- #1

- 8

- 0

## Main Question or Discussion Point

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

- Thread starter crazylum
- Start date

- #1

- 8

- 0

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

- #2

- 1,250

- 2

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

- #3

CRGreathouse

Science Advisor

Homework Helper

- 2,820

- 0

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

- #4

- 119

- 0

- #5

- 14

- 0

- #6

- 79

- 0

[tex]\frac{\mathrm{e}^{2}}{\mathrm{e}^{\omega\!\left(\ln\!\left(\frac{2}{3}\right) + 2\right)}}[/tex]

Where [tex]\omega[/tex] 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.

- Last Post

- Replies
- 2

- Views
- 1K

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 1K

- Replies
- 2

- Views
- 1K

- Replies
- 22

- Views
- 31K

- Last Post

- Replies
- 5

- Views
- 2K

- Replies
- 3

- Views
- 614

- Replies
- 3

- Views
- 1K