Solving 3^x - 2 = Log(2x + 6) With Successive Approximation

Successive approximation, also known as the method of iterations, is a powerful numerical technique for finding solutions to equations that cannot be solved algebraically. It's like a detective's approach to solving a mystery – we start with a clue (an initial guess) and iteratively refine it until we get closer and closer to the solution. In this article, we'll break down the process of solving the equation 3^x - 2 = log(2x + 6) using successive approximation. We'll explore how to use a graph as a starting point and walk through the iterations, making this method crystal clear for you guys.

Understanding Successive Approximation

Before diving into the specifics of our equation, let's grasp the core idea behind successive approximation. Imagine you're trying to find the root of an equation, which is the value of x that makes the equation equal to zero. Successive approximation involves rearranging the equation into the form x = g(x), where g(x) is some function of x. We then start with an initial guess, x₀, and plug it into the right side of the equation to get a new value, x₁ = g(x₀). This new value becomes our next guess, and we repeat the process, generating a sequence of approximations: x₂ = g(x₁), x₃ = g(x₂), and so on. If this sequence converges, it will approach the solution of the original equation. Think of it like zooming in on the solution with each iteration, refining our estimate until it's accurate enough. The beauty of this method is its ability to tackle equations that are difficult or impossible to solve directly. It's a workhorse in numerical analysis, used across various fields of science and engineering.

The Art of Rearranging Equations

The key to successful successive approximation lies in rearranging the original equation into the x = g(x) form. This isn't always straightforward, and sometimes there are multiple ways to rearrange the equation. The choice of rearrangement can significantly impact the convergence of the method. A poorly chosen g(x) might lead to a sequence of approximations that diverge, meaning they move further away from the solution. Ideally, we want a g(x) that is well-behaved and leads to a stable convergence. There's no one-size-fits-all rule for rearranging, and it often involves some trial and error. However, a good starting point is to isolate x on one side of the equation as much as possible. For example, if we have an equation like f(x) = h(x), we might try rearranging it to x = f^-1(h(x)), where f^-1 is the inverse of f. It's like untangling a knot – we need to manipulate the equation in a way that makes the iterative process converge towards the solution. In our specific case, we'll explore different rearrangements to find one that suits our needs.

The Significance of the Starting Point

The initial guess, x₀, plays a crucial role in the success of successive approximation. A good starting point can significantly reduce the number of iterations needed to reach a satisfactory solution, while a poor starting point might lead to slow convergence or even divergence. This is where the provided graph comes into play. Graphs provide a visual representation of the equation, allowing us to identify approximate solutions where the curves intersect. These intersection points are the roots of the equation, and they serve as excellent starting points for our iterative process. Choosing a starting point close to the actual solution helps the approximation converge quickly. It's like having a treasure map – the graph guides us to the vicinity of the treasure (the solution), and successive approximation helps us pinpoint the exact location. The graph provides a global view of the equation's behavior, while the iterative method zooms in on the specific solution we're looking for. This combination of graphical insight and numerical refinement is a powerful tool in solving equations.

Applying Successive Approximation to 3^x - 2 = log(2x + 6)

Now, let's tackle our specific equation: 3^x - 2 = log(2x + 6). This equation is a mix of exponential and logarithmic functions, making it difficult to solve algebraically. That's where successive approximation shines. We'll start by rearranging the equation and then use the graph to find a suitable initial guess.

Step 1 Rearranging the Equation

First, we need to rearrange the equation into the form x = g(x). There are a couple of ways we could do this. One option is to isolate the exponential term:

3^x = log(2x + 6) + 2

Then, taking the logarithm base 3 of both sides, we get:

x = log₃(log(2x + 6) + 2)

This gives us our first possible g(x):

g₁(x) = log₃(log(2x + 6) + 2)

Another option is to isolate the logarithmic term:

log(2x + 6) = 3^x - 2

Then, exponentiating both sides (using base 10, since the logarithm is assumed to be base 10), we get:

2x + 6 = 10^{(3x - 2)}

Isolating x, we have:

x = (10^{(3x - 2)} - 6) / 2

This gives us a second possible g(x):

g₂(x) = (10^{(3x - 2)} - 6) / 2

It's important to note that the choice of g(x) can affect the convergence of the method. We'll start with g₁(x) and see how it works. If it doesn't converge well, we can always try g₂(x) or another rearrangement. It's like choosing the right tool for the job – some rearrangements are more effective than others.

Step 2 Using the Graph as a Starting Point

The problem mentions using a provided graph as a starting point. Assuming we have a graph of y = 3^x - 2 and y = log(2x + 6), we can look for the point(s) where the two curves intersect. The x-coordinate of the intersection point(s) will be the solution(s) to our equation. Let's say, from the graph, we estimate that the positive solution is approximately x = 2. This will be our initial guess, x₀. The graph acts as a visual guide, helping us to zero in on the region where the solution lies. It's like having a map that points to the general location of the treasure – now we need to dig to find the exact spot. This initial estimate is crucial for the efficiency of the successive approximation method.

Step 3 Iterations of Successive Approximation

Now, we'll perform the iterations using our chosen g(x) and our initial guess. We'll use g₁(x) = log₃(log(2x + 6) + 2) and x₀ = 2.

• Iteration 1:

  x₁ = g₁(x₀) = log₃(log(2(2) + 6) + 2) = log₃(log(10) + 2) ≈ log₃(1 + 2) = log₃(3) = 1

• Iteration 2:

  x₂ = g₁(x₁) = log₃(log(2(1) + 6) + 2) = log₃(log(8) + 2) ≈ log₃(0.903 + 2) = log₃(2.903) ≈ 0.96

• Iteration 3:

  x₃ = g₁(x₂) = log₃(log(2(0.96) + 6) + 2) = log₃(log(7.92) + 2) ≈ log₃(0.899 + 2) = log₃(2.899) ≈ 0.958

After three iterations, the positive solution for x is approximately 0.958.

Step 4: Iteration 4 and the Approximate Solution

Let's continue for one more iteration to see if our approximation stabilizes:

• Iteration 4:

  x₄ = g₁(x₃) = log₃(log(2(0.958) + 6) + 2) = log₃(log(7.916) + 2) ≈ log₃(0.8985 + 2) = log₃(2.8985) ≈ 0.958

After four iterations, the positive solution for x is still approximately 0.958. The values seem to be converging, giving us confidence in our solution. This fourth iteration acts as a check, confirming the stability of our approximation. It's like double-checking your work – ensuring that the solution is consistent and reliable.

Answer to the Question

The question asks for the approximate positive solution after three iterations. Based on our calculations, the answer is approximately 0.958.

Key Takeaways

• Successive approximation is a powerful technique for solving equations that are difficult to solve algebraically.
• Rearranging the equation into the form x = g(x) is crucial, and the choice of g(x) can affect convergence.
• A graph provides a visual starting point for the iterations.
• Each iteration refines the approximation, getting closer to the solution.
• The more iterations you perform, the more accurate your solution is likely to be, up to a point.

This method, my friends, is a cornerstone of numerical analysis, providing a way to tackle complex problems with a systematic approach. It's like having a versatile tool in your problem-solving toolkit, ready to be deployed whenever analytical methods fall short. So, keep practicing and exploring its applications, and you'll become a master of successive approximation!