# tough logarithmic equation

• Jun 15th 2011, 02:13 AM
elieh
tough logarithmic equation
This problem so far seems impossible,

$3x^2=lnx$

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

and I can't really move forward from there.
any suggestions?
• Jun 15th 2011, 02:31 AM
Prove It
Re: tough logarithmic equation
It can't be solved exactly. Use a numerical method and/or technology to get an approximate answer.
• Jun 15th 2011, 02:31 AM
HallsofIvy
Re: 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, 02:48 AM
elieh
Re: 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, 04:31 AM
e^(i*pi)
Re: tough logarithmic equation
Quote:

Originally Posted by elieh
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

Pretty much although don't forget that $x > 0$. If you plot the graphs you will see that they never intersect and so there are no solutions