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

- Thread starter crazylum
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- #1

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

- #2

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

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CRGreathouse

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

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- #6

clustro

[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.

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