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

Barton Willis willisb at unk.edu
Sun Nov 15 16:05:19 CST 2009

```(%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))])

(%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
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