Find the value x

y = x ^ e - e ^ x is [0 5] when x^e=e^x

Answer 1

[Matt J]

>> [x,fval]=fminbnd(@(x) abs(exp(x)-x^exp(1)),0,5)

x =

    2.7183


fval =

   4.4743e-11

[Pichaya Thochampa]

Thank you so much:)

[Matt J]

You're welcome, but please Accept-click whichever answer worked best for you.

[Pichaya Thochampa]

Excuse me, ask me why. When issuing a command of a program that is

>> x = fminbnd (@ (x) (x ^ (exp (1)) - (exp (1) ^ x)), 0,5) 

to get the answer 1.0000, but when using

>> x = fminbnd (@ (x) ((exp (1) ^ x) -x ^ exp (1)), 0,5)

is only 2.7183.

[Matt J]

Because the first version has a local minimum at x=1. (Note: fminbnd may find a local, rather than a global minimum).

fplot(@ (x) (x ^ (exp (1)) - (exp (1) ^ x)), [0,5]);

whereas the second version has a minimum at x=exp(1).

fplot(@ (x) ((exp (1) ^ x) -x ^ exp (1)), [0,5])

Answer 2

[hosein Javan]

you can rewrite the equation in the form of "x^(1/x) = exp(1/exp(1))". this means that if "f(z)=z^(1/z)" then we have "f(x) = f(exp(1))". giving an inverse function, then it is "x=exp(1)".

 syms x
f(x) = x^(1/x) - exp(1/exp(1));
x_ex = solve(f(x),x) % exact value
x_app = vpa(x_ex) % approximation
err_f = subs(f(x),x,x_app) % error

x_ex =

lambertw(0, 51*log(2) - log(3253102820258857))/(51*log(2) - log(3253102820258857))

x_app =

2.7182817878772095927811977987387

err_x =

2.9387358770557187699218413430556e-39

x = 0:0.001:5;
y = x.^exp(1) - exp(x);
plot(x,y)

[Pichaya Thochampa]

Thank you so much:)

[hosein Javan]

don't mention it.

Answer 3

[KSSV]

The value of x would be e i.e exp(1). Check the below:

x = linspace(0,5,10^6) ;    % as you said x lies in [0 5]
f =  x.^exp(1)-exp(x) ; 

[val,idx] = min(abs(f)) ;  % pick the value for which f is minimum i.e f = 0 
x0 = x(idx) ;   % the value of x 
f = x0^exp(1)-exp(x0)   % check 

[Pichaya Thochampa]

Thank you so much:)

Answer 4

[Matt J]

x=exp(1)