On Sun, Oct 28, 2012 at 04:05:08PM +0000, Dan wrote:
Hi DAn,
> On Sun, 28 Oct 2012, Daniel Dalton wrote:
>
> >2. I've had a problem with simplification. The function f(x)=log(x) has
> >been defined.
> >Then I must set up the equation:
> >2*f(u)=f(2*v)+f(3*v)
> >Then solve in terms of u:
> >solve(2*f(u)=f(2*v)+f(3*v),u);
> >[u = %e^(log(3*v)/2+log(2*v)/2)]
> >I ratsimp (%) this, but it remains the same.
> >Shouldn't it simplify down to sqrt(6)v?
>
> Using radcan(%), instead of ratsimp(%), will give sqrt(2) sqrt(3) v.
It works, thank you. Any reason why sqrt(2)*sqrt(3) isn't simplified?
Though it is not a problem I can manage this myself.
Also, can I use radcan () as a replacement for ratsimp or is it still
best to use ratsimp when you aren't dealing with exponents and logs?
> >3. I have two integrals, which must equal 3 and 2 respectively. So I
> >assign the integrated equation to f(x) in maxima and attempt to solve
> >for the two integrals like this simultaneously:
> >f(x):=7*log(x);
> >
> >(%o3) f(x):=7*log(x)
> >(%i4) solve([f(m*n)-f(1)=3, f(m/n)-f(1)=2],[m,n]);
> >(%o4) []
> >
> >How do I find the solution to this? (I need an exact value)
>
> I think you said you'd already loaded the to_poly_solve package, in
> which case using the to_poly_solve function instead of the solve
> function should do the trick here.
It works, thank you.
Again should the to_poly_solve function be used instead of solve? Or is
it only useful in some cases?
Thanks for your help.
Daniel