This problem so far seems impossible,

$\displaystyle 3x^2=lnx$

so far I have e^(3x^2) =x

and I can't really move forward from there.

any suggestions?

Printable View

- Jun 15th 2011, 01:13 AMeliehtough logarithmic equation
This problem so far seems impossible,

$\displaystyle 3x^2=lnx$

so far I have e^(3x^2) =x

and I can't really move forward from there.

any suggestions? - Jun 15th 2011, 01:31 AMProve ItRe: tough logarithmic equation
It can't be solved exactly. Use a numerical method and/or technology to get an approximate answer.

- Jun 15th 2011, 01:31 AMHallsofIvyRe: tough logarithmic equation
There is not going to be any "algebraic" solution to that equation. That is generally true of equations that involve x both inside and outside a transcendental function.

- Jun 15th 2011, 01:48 AMeliehRe: tough logarithmic equation
Right so I can have e^(3x^2)-x = 0

and guess values which get me closer to 0 till I get the accuracy desired.

Thanks - Jun 15th 2011, 03:31 AMe^(i*pi)Re: tough logarithmic equation