[Maxima] Easy Problem. Can this be done in Maxima?

Stavros Macrakis macrakis at alum.mit.edu
Sun Nov 15 17:15:38 CST 2009


On Sun, Nov 15, 2009 at 6:02 PM, Richard Hennessy <rich.hennessy at verizon.net
> wrote:

> This is far from obvious.  I guess there are infinitely many complex
> solutions.  The test was multiple choice with only real answers to pick
> from.
>
> I came up with a third way.
>

Not really a third way --  map(log,ex) has the same effect as
log(lhs(ex))=log(rhs(ex))

>
>
(%i1) solve((3/7)^(4*x-5)*(7/3)^(2*x-7)=1,x);
> (out1) [3^(2*x+2) = 7^(2*x+2)]
> (%i2) log(rhs(%[1]))=log(lhs(%[1]));
> (out2) log(7)*(2*x+2) = log(3)*(2*x+2)
> (%i3) solve(%,x);
> (out3) [x = -1]
>
> It's very hard to get my head around this type of problem, but Maxima does
> not try step 2.
>
> Rich
>
>
> ----- Original Message -----
> From: "Barton Willis" <willisb at unk.edu>
> To: "Richard Hennessy" <rich.hennessy at verizon.net>
> Cc: "Maxima List" <maxima at math.utexas.edu>; <
> maxima-bounces at math.utexas.edu>
> Sent: Sunday, November 15, 2009 5:05 PM
> Subject: Re: [Maxima] Easy Problem. Can this be done in Maxima?
>
>
> > (%i57) load(to_poly_solver)$
> >
> > (%i58)  nicedummies(to_poly_solve((3/7)^(4*x-5)*(7/3)^(2*x-7)=1,x));
> > (%o58) %union([x=(2*%i*%pi*%z0+log(9/49))/(log(49)-log(9))])
> >
> > (%i59) subst(%z0=0,%);
> >
> > (%o59) %union([x=log(9/49)/(log(49)-log(9))])
> >
> > (%i60) radcan(logcontract(%));
> > (%o60) %union([x=-1])
> >
> > If solve means solve over the reals, then the solution set is {x = -1}.
> >
> > Barton
> >
> > maxima-bounces at math.utexas.edu wrote on 11/15/2009 03:41:41 PM:
> >
> >> [image removed]
> >>
> >> [Maxima] Easy Problem. Can this be done in Maxima?
> >>
> >> Richard Hennessy
> >>
> >> to:
> >>
> >> Maxima List
> >>
> >> 11/15/2009 03:43 PM
> >>
> >> Sent by:
> >>
> >> maxima-bounces at math.utexas.edu
> >>
> >> Hi List,
> >>
> >> I recently took a placement test with this problem in it.  It was
> >> very easy to solve but after getting home I suspected Maxima could
> >> not do it.  At least not directly.
> >>
> >> (%i1) solve((3/7)^(4*x-5)*(7/3)^(2*x-7)=1,x);
> >>
> >> (%o1) [3^(2*x+2) = 7^(2*x+2)]
> >>
> >> Is there an easy way Maxima can do it?  The answer is -1.
> >>
> >> Rich
> >>
> >>  _______________________________________________
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> >> Maxima at math.utexas.edu
> >> http://www.math.utexas.edu/mailman/listinfo/maxima
> >
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> >
>
>
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