Re: [Maxima-discuss] Why can't Maxima solve 5^(3*x+2) = 7^(3*x+2) symbolically? PEBCAK or a limitation?

[Jaime V. <vi...@fe...>]

Hello Barton,

Very nice solution. I have a couple of comments:


1. If one uses radcan right after %o10, me can see the "naive" solution 
x=-2/3


(%o10) %union([x = -(log(49/25)/log(343/125))])
(%i11) radcan(%);
(%o11) %union([x = -(2/3)])


2. I wish we also had an "equation_convert" that would transform the 
original equation (eq1) into another equation (eq3) that %solve can 
solve right away, like this:


(%i1) eq1: 5^(3*x+2) = 7^(3*x+2)$
(%i2) eq3: lhs(eq1)/rhs(eq1)=1$
(%i3) load(to_poly_solve)$
(%i4) %solve(eq3,x);
(%o4) %union([x = (2*%i*%pi*%z802+log(25/49))/log(343/125)])

Regards,

Jaime


On 3/3/26 22:54, Barton Willis via Maxima-discuss wrote:
> Here is my solution—I'll look at your solution in a bit.  To try this 
> for yourself,  you'll need a recent version of function_convert 
> <https://github.com/barton-willis/function_convert>
>
> (%i1) load(function_convert)$
>
> Define a converter that standardizes the base of an exponential to 
> %e.   The repeated "^" = standardize_base looks weird --- the second 
> is an alias for the first:
>
> (%i2) register_converter("^" = standardize_base, "^" = 
> standardize_base, lambda([a,b], exp(b*log(a))));
> (%o2) done
>
> Standardize the base to %e:
>
> (%i3) eq1: 5^(3*x+2) = 7^(3*x+2)$
>
> (%i4) eq2 : function_convert("^" = standardize_base, eq1);
> (eq2) %e^(log(5)*(3*x+2))=%e^(log(7)*(3*x+2))
>
> And solve with to_poly_solve:
>
> (%i5) load(to_poly_solve)$
>
> (%i8) sol : %solve(eq2,x);
> (sol) %union([x=-((2*%i*%pi*%z927+log(49/25))/log(343/125))])
>
> Reset the pesky counter to zero:
>
> (%i9) sol : nicedummies(sol);
> (sol) %union([x=-((2*%i*%pi*%z0+log(49/25))/log(343/125))])
>
> Let's find a real solution --- we need to use carbon-based computing 
> (our brains) to see that we need to set %z0 to zero:
>
> (%i10)      solX : subst(%z0=0,sol);
> (solX)      %union([x=-(log(49/25)/log(343/125))])
>
> Let's check --- always check:
>
> (%i11)      subst(first(solX),eq1);
> (%o11)      5^(2-(3*log(49/25))/log(343/125))=7^(2-(3*log(49/25))/log(343/125))
>
> (%i12)      radcan(%);
> (%o12)      1=1
>
> There are light-weight methods to standardize the base to %e, using 
> function_convert isn't needed.  Something like  subst("^" =  
> lambda([a,b], exp(b*log(a)), xxx)) will work.
>
> Maybe I will find a new syntax for register_converter that allows the 
> alias to be optional.  Also, for now , the third argument to 
>  register_converter must be a Maxima lambda form. I think I can insert 
> a few lines of code and allow the last argument to be a Maxima 
> function.  Maybe I'll try to do that too.
>
> --Barton
>
>
>
>
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