How do I find x for this equation? I think I should get e^(something).
Thanks in advance!
3x^2+4lnx+4=0
There's no algebraic way to do it. A numeric method or a graphical analysis will be required. See
http://www.wolframalpha.com/input/?i=solve+3x^2%2B4lnx%2B4%3D0
For a complicated equation such as yours, the most powerful method is probably Newton's Method.
let y = 3x^2+4lnx+4
dy/dx = 6x + 4/x
newx = x - y/(dy/dx)
= x - (3x^2+4lnx+4)/(6x + 4/x)
which simplifies to
newx = (3x^2 - 4lnx)/(6x + 4/x)
now use an iteration process, I started with x = 1
x | newx
1 -- .3
.3 -- .336072..
.336072 -- .337716
.337716 -- .337719143
.337719143 -- .337719143
since the last x used returned the same value of x, we cannot improve on that answer.
Amazing that after only 4 interations, I got
x =.337719143 correct to 9 decimal places
To find the value of x in the equation 3x^2 + 4ln(x) + 4 = 0, you can use a numerical method called the Newton-Raphson method or an approximate method called the graphical method.
The Newton-Raphson method involves recursively approximating the solution by improving an initial guess. Here's how you can use it:
1. Start by finding the derivative of the equation with respect to x. In this case, the derivative of 3x^2 + 4ln(x) + 4 is 6x + 4/x.
2. Make an initial guess for the value of x. Let's say x₀ is your initial guess.
3. Use the Newton-Raphson formula to improve the approximation:
x₁ = x₀ - (3x₀^2 + 4ln(x₀) + 4) / (6x₀ + 4/x₀)
4. Repeat step 3 until you reach an approximation of x that is accurate enough for your requirements.
Alternatively, if you expect the solution to be of the form e^something, you can try a graphical approach:
1. Rewrite the equation: 3x^2 + 4ln(x) + 4 = 0 as ln(x) = -(3x^2 + 4) / 4.
2. Plot the graph of y = ln(x) and y = -(3x^2 + 4) / 4 on the same graphing calculator or software.
3. Find the point(s) of intersection between the two graphs. These points correspond to the solution(s) of the equation.
Once you find the value of x using either method, you can check if it satisfies the equation by substituting it back into the original equation.