find exponential function given two points

Question

I know I can solve equations in maxima using the commands below but how do I solve for two different equations.

kill(all);
r:.5; a:1; b:-5.7; theta:theta; solve(a*e^(b*theta)=r,theta);
tex(''%);

I'm trying to get the equation of a exponetial function given two points. How do I go about doing this. Example point 1 is at (2,12) and point 2 is at (8,768)

Answer 1

Maxima needs some help to solve this problem, but it can be done. Start by expression the problem data.

(%i1) [x1,y1]:[2,12];
(%o1)                               [2, 12]
(%i2) [x2,y2]:[8, 768];
(%o2)                              [8, 768]
(%i3) eq1:y1 = a*exp(b*x1);
                                          2 b
(%o3)                            12 = a %e
(%i4) eq2:y2 = a*exp(b*x2);
                                           8 b
(%o4)                            768 = a %e

Now try to solve eq1 and eq2 for a and b.

(%i5) solve([eq1, eq2], [a, b]);
(%o5)                                 []

Hmm, that's unsatisfying. I'm guessing that Maxima could solve it if we take logarithms which make it linear.

(%i6) log([eq1, eq2]);
                                  2 b                      8 b
(%o6)          [log(12) = log(a %e   ), log(768) = log(a %e   )]

Apply the logexpand flag to simplify. Note that % means the previous result.

(%i7) %, logexpand;
                                  2 b                      8 b
(%o7)          [log(12) = log(a %e   ), log(768) = log(a %e   )]

Hmm, that didn't do it. There are different forms of logexpand, try another.

(%i8) %, logexpand=super;
(%o8)          [log(12) = 2 b + log(a), log(768) = 8 b + log(a)]

OK, good. Now try to solve it.

(%i9) solve (%, [a, b]);
(%o9)                                 []

Well, that still didn't work. But I see it's linear in log(a) so solve for that instead.

(%i10) solve (%o8, [log(a), b]);
                     4 log(12) - log(768)        log(12) - log(768)
(%o10)    [[log(a) = --------------------, b = - ------------------]]
                              3                          6

Great. Here are the numerical values:

(%i11) float (%);
(%o11)       [[log(a) = 1.09861228866811, b = 0.6931471805599454]]

I'll try to simplify the exact values.

(%i12) %o10, logexpand=super;
                     4 log(12) - log(768)        log(12) - log(768)
(%o12)    [[log(a) = --------------------, b = - ------------------]]
                              3                          6

Hmm, that didn't work. I'll try another function:

(%i13) radcan(%);
(%o13)                  [[log(a) = log(3), b = log(2)]]

OK, that was a little bit of work, but anyway maybe it helps.