Thread: tough logarithmic equation

1. tough 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?

2. Re: tough logarithmic equation

It can't be solved exactly. Use a numerical method and/or technology to get an approximate answer.

3. 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.

4. 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

5. Re: tough logarithmic equation

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 $\displaystyle x > 0$. If you plot the graphs you will see that they never intersect and so there are no solutions