solve for x.
xe^x = 9
please help i am stuck.
i keep going in circles.
Hmmmm....
If the equation was
e^x = 9
I would natural log bothsides
x = ln(9)
thats without the first x in the problem.. I don't know if that helps you or not. Maybe you have to divide that first x to both sides first?
This has been answered elsewhere; it was posted twice
To solve for x in the equation xe^x = 9, we can use a numerical or graphical method to approximate the solution. Here, we will use an iterative method called the Newton-Raphson method.
First, let's define a function f(x) = xe^x - 9. The solution to the equation xe^x = 9 will be the value of x for which f(x) = 0.
To use the Newton-Raphson method, we need to make an initial guess for the value of x. Let's start with x = 1.
1. Calculate f(x) and its derivative f'(x) for the given value of x.
f(x) = xe^x - 9
f'(x) = e^x + xe^x
2. Substitute the value of x into f(x) and f'(x).
f(1) = (1)(e^1) - 9 = e - 9
f'(1) = e^1 + (1)(e^1) = 2e
3. Use the formula of the Newton-Raphson method to update the value of x.
x_new = x - f(x) / f'(x)
x_new = 1 - (e - 9) / (2e)
4. Calculate x_new and repeat steps 2-3 until you get an accurate enough approximation.
For example, let's calculate x_new as follows:
x_new = 1 - (e - 9) / (2e)
= 1 - (2.71828 - 9) / (2 * 2.71828)
= 1 - (-6.28172) / (5.43656)
= 2.15307
5. Repeat steps 2-4 with the updated value of x_new until you reach a desired level of accuracy. For instance, you can keep calculating x_new and updating the value of x until the difference between consecutive approximations is less than a predetermined threshold (e.g., 0.00001).
Using this method, you can continue iterating until you find a value of x that gives f(x) close to zero. In this case, the solution is approximately x ≈ 2.15307.