I'm trying to use solve() to get two variables for a power function.
It looks to me, according to the documentation, that I can do the following:
solve ([b*8^a=6,b*18^a=2], [a,b]);
I use the following notation for a power function:
f(x) = b * x^a
Let's say that I have two points (8, 6) and (18, 2) and the graph goes through both points.
In wxMaxima I tried to do:
%i1 solve ([b*8^a=6,b*18^a=2], [a,b]);
%o1 []
Now, I know that I can use define formulas to find a and b, but that is rather long compared to the above. Am I missing some setting for solve()?
Below is my current way to get to get the calculation rule for a power function:
findA: a=(log(y_2/y_1)/log(x_2/x_1));
findB: b = y_1/x_1^a;
[x_1=8,y_1=6,x_2=18,y_2=2];
subst (%, findA), numer;
a=-1.354755645675727
subst ([x_1=8,y_1=6,x_2=18,y_2=2,a=-1.354755645675727], findB), numer;
b=100.37313960722
I'll use the log function to transform the equations into something solve can handle.
(%i2) e: [b*8^a=6, b*18^a=2];
a a
(%o2) [8 b = 6, 18 b = 2]
(%i3) log(e);
a a
(%o3) [log(8 b) = log(6), log(18 b) = log(2)]
I need to encourage Maxima to simplify log further.
(%i4) log(e), logexpand = true;
a a
(%o4) [log(8 b) = log(6), log(18 b) = log(2)]
Hmm, need to try still harder.
(%i5) log(e), logexpand = super;
(%o5) [log(b) + log(8) a = log(6), log(b) + log(18) a = log(2)]
Okay, that looks solvable. I'll try solve again.
(%i6) solve (%o5, [a, b]);
(%o6) []
Oh, I see it's a function of log(b). I can tell solve to consider log(b) as a variable to solve for.
(%i7) solve (%o5, [a, log(b)]);
log(6) - log(2)
(%o7) [[a = ----------------, log(b) =
log(8) - log(18)
log(2) log(8) - log(6) log(18)
------------------------------]]
log(8) - log(18)
Great, I got a solution, now I'll look at the numerical value.
(%i8) float (%);
(%o8) [[a = - 1.354755645675727, log(b) = 4.608894637671451]]