ADMIN 62 views

Introduction

In this article, we will delve into the fascinating realm of exponential equations, specifically focusing on the equation 2^x + 1/(2^x) = 17/4. This equation, seemingly simple at first glance, requires a clever approach to unravel its solution. We will explore various techniques, including algebraic manipulation and substitution, to arrive at the final answer. Understanding how to solve such equations is crucial in many areas of mathematics, physics, and engineering, where exponential relationships frequently arise. This comprehensive guide will not only provide the solution but also elucidate the underlying concepts and methods, making it a valuable resource for students and enthusiasts alike.

Problem Statement: Unveiling the Exponential Mystery

The equation we aim to solve is:

2^x + 1/(2^x) = 17/4

This equation involves an exponential term (2^x) and its reciprocal, which adds an interesting twist to the problem. Our goal is to find the value(s) of 'x' that satisfy this equation. To do so, we'll embark on a step-by-step journey, employing algebraic techniques to simplify and ultimately solve for 'x'. This problem exemplifies the importance of recognizing patterns and employing appropriate substitutions to transform complex equations into more manageable forms. The beauty of mathematics lies in its ability to reveal hidden structures, and this problem is a perfect illustration of that principle. We'll begin by exploring the initial steps in solving this exponential puzzle.

Transforming the Equation: A Step-by-Step Approach

To tackle this exponential equation effectively, we'll first aim to simplify it. The presence of a fraction makes the equation slightly cumbersome, so our initial strategy will be to eliminate the fraction. We can achieve this by multiplying both sides of the equation by 2^x. This is a common technique used in solving equations involving fractions, as it clears the denominator and makes the equation easier to manipulate. Let's see how this transformation unfolds:

Original equation:

2^x + 1/(2^x) = 17/4

Multiply both sides by 2^x:

(2^x) * [2^x + 1/(2^x)] = (17/4) * (2^x)

Distributing 2^x on the left side, we get:

(2x)2 + 1 = (17/4) * (2^x)

This transformation has led us to a new equation that looks more manageable. We now have a quadratic-like term in (2x)2, which hints at the possibility of using a substitution to simplify the equation further. The next step involves rearranging the terms to prepare for this substitution, bringing us closer to the solution. By carefully applying algebraic manipulations, we are gradually unraveling the complexity of the original equation.

Substitution: Introducing a New Variable

The transformed equation, (2x)2 + 1 = (17/4) * (2^x), now resembles a quadratic equation. To make this resemblance more explicit and facilitate the solution process, we will introduce a substitution. Let's substitute y = 2^x. This substitution is a powerful technique that allows us to replace a complex expression with a single variable, making the equation easier to handle. With this substitution, our equation transforms into:

y^2 + 1 = (17/4) * y

This new equation is a quadratic equation in 'y'. Quadratic equations are well-studied, and we have a variety of methods at our disposal to solve them, such as factoring, completing the square, or using the quadratic formula. The substitution has effectively converted our exponential equation into a more familiar form, paving the way for us to find the values of 'y'. Once we determine the values of 'y', we can then reverse the substitution to find the corresponding values of 'x'. This strategic use of substitution demonstrates a key problem-solving technique in mathematics.

Solving the Quadratic Equation: Finding the Values of 'y'

Now that we have the quadratic equation y^2 + 1 = (17/4) * y, our next step is to solve for 'y'. To do this, we'll first rearrange the equation into the standard quadratic form: ay^2 + by + c = 0. This standard form allows us to easily apply methods like factoring or the quadratic formula. Let's rearrange the equation:

Subtract (17/4)y from both sides:

y^2 - (17/4)y + 1 = 0

To eliminate the fraction, we can multiply the entire equation by 4:

4y^2 - 17y + 4 = 0

Now we have a quadratic equation with integer coefficients. We can solve this equation by factoring. We are looking for two numbers that multiply to (4 * 4 = 16) and add up to -17. These numbers are -16 and -1. So, we can factor the quadratic as follows:

(4y - 1)(y - 4) = 0

Setting each factor equal to zero, we get two possible solutions for 'y':

4y - 1 = 0 => y = 1/4

y - 4 = 0 => y = 4

Thus, we have found two values for 'y': 1/4 and 4. These values are crucial as they will lead us to the solutions for 'x' once we reverse the substitution. The process of solving this quadratic equation highlights the importance of algebraic manipulation and factoring techniques in solving mathematical problems.

Back-Substitution: Unveiling the Solutions for 'x'

We have successfully found the values of 'y' as 1/4 and 4. Now, we need to reverse the substitution y = 2^x to find the corresponding values of 'x'. This step is crucial as it connects our intermediate solutions in terms of 'y' back to the original variable 'x'. Let's consider each value of 'y' separately:

Case 1: y = 1/4

Substituting back into y = 2^x, we get:

2^x = 1/4

We can rewrite 1/4 as 2^(-2):

2^x = 2^(-2)

Since the bases are equal, we can equate the exponents:

x = -2

Case 2: y = 4

Substituting back into y = 2^x, we get:

2^x = 4

We can rewrite 4 as 2^2:

2^x = 2^2

Equating the exponents:

x = 2

Therefore, we have found two solutions for 'x': -2 and 2. These solutions satisfy the original equation, completing our journey to solve the exponential equation. The process of back-substitution is a vital step in solving equations involving substitutions, ensuring that we express the solutions in terms of the original variables.

Verification: Ensuring the Accuracy of Our Solutions

To ensure the accuracy of our solutions, we must verify them by substituting them back into the original equation: 2^x + 1/(2^x) = 17/4. This step is crucial in any problem-solving process, as it helps us catch any potential errors made during the solution process. Let's verify our solutions:

Case 1: x = -2

Substitute x = -2 into the original equation:

2^(-2) + 1/(2^(-2)) = 17/4

1/4 + 1/(1/4) = 17/4

1/4 + 4 = 17/4

1/4 + 16/4 = 17/4

17/4 = 17/4

The equation holds true for x = -2.

Case 2: x = 2

Substitute x = 2 into the original equation:

2^2 + 1/(2^2) = 17/4

4 + 1/4 = 17/4

16/4 + 1/4 = 17/4

17/4 = 17/4

The equation also holds true for x = 2.

Both solutions, x = -2 and x = 2, satisfy the original equation. This verification step provides us with confidence in the correctness of our solutions. The importance of verification cannot be overstated, as it is the final step in ensuring the accuracy of our results.

Conclusion: Summarizing the Solution and Key Takeaways

In this comprehensive analysis, we successfully solved the exponential equation 2^x + 1/(2^x) = 17/4. Our journey involved several key steps, each highlighting important mathematical techniques. We began by transforming the equation to eliminate the fraction, followed by a crucial substitution to convert the equation into a quadratic form. We then solved the quadratic equation, found the values of the substituted variable, and reversed the substitution to obtain the solutions for 'x'. Finally, we verified our solutions to ensure their accuracy.

The solutions we found are x = -2 and x = 2. These values satisfy the original equation, demonstrating the power of algebraic manipulation and strategic problem-solving techniques.

Key takeaways from this analysis include:

  • The importance of algebraic manipulation in simplifying equations.
  • The effectiveness of substitution in transforming complex equations into more manageable forms.
  • The application of quadratic equation solving methods.
  • The necessity of verification to ensure the accuracy of solutions.

This problem serves as a valuable example of how exponential equations can be solved using a combination of algebraic techniques and strategic thinking. The methods employed here can be applied to a wide range of similar problems, making this analysis a valuable resource for students and anyone interested in mathematical problem-solving.