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Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

Determine if there exist a function f:(0;1)→(2018;+∞)f : (0; 1) \rightarrow (2018; +\infty), so that the following conditions hold:
* f(x⋅y)=f(x)⋅f(y)f(x \cdot y) = f(x) \cdot f(y) for any x,y∈(0;1)x, y \in (0; 1);
* for any y∈(2018;+∞)y \in (2018; +\infty) there exists x∈(0;1)x \in (0; 1) such that f(x)=yf(x) = y?

Solution

Suppose such a function exists. Since for any x∈(0;1)x \in (0; 1) equation x=x⋅xx = \sqrt{x} \cdot \sqrt{x} holds, then f(x)=f(x⋅x)=f(x)⋅f(x)>20182f(x) = f(\sqrt{x} \cdot \sqrt{x}) = f(\sqrt{x}) \cdot f(\sqrt{x}) > 2018^2, that contradicts the fact that Ef=(2018,+∞)E_f = (2018, +\infty).

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