How can I solve this equation in which variable is in power

I tried this but getting wrong solution

syms a b c d

eqn1 = 2.1544 == a*(5.95^b) + c*(5.95^d);

eqn2 = 1.71 == a*(6.72^b) + c*(6.72^d);

eqn3 = 1.99 == a*(6.35^b) + c*(6.35^d);

eqn4 = 1.59 == a*(6.98^b) + c*(6.98^d);

eqns = [eqn1 eqn2 eqn3 eqn4];

vars = [a b c d];

R = vpasolve(eqns,vars);

[AJMIT KUMAR]

I got these values by using vpasolve which is wrong (Put back these value to equation, it will not satisfy)

a = -4.5917748078995605780028770985244e-41;

b = 98.395408219472484524557308589775;

c = 61.139888401602841212761952348352;

d = 23.111701126115183488493226207409;

Answer 1

[Star Strider]

One option is to use GlobalSearch:

y = [2.1544; 1.71; 1.99; 1.59];
rhs = @(a,b,c,d) [a*(5.95.^b) + c.*(5.95.^d); a.*(6.72.^b) + c.*(6.72.^d); a.*(6.35.^b) + c.*(6.35.^d); a.*(6.98.^b) + c.*(6.98.^d)];

gs = GlobalSearch;
fcn = @(b) norm(y - rhs(b(1),b(2),b(3),b(4)));
Problem = createOptimProblem('fmincon', 'x0',randn(4,1), 'objective',fcn)
B = run(gs, Problem)
fval = fcn(B)

producing (with reasonable consistency):

B =
  67.619440740979044
  -1.924836292732742
  81.251260559584935
 -43.549761874231805

and:

fval =
   0.073738107681070

That is likely as good as it can get, although other values may be possible, with:

B =
  46.932805349518709
 -15.145390708060569
  67.619902371215474
  -1.924839968141592

producing approximately the same value for ‘fval’.

Note that this is not a particularly well-posed problem, so the individual parameter values will vary, and there could be a wide range of possible solutions.

[AJMIT KUMAR]

@Star Strider Thanks for above code, I got lesser fval by running above matlab code for many times.

Kindly tell me how to fix the condition(Range) for values of a,b,c,d like matlab find a value in such a way that b & d must come negative (define a range like 0 to -5) and a & c must come positive (range like 0 to 5000)

[Star Strider]

As always, my pleasure!

See the documentation for fmincon and GlobalSearch to set the parameter limits.

This sets the upper and lower bounds on the estimated parameters:

y = [2.1544; 1.71; 1.99; 1.59];
rhs = @(a,b,c,d) [a*(5.95.^b) + c.*(5.95.^d); a.*(6.72.^b) + c.*(6.72.^d); a.*(6.35.^b) + c.*(6.35.^d); a.*(6.98.^b) + c.*(6.98.^d)];

gs = GlobalSearch;
fcn = @(b) norm(y - rhs(b(1),b(2),b(3),b(4)));
Problem = createOptimProblem('fmincon', 'x0',randn(4,1), 'objective',fcn, 'ub',[Inf 0 Inf -5], 'lb',[0 -Inf 0 -Inf])
B = run(gs, Problem)
fval = fcn(B)

Change the bounds to get the result you want.

One result using that was:

B =
  67.619855188755324
  -1.924839592284333
  41.807642742258828
 -46.760550636559877

fval =
   0.073738107644516

.

[AJMIT KUMAR]

@ Star Strider Many many thanks to you, I got the values as per my requirement. There is so much to learn about Matlab function.

[Star Strider]

As always, my pleasure!

Quite definitely, expecially since new functions continue to be introduced, and older ones are upgraded.