Romanian Mathematical Olympiad - District Round

We notice that $x > 0$. The given equation can be written as $2^x + 2 \cdot 2^{\frac{1}{\sqrt{x}}} = 6$.

From the AM-GM inequality we have:
$$2^x + 2^{\frac{1}{\sqrt{x}}} + 2^{\frac{1}{\sqrt{x}}} \ge 3 \cdot \sqrt[3]{2^x \cdot 2^{\frac{1}{\sqrt{x}}} \cdot 2^{\frac{1}{\sqrt{x}}}} = 3 \cdot 2^{\frac{1}{3}(x+\frac{2}{\sqrt{x}})}, \forall x > 0.$$
Also, using the AM-GM inequality, we have:
$$\frac{1}{3} \left( x + \frac{2}{\sqrt{x}} \right) \ge \sqrt[3]{x \cdot \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}} = 1,$$
which implies that $2^x + 2 \cdot 2^{\frac{1}{\sqrt{x}}} \ge 6$. Therefore, we need equality in all AM-GM inequalities applied above, which is equivalent to $x = \frac{1}{\sqrt{x}}$, so $x = 1$ is the unique solution of the given equation.