Where to start?

[crazylum]

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

 

[Crosson]

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

 

[CRGreathouse]

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

 

[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 $x^{3}e^{2x}-e^{6}=0$

 

[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

 

[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.