unexpected parameter z3 when using solve command

Hi, I am new to matlab and want to solve an eqution.

syms x
eqn = x-((1-1/0.9925)*0.65-0.003)*x^((1/0.66)/(1/0.97))== 20+((1-1/0.9925)*0.65-0.003)*((1/0.97)^(-0.66)*(1-0.97*(1-0.003)-(1-0.97/0.9925)*0.65)^(-0.66));

Then, the following warning pops up.

Warning: Solutions are parameterized by the symbols: z2. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'. 
> In sym/solve>warnIfParams (line 475)
In sym/solve (line 357)
In Untitled (line 4) 
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'. 
> In sym/solve>warnIfParams (line 478)
In sym/solve (line 357)
In Untitled (line 4) 

I edited my code and ran again, then unexpected parameter z3 came up.

syms x
eqn = x-((1-1/0.9925)*0.65-0.003)*x^((1/0.66)/(1/0.97))== 20+((1-1/0.9925)*0.65-0.003)*((1/0.97)^(-0.66)*(1-0.97*(1-0.003)-(1-0.97/0.9925)*0.65)^(-0.66));
[solx,parameters,conditions] = solve(eqn,x,'ReturnConditions',true)
assume(conditions);
restriction = [100>solx , -100 <solx ];
solz = solve(restriction,parameters)
valx = subs(solx,parameters,solz)

I don't understand what z3 means here and how to get my solution valx. Even the equation inside the root parenthesis seems illogical. Please help me with this.

solx =
 
z3^66
 
 
parameters =
 
z3
 
 
conditions =
 
-pi/66 < angle(z3) & angle(z3) <= pi/66 & z3^97 + (576460752303423488*z3^66)/4560864541524069 - 3822084685500913664/1520288180508023 == 0
 
 
solz =
 
root(z^97 + (576460752303423488*z^66)/4560864541524069 - 3822084685500913664/1520288180508023, z, 1)
 
 
valx =
 
root(z^97 + (576460752303423488*z^66)/4560864541524069 - 3822084685500913664/1520288180508023, z, 1)^66

Answer 1

[Walter Roberson]

Should the 0.66 instead be exactly 2/3 ? Should the 0.65 be 2/3 - 1/100 ?

Should the 0.97 be 1 - 0.003*10 ?

Is there a relationship between the 0.97 and the 0.9925 ?

Q = @(v) sym(v);
F = sym(2)/sym(3);
Fm = F - 1/sym(100);
syms x
eqn = x-((1-1/Q(0.9925))*Fm-Q(0.003))*x^((1/F)/(1/Q(0.97))) == 20+((1-1/Q(0.9925))*Fm-Q(0.003))*((1/Q(0.97))^(-F)*(1-Q(0.97)*(1-Q(0.003))-(1-Q(0.97)/Q(0.9925))*Fm)^Q(-F))

eqn = 

%sol = solve(eqn, x, 'returnconditions', true)
vpa(eqn, 10)

ans = 

vpasolve(eqn, 10)

ans = 

19.295781077501889097831548029764

[Seraphic]

Hi Walter,

no, they are exogenously given. 0.97 has nothing to do with 0.9925 and 0.65 is also exogenously given number (It is a specific ratio, 65%).

[Seraphic]

Hi again Walter,

I just found this post: https://www.mathworks.com/matlabcentral/answers/340330-solve-returns-term-with-z-how-do-i-get-a-real-result

Can I use vpasolve command here?

[Walter Roberson]

Sure, I show using vpasolve here. The Q() calls are there to convert coefficients to rationals because solve() is intended to find exact closed form solutions whenever possible, and converting to rational reduces floating point round off problems. But most importantly if you want to use solve then you need better precision about exactly what you mean by taking a variable to the power of 0.66