Mark Clements wrote:
The following problem came up on the Maxima mail-list (translated here
into Axiom):
-- solve for x (over the Reals?)
ex := (3/7)^(4*x-5)*(7/3)^(2*x-7)=1
I came up with several overly long solutions (using Fricas 1.0.3):
rule1 := (rule log(7/3)==-log(3/7))
rule1 rhs solve(map(expandLog,map(log,ex)), x).1
-- or, similarly
rule1 rhs solve(expandLog log lhs ex, x).1
rule2 := rule((a/b)^c*(b/a)^d==(a/b)^(c-d))
solve((rule2 lhs ex)=(rhs ex),x)
Can anyone come up with a better solution?
Kindly, Mark Clements.
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a:=(3/7)*(4*x-5)*(7/3)*(2*x-7)-1
b:=zerosOf(a)
+--+ +--+
\|89 + 19, - \|89 + 19
[---------, -----------]
8 8
Type: List AlgebraicNumber
eval(a,x=b.1)
0
eval(a,x=b.2)
0
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