Solve the following equation:

$$
\lg \left(3^{x^{\sqrt{x}}-(\sqrt{x})^{x}+2}+1\right)=1
$$
Since $\lg 10=1$, and the function $y=\lg x$ takes every value in its domain only once, the equation to be solved is equivalent to

$$
3^{x^{\sqrt{x}}-(\sqrt{x})^{x}+2}+1=10
$$

from which, after rearrangement, we get

$$
3^{x^{\sqrt{x}}-(\sqrt{x})^{x}+2}=9=3^{2}
$$

Since the bases are equal in (1), the exponents must also be equal, that is,

$$
x^{\sqrt{x}}-(\sqrt{x})^{x}+2=2
$$

from which

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

The value $x=0$ must be excluded in (2) because $\gg 0^{\circ} \ll$ is not defined. Taking the logarithm of both sides of (2):

$$
\sqrt{x} \lg x=\frac{x}{2} \lg x
$$

After rearrangement and factoring, we get

$$
(\lg x)\left(\frac{x}{2}-\sqrt