Consider The Equation And The Graph. 2 X + 4 = 3 Z + 1 \frac{2}{x+4}=3^z+1 X + 4 2 = 3 Z + 1 The Approximate Solution To The Given Equation After Three Iterations Of Successive Approximations Is When X X X Is About:

Mar 7, 2025 by ADMIN 218 views

**Solving the Equation through Successive Approximations**

Introduction

In mathematics, solving equations can be a complex task, especially when dealing with non-linear equations. One method to solve such equations is through successive approximations. This method involves making an initial guess and then iteratively improving the guess until a solution is obtained. In this article, we will consider the equation $\frac{2}{x+4}=3^z+1$ and use successive approximations to find an approximate solution for $x$ after three iterations.

Understanding the Equation

The given equation is $\frac{2}{x+4}=3^z+1$ . To solve this equation, we need to isolate $x$ . However, the equation is not in a straightforward form, and we cannot easily isolate $x$ . Therefore, we will use successive approximations to find an approximate solution for $x$ .

Successive Approximations

Successive approximations involve making an initial guess and then iteratively improving the guess until a solution is obtained. The process involves the following steps:

Make an initial guess for $x$ .
Substitute the initial guess into the equation and solve for $z$ .
Substitute the value of $z$ back into the equation and solve for $x$ .
Repeat steps 2 and 3 until a solution is obtained.

Initial Guess

Let's make an initial guess for $x$ . Since we are not given any specific information about the value of $x$ , we will make a random guess. Let's say our initial guess for $x$ is $x=1$ .

First Iteration

Substitute the initial guess into the equation and solve for $z$ .

\frac{2}{1+4}=3^z+1

Simplifying the equation, we get:

\frac{2}{5}=3^z+1

Subtracting 1 from both sides, we get:

\frac{2}{5}-1=3^z

Simplifying further, we get:

-\frac{3}{5}=3^z

Taking the logarithm of both sides, we get:

\log(-\frac{3}{5})=z\log(3)

Solving for $z$ , we get:

z=\frac{\log(-\frac{3}{5})}{\log(3)}

Substituting the value of $z$ back into the equation, we get:

\frac{2}{x+4}=3^{\frac{\log(-\frac{3}{5})}{\log(3)}}+1

Simplifying the equation, we get:

\frac{2}{x+4}=-0.051+1

Simplifying further, we get:

\frac{2}{x+4}=0.949

Multiplying both sides by $x+4$ , we get:

2=0.949(x+4)

Dividing both sides by 0.949, we get:

2.11=x+4

Subtracting 4 from both sides, we get:

-1.89=x

Second Iteration

Substitute the value of $x$ from the first iteration into the equation and solve for $z$ .

\frac{2}{-1.89+4}=3^z+1

Simplifying the equation, we get:

\frac{2}{2.11}=3^z+1

Simplifying further, we get:

0.946=3^z+1

Subtracting 1 from both sides, we get:

-0.054=3^z

Taking the logarithm of both sides, we get:

\log(-0.054)=z\log(3)

Solving for $z$ , we get:

z=\frac{\log(-0.054)}{\log(3)}

Substituting the value of $z$ back into the equation, we get:

\frac{2}{x+4}=3^{\frac{\log(-0.054)}{\log(3)}}+1

Simplifying the equation, we get:

\frac{2}{x+4}=-0.00017+1

Simplifying further, we get:

\frac{2}{x+4}=0.99983

Multiplying both sides by $x+4$ , we get:

2=0.99983(x+4)

Dividing both sides by 0.99983, we get:

2.002=x+4

Subtracting 4 from both sides, we get:

-1.998=x

Third Iteration

Substitute the value of $x$ from the second iteration into the equation and solve for $z$ .

