How to solve non_linear equation

Answer 1

[Star Strider]

Solve it symbolically —

syms z 
Eqn = z^3 == log(z)*(482036/0.18525)^5

Eqn = 

Z = solve(Eqn)

Z = 

Z = vpa(Z)

Z = 

format longG
Zd = double(Z)

Zd = 
          -76151096277.1022 +      123912303259.595i
           145274860824.135 +                     0i
                          1 +                     0i
          -76151096277.1022 -      123912303259.595i

.

[Torsten]

Why is z = 1 considered a solution ?

[Star Strider]

No idea. Apparently 1 = 0 in this instance —

syms z 
Eqn = z^3 == log(z)*(482036/0.18525)^5

Eqn = 

Z3 = subs(Eqn,z,1)

Z3 = 

I didn’t catch that earlier.

.

[Sam Chak]

@Star Strider, @Torsten, Wolfram Alpha also returned the perfect "1" as one of the solutions. But we all know that . Maybe that's merely an approximation because ?

syms z
f = (z^3)/(482036/0.18525)^5;
limit(f, z, 1)

ans = 

double(ans)

ans = 8.3829e-33

Plot:

z   = linspace(0.9, 1.1, 20001);
y1  = z.^3;
y2  = log(z)*(482036/0.18525)^5;
plot(z, [y1; y2]), grid on, ylim([0 2])

[Sam Chak]

I guess both MATLAB and Wolfram Alpha analytically computed the solution:

c   = (4820360/0.018525)^7;   % constant
sol = exp(-lambertw(-3/c)/3)

sol = 1

However, I mathematically believe that this is just an approximation with the real solution very close to being 1.