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

[Barton W. <wi...@un...>]

I think your code should quote k, kk, kkk  and unsolved.  Otherwise:

%i5) k : %pi/6$

(%i6) eq1 : 5^(3*x+2) = 7^(3*x+2);
(eq1) 5^(3*x+2)=7^(3*x+2)

(%i7) complex_solution(eq1, x);
(%o7) x=-((%i*%pi^2)/(9*log(7/5)))-2/3

Alternatively, you might like to consider using a gensym variable (see user documentation).

________________________________
From: Matthias A.Steiner <mat...@we...>
Sent: Wednesday, March 4, 2026 11:26 PM
To: Barton Willis <wi...@un...>; Max...@li... <Max...@li...>
Subject: Re: [Maxima-discuss] Why can't Maxima solve 5^(3*x+2) = 7^(3*x+2) symbolically? PEBCAK or a limitation?

Caution: Non-NU Email

Dear Barton,

Thanks for your spot-on comments, which I tried to apply in v1.0.2. I’ve attached the file below. Here's what it does.

(%i1)  kill(all)$
(%i1)  load("all_exp_eq_solutions_v1.0.2.mac")$

(%i2)  eq1: 5^(3*x+2) = 7^(3*x+2)$
(%i3)  sols1: all_solutions(eq1, x)$
(%i4)  real(eq1, x);
(%o4) x=-(2/3)
(%i5)  complex(eq1, x);
(%o5) x=-((2*i*π*k)/(3*log (7/5)))-2/3
(%i6)  complex_report(eq1, x);
(%o6) [x=-((2*i*π*k+2*log (7)-2*log (5))/(3*log (7)-3*log (5))),x=-((2*i*π*k)/(3*log (7/5)))-2/3]

(%i7)  eq2: 5^(3*k+2) = 7^(2*k+2)$
(%i8)  sols2: all_solutions(eq2, k)$
(%i9)  real_solution(eq2, k);
(%o9) k=-((2*log (7)-2*log (5))/(2*log (7)-3*log (5)))
(%i10)  complex(eq2, k);
(%o10) k=-((2*i*π*kk+2*log (7)-2*log (5))/(2*log (7)-3*log (5)))
(%i11)  complex_report(5^(3*k+2) = 7^(3*k+2), k);
%o11) [k=-((2*i*π*kk+2*log (7)-2*log (5))/(3*log (7)-3*log (5))),k=-((2*i*π*kk)/(3*log (7/5)))-2/3]

(%i12)  eq3: 5^(3*x+2) = 7^(2*x+1)$
(%i13)  sols3: all_solutions(eq3, x)$
(%i14)  real_solution(eq3, x);
(%o14) x=-((log (7)-2*log (5))/(2*log (7)-3*log (5)))
(%i15)  complex_solution(eq3, x);
(%o15) x=-((2*i*π*k+log (7)-2*log (5))/(2*log (7)-3*log (5)))
(%i16)  complex_report(eq3, x);
(%o16) x=-((2*i*π*k+log (7)-2*log (5))/(2*log (7)-3*log (5)))

(%i17)  complex(5^(3*x+2) + 1 = 7^(3*x+2), x);
#0: all_solutions_k(eq=5^(3*x+2)+1 = 7^(3*x+2),x=x,ksym=k) (all_exp_eq_solutions_v1.0.2.mac line 95)
#1: all_solutions(eq=5^(3*x+2)+1 = 7^(3*x+2),x=x) (all_exp_eq_solutions_v1.0.2.mac line 121)
#2: complex_solution(eq=5^(3*x+2)+1 = 7^(3*x+2),x=x) (all_exp_eq_solutions_v1.0.2.mac line 141)
all_solutions: expected LHS to be a power a^u. Got LHS =  5^(3*x+2)+1
 -- an error. To debug this try: debugmode(true);

(%i18)  try_complex(5^(3*x+2) + 1 = 7^(3*x+2), x);
all_solutions: expected LHS to be a power a^u. Got LHS =  5^(3*x+2)+1
(%o18) unsolved

Any suggestions or comments are more than welcome.

Cheers, Tilda



Am 04.03.2026 um 16:03 schrieb Barton Willis <wi...@un...>:

I tried the code in  all_exp_eq_solutions_v0.2.6.mac. Some comments:

When the solve variable is k, the result is an expression that has k with two meanings:

(%i7) complex(5^(3*k+2) = 7^(3*k+2),k);
(%o7) k=-((2*%i*%pi*k)/(3*log(7/5)))-2/3

When the base of the exponential isn't a constant, the result isn't wrong, but the equation isn't solved:

(%i8) complex(x^(3*x+2) = 7^(3*x+2),x);
(%o8) x=-((2*%i*%pi*k)/(3*log(7/x)))-2/3

The code assumes a very specific set of equations;  when  it isn't, the user gets a hard to understand error message:

(%i11)      complex(5^(3*x+2) + 1 = 7^(3*x+2),x);
pow_parts: expected a power a^e, got:  5^(3*x+2)+1

The function complex_report  uses simp as a local variable. But simp is a Maxima option variable—when it is false, Maxima doesn't simplify expressions, and much of Maxima will not work as expected.  Likely, you should name this local variable something else.

I hope these comments might help make your code better -- please keep us up on your efforts.

--Barton