Question about the solve function

Let's say I have:

A = [ 1 2 ; 3 4 ]

e1='c*A(1,1)^b=A(1,2)'
e2='c*A(2,1)^b=A(2,2)'

I'm trying to use "solve" to find c and b: solve(e1,e2)

The answer looks like: [ log(A(1, 2)/A(2, 2))/(log(A(1, 1)) - log(A(2, 1))), A(2, 2)/A(2, 1)^(log(A(1, 2)/A(2, 2))/(log(A(1, 1)) - log(A(2, 1))))]

I want it to look like: 0.6309 2.0000

Is "solve" the wrong function for this? How do I do to get a numerical answer?

Best Regards Peter

Answer 1

[Sean de Wolski]

One way:

S = solve(e1,e2)
eval(S.b)

[Peter]

Thanks!

Answer 2

[Walter Roberson]

syms b c
A = [ 1 2 ; 3 4 ]
e1 = (c*A(1,1)^b) - (A(1,2));
e2 = (c*A(2,1)^b) - (A(2,2));
double(solve(e1, e2))

[Sean de Wolski]

Error using double

Conversion to double from struct is not possible.

[Sean de Wolski]

S = solve(e1,e2)

double(S.b)

double(S.c)

@Peter: Walter's method is the better way to do it!

Answer 3

[Peter]

Next question:

A = [ 1 2 ; 3 4 ; 5 6 ; 7 8 ]

When trying to implement it into a loop:

for k= 1:size(A,1)-1
  S=solve('c*A(k,1)^b=A(k,2)','c*A(k+1,1)^b=A(k+1,2)');
  b=eval(S.b);
  C=eval(S.c);
end

it seems like it handles the variable k as an unknown. How do I solve this?

[Walter Roberson]

Either don't solve() on quoted strings, or learn to use subs()

syms b c

A = [ 1 2 ; 3 4 ; 5 6 ; 7 8 ]

b = zeros(1,size(A,1)-1);

c = b;

for k= 1:size(A,1)-1

S = solve(c*A(k,1)^b-A(k,2),c*A(k+1,1)^b-A(k+1,2));

b(k) = double(S.b);

C(k) = double(S.c);

end