We can solve exactly the given equation in terms of the Root objects.

xs = x /. First @ Solve[ x^x^x == 36 && x > 0, x]

Root[{-36 + #1^#1^#1 & ,
       2.10703646395674928520879246974369419433`20.601147030102787}]

This solution is not an algebraic number and cannot be represented as a root of polynomial with rational coefficients. Nonetheless solutions of transcendental equations represented with Root objects are exact as well even though they cannot be represented in terms of radicals. For more detailed discussion of Root objects see e.g. How do I work with Root objects?

Our solution xs can be numerically approximated with arbitrary precision, e.g.

N[ xs, 50]

2.1070364639567492852140489758794729541219024075415

We can also find an algebraic approximation of the solution xs, e.g. in terms of a root of third order polynomial:

RootApproximant[xs, 3] // InputForm

Root[-9946 + 4127*#1 - 262*#1^2 + 258*#1^3 & , 1, 0] 

ToRadicals @ %

1/387 (131 + (3850120229 + 15093 Sqrt[98585357943])^(1/3)/2^(2/3) - 
      1562827/(2 (3850120229 + 15093 Sqrt[98585357943]))^(1/3))

qiute good approximation

N[{xs, RootApproximant[xs, 3]}, 17]

 {2.1070364639567493, 2.1070364639567491}

Let's represent the solution graphically:

Plot[ -Log[36] + Log[x] x^x, {x, 0, 2.5}, PlotStyle -> Thick]

Warning

Equation x^x^x == 36 can be rewritten equivalently as Exp[x Log[x]] Log[x] - Log[36] == 0. Solving it with Reduce or Solve yields another Root object which should be the same solution even though Mathematica might have difficulties with automatic demonstrating that they are mathematically equal.

Symbolic functions like Reduce and Solve provide an exact solution of the given equation in terms of the Root objects.

xs = x /. First @ Solve[ x^x^x == 36 && x > 0, x]

Root[{-36 + #1^#1^#1 & ,
       2.10703646395674928520879246974369419433`20.601147030102787}]

This is not an algebraic number and cannot be represented as a root of polynomial with rational coefficients. Nonetheless solutions of transcendental equations represented with Root objects are exact as well even though they cannot be represented in terms of radicals. For more detailed discussion of Root objects see e.g. How do I work with Root objects?

Our solution xs can be numerically approximated with arbitrary precision, e.g.

N[ xs, 50]

2.1070364639567492852140489758794729541219024075415

We can also find an algebraic approximation of the solution xs with RootApproximant, e.g. in terms of a root of third order polynomial:

RootApproximant[xs, 3] // InputForm

Root[-9946 + 4127*#1 - 262*#1^2 + 258*#1^3 & , 1, 0] 

ToRadicals @ %

1/387 (131 + (3850120229 + 15093 Sqrt[98585357943])^(1/3)/2^(2/3) - 
      1562827/(2 (3850120229 + 15093 Sqrt[98585357943]))^(1/3))

this is quite a good approximation

N[{xs, RootApproximant[xs, 3]}, 17]

 {2.1070364639567493, 2.1070364639567491}

Let's represent the solution graphically:

Plot[ -Log[36] + Log[x] x^x, {x, 0, 2.5}, PlotStyle -> Thick]

Warning

Equation x^x^x == 36 can be rewritten equivalently as Exp[x Log[x]] Log[x] - Log[36] == 0. Solving it with Reduce or Solve yields another Root object which is the same solution even though Mathematica does not demonstrate automatically that they are mathematically equal.