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

[Matthias A. S. <mat...@we...>]

Dear Barton,

thanks for your suggestion. I tried to apply the first in 1.0.6 (see below). For aesthetic reasons, formatting issues in the TeXmacs maxima-plugin, I didn’t go for the alternative. Here’s how the program behaves now:

(%i1)  kill(all)$
(%i1)  load("all_exp_eq_solutions_v1.0.6.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: complex_solution(eq=5^(3*x+2)+1 = 7^(3*x+2),x=x) (all_exp_eq_solutions_v1.0.6.mac line 161)
#1: complex(eq=5^(3*x+2)+1 = 7^(3*x+2),x=x) (all_exp_eq_solutions_v1.0.6.mac line 181)
all_solutions: expected LHS to be a power of 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 of a^u. Got LHS = 5^(3*x+2)+1 
(%o18) unsolved

(%i19)  k: %pi/6$
(%i20)  eq1: 5^(3*x+2) = 7^(3*x+2)$
(%i21)  complex_solution(eq1, x);
(%o21) x=-((2*i*π*k)/(3*log (7/5)))-2/3

(%i22)  try_complex(5^(3*x+2)+1 = 7^(3*x+2), x);
all_solutions: expected LHS to be a power of a^u. Got LHS = 5^(3*x+2)+1 
(%o22) unsolved

Cheers, Tilda



> Am 05.03.2026 um 18:44 schrieb Barton Willis <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).
>