Easy Problem. Can this be done in Maxima?

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.

(%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
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>
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