\frac{2}{-1.998+4}=3^z+1

Simplifying the equation, we get:

\frac{2}{2.002}=3^z+1

Simplifying further, we get:

0.999=3^z+1

Subtracting 1 from both sides, we get:

-0.001=3^z

Taking the logarithm of both sides, we get:

\log(-0.001)=z\log(3)

Solving for $z$ , we get:

z=\frac{\log(-0.001)}{\log(3)}

Substituting the value of $z$ back into the equation, we get:

\frac{2}{x+4}=3^{\frac{\log(-0.001)}{\log(3)}}+1

Simplifying the equation, we get:

\frac{2}{x+4}=-0.0000003+1

Simplifying further, we get:

\frac{2}{x+4}=0.9999997

Multiplying both sides by $x+4$ , we get:

2=0.9999997(x+4)

Dividing both sides by 0.9999997, we get:

2.000002=x+4

Subtracting 4 from both sides, we get:

-1.999998=x

Conclusion

After three iterations of successive approximations, we have obtained an approximate solution for $x$ . The approximate solution is $x=-1.999998$ . This solution is very close to the exact solution, which is not obtainable through successive approximations.

Limitations of Successive Approximations

Successive approximations is a powerful method for solving equations, but it has some limitations. One of the main limitations is that it may not converge to the exact solution. In this case, the successive approximations converged to an approximate solution that is very close to the exact solution. However, in some cases, the successive approximations may converge to a different solution or may not converge at all.

Conclusion

Q: What is successive approximations?

A: Successive approximations is a method for solving equations that involves making an initial guess and then iteratively improving the guess until a solution is obtained.

Q: How does successive approximations work?

A: Successive approximations works by substituting the initial guess into the equation and solving for the unknown variable. The value of the unknown variable is then substituted back into the equation, and the process is repeated until a solution is obtained.

Q: What are the advantages of successive approximations?

A: The advantages of successive approximations include:

It is a simple and easy-to-use method for solving equations.
It can be used to solve a wide range of equations, including non-linear equations.
It can be used to find approximate solutions to equations that do not have an exact solution.

Q: What are the disadvantages of successive approximations?

A: The disadvantages of successive approximations include:

It may not converge to the exact solution.
It may converge to a different solution or may not converge at all.
It requires an initial guess, which can be difficult to obtain.

Q: When should I use successive approximations?

A: You should use successive approximations when:

You are trying to solve a non-linear equation.
You are trying to find an approximate solution to an equation that does not have an exact solution.
You are trying to solve an equation that is difficult to solve using other methods.

Q: How do I choose an initial guess for successive approximations?

A: Choosing an initial guess for successive approximations can be difficult. Here are some tips to help you choose a good initial guess:

Start with a simple guess, such as 0 or 1.
Use a guess that is close to the expected solution.
Use a guess that is easy to calculate.

Q: How many iterations should I perform for successive approximations?

A: The number of iterations you should perform for successive approximations depends on the equation you are trying to solve and the desired level of accuracy. Here are some general guidelines:

For simple equations, 1-2 iterations may be sufficient.
For more complex equations, 5-10 iterations may be necessary.
For very complex equations, 20-50 iterations or more may be required.

Q: How do I know when to stop iterating for successive approximations?

A: You can stop iterating for successive approximations when:

The solution converges to a stable value.
The solution converges to a value that is close to the expected solution.
The solution converges to a value that is within a certain tolerance of the expected solution.

Q: What are some common mistakes to avoid when using successive approximations?

A: Here are some common mistakes to avoid when using successive approximations:

Not choosing a good initial guess.
Not iterating enough.
Not checking for convergence.
Not using a stable algorithm.

Q: Can I use successive approximations to solve systems of equations?

A: Yes, you can use successive approximations to solve systems of equations. However, you will need to modify the algorithm to accommodate the multiple unknown variables.

Q: Can I use successive approximations to solve differential equations?

A: Yes, you can use successive approximations to solve differential equations. However, you will need to modify the algorithm to accommodate the differential equation.

Conclusion

In conclusion, successive approximations is a powerful method for solving equations. It involves making an initial guess and then iteratively improving the guess until a solution is obtained. By understanding the advantages and disadvantages of successive approximations, you can use it to solve a wide range of equations, including non-linear equations.