Rewrite the bases 9 and 27 as powers of 3, and express the square root on the right side as a fractional exponent.

We know that $9 = 3^2$ and $27 = 3^3$.
Also, $\sqrt{27^x} = (27^x)^{1/2}$.

Substitute these into the equation:

$$(3^2)^{\sqrt{x}} = ((3^3)^x)^{1/2}$$

$$3^{2\sqrt{x}} = (3^{3x})^{1/2}$$ $$3^{2\sqrt{x}} = 3^{\frac{3x}{2}}$$

Multiply both sides by 2:

$$4\sqrt{x} = 3x$$

Square both sides to eliminate the square root:

$$(4\sqrt{x})^2 = (3x)^2$$ $$16x = 9x^2$$

Rearrange the equation to form a quadratic equation:

$$9x^2 - 16x = 0$$

Factor out $x$:

$$x(9x - 16) = 0$$

This gives two possible solutions:

$$x = 0 \quad \text{or} \quad 9x - 16 = 0$$

If $9x - 16 = 0$:

$$9x = 16$$ $$x = \frac{16}{9}$$

For $x=0$:
$9^{\sqrt{0}} = \sqrt{27^0}$
$9^0 = \sqrt{1}$
$1 = 1$
So, $x=0$ is a valid solution.

For $x=\frac{16}{9}$:
Left side: $9^{\sqrt{\frac{16}{9}}} = 9^{\frac{4}{3}} = (3^2)^{\frac{4}{3}} = 3^{\frac{8}{3}}$
Right side: $\sqrt{27^{\frac{16}{9}}} = (27^{\frac{16}{9}})^{1/2} = ((3^3)^{\frac{16}{9}})^{1/2} = (3^{3 \cdot \frac{16}{9}})^{1/2} = (3^{\frac{16}{3}})^{1/2} = 3^{\frac{16}{3} \cdot \frac{1}{2}} = 3^{\frac{8}{3}}$
Since $3^{\frac{8}{3}} = 3^{\frac{8}{3}}$, $x=\frac{16}{9}$ is also a valid solution.

The values of $x$ are:

$$x = 0, \frac{16}{9}$$