# Where to start?

1. May 18, 2008

### crazylum

Where does one begin to solve the equation 3log(x)=6-2x?

2. May 18, 2008

### Crosson

The solution to that equation is not expressible in terms of elementary functions.

3. May 18, 2008

### CRGreathouse

Looks like material for Lambert's W (or Newton's method).

4. May 18, 2008

### exk

you could use a numerical approach like newton's method to find a solution. if I am not mistaken you can say that x>0 since ln(x) is undefined otherwise and then rewrite your equation as $x^{3}e^{2x}-e^{6}=0$

5. Mar 15, 2010

### DV10

an approximate solution is possible through pt of intersection of the 2 graphs...but i dont think theres any way to find an exact solution save hit n trial

6. Mar 15, 2010

### clustro

MATLAB's solver gave:

$$\frac{\mathrm{e}^{2}}{\mathrm{e}^{\omega\!\left(\ln\!\left(\frac{2}{3}\right) + 2\right)}}$$

Where $$\omega$$ is given by: http://en.wikipedia.org/wiki/Wright_Omega_function.

Good call by whoever said it needed the Lambert W-function ^_^

But for practical purposes, just use bisection+newton.