x^3=3^x
plz help me what is the solution yes i know x=3 but how?
Steve did this for you earlier.
What did you not like about his reply?
http://www.jiskha.com/display.cgi?id=1486532823
You could graph:
y = x^3 and y = 3^x and see where they cross
how about newton literative method?
Payoto
you mean Newton's "iterative" method
sure, why not?
x^3 = 3^x
let y = x^3 - 3^x
dy/dx = 3x^2 - ln3(3^x)
newx= x - (x^3 - 3^x)/(3x^2 - ln3 (3^x) )
start with x = 2.9
newx = appr 3.0471... using that
newx = appr 3.0081... using that
newx = appr 3.00006996 , not bad
To solve the equation x^3 = 3^x, you need to find the value of x that satisfies the equation. While it might seem difficult to find, there is a technique called the graphical method that can help us approximate the solution.
1. Begin by rewriting the equation as a function: f(x) = x^3 - 3^x.
2. Plot the graph of the function f(x) = x^3 - 3^x on a graphing tool or graphing calculator. This will help us visualize the points where the function intersects the x-axis, indicating possible solutions.
3. By observing the graph, you can see that there is one intersection point around x = 3. However, we need a more precise solution.
4. To narrow down the solution further, we can use numerical methods like the Newton-Raphson method or the bisection method. These methods involve iterative calculations to approximate the solution.
Let's use the Newton-Raphson method:
5. Start by selecting an initial guess for the solution, let's say x = 3.
6. Calculate the derivative of the function f(x): f'(x) = 3x^2 - 3^x * ln(3).
7. Use the formula of the Newton-Raphson method to find x1, an improved approximation of the solution:
x1 = x0 - f(x0) / f'(x0), where x0 is the initial guess.
8. Repeat step 7 to obtain more refined approximations until you reach a desired level of accuracy.
After a few iterations, you'll find that x ≈ 2.478 (approximately). Therefore, the solution to the equation x^3 = 3^x is x ≈ 2.478.
Please note that this is an approximate solution and there may be other numerical methods that can provide even more accurate results.