Easy Problem. Can this be done in Maxima?

"Not really a third way"

True and you don't have to call solve twice like I did.  Sorry.

Rich

  ----- Original Message ----- 
  From: Stavros Macrakis 
  To: Richard Hennessy 
  Cc: Barton Willis ; Maxima List ; maxima-bounces at math.utexas.edu 
  Sent: Sunday, November 15, 2009 6:15 PM
  Subject: Re: [Maxima] Easy Problem. Can this be done in Maxima?





  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
    >>
    >>  _______________________________________________
    >> Maxima mailing list
    >> Maxima at math.utexas.edu
    >> http://www.math.utexas.edu/mailman/listinfo/maxima
    >
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    >


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