Solve this equation using successive approximation. Use the provided graph as a starting point:
[tex]3^x-2=\log (2 x+6)[/tex]
Select the correct answer from each drop-down menu.
After three iterations of successive approximation, the positive solution for [tex]x[/tex] is approximately ____.
After four iterations of successive approximation, the positive solution for [tex]x[/tex] is approximately ____.
An approximate solution from four iterations is always ____ an approximate solution from three iterations.
Answer (2)
Rewrite the equation as x = f ( x ) , where f ( x ) = lo g 3 ( lo g ( 2 x + 6 ) + 2 ) .
Start with an initial guess x 0 = 1.0 from the graph.
Iterate using the formula x n + 1 = f ( x n ) for n = 0 , 1 , 2 , 3 .
After three iterations, x 3 ≈ 1.2955 , and after four iterations, x 4 ≈ 1.2955 . An approximate solution from four iterations is always more accurate than an approximate solution from fewer iterations. The final answer is 1.2955 .
Explanation
Understanding the Problem We are given the equation 3 x − 2 = lo g ( 2 x + 6 ) and we want to solve it using successive approximation. The successive approximation method involves rewriting the equation in the form x = f ( x ) and then iteratively applying the function f to an initial guess x 0 until the solution converges.
Rewriting the Equation and Initial Guess First, we need to rewrite the given equation in the form x = f ( x ) .
3 x − 2 = lo g ( 2 x + 6 ) 3 x = lo g ( 2 x + 6 ) + 2 x = lo g 3 ( lo g ( 2 x + 6 ) + 2 ) So, f ( x ) = lo g 3 ( lo g ( 2 x + 6 ) + 2 ) .
From the graph (not provided here, but assumed to be available), we can make an initial guess for x . Let's assume the initial guess is x 0 = 1.0 .
Performing Iterations Now, we iterate using the formula x n + 1 = f ( x n ) .
x 1 = f ( x 0 ) = f ( 1 ) = lo g 3 ( lo g ( 2 ( 1 ) + 6 ) + 2 ) = lo g 3 ( lo g ( 8 ) + 2 ) ≈ lo g 3 ( 0.903 + 2 ) = lo g 3 ( 2.903 ) ≈ 1.2798 x 2 = f ( x 1 ) = f ( 1.2798 ) = lo g 3 ( lo g ( 2 ( 1.2798 ) + 6 ) + 2 ) = lo g 3 ( lo g ( 8.5596 ) + 2 ) ≈ lo g 3 ( 0.932 + 2 ) = lo g 3 ( 2.932 ) ≈ 1.2947 x 3 = f ( x 2 ) = f ( 1.2947 ) = lo g 3 ( lo g ( 2 ( 1.2947 ) + 6 ) + 2 ) = lo g 3 ( lo g ( 8.5894 ) + 2 ) ≈ lo g 3 ( 0.934 + 2 ) = lo g 3 ( 2.934 ) ≈ 1.2955 $x_4 = f(x_3) = f(1.2955) = \log_3(\log(2(1.2955) + 6) + 2) = \log_3(\log(8.591) + 2) \approx \log_3(0.934 + 2) = \log_3(2.934) \approx 1.2955
Determining the Approximate Solutions and Comparison After three iterations, the approximate solution for x is x 3 ≈ 1.2955 .
After four iterations, the approximate solution for x is x 4 ≈ 1.2955 .
Since the values are converging, an approximate solution from four iterations is always more accurate than an approximate solution from fewer iterations (assuming the method converges).
Examples
Successive approximation is a powerful tool used in various fields, such as engineering and computer science, to find approximate solutions to equations that are difficult or impossible to solve analytically. For example, in circuit analysis, it can be used to determine the voltage or current in a complex circuit by iteratively refining an initial guess until a stable solution is reached. This method is also used in optimization problems, where finding the exact minimum or maximum of a function is computationally expensive, and an approximate solution is sufficient.
After three iterations, the positive solution for x is approximately 1.2955. After four iterations, the solution remains approximately 1.2955. An approximate solution from four iterations is always more accurate than that from three iterations.
;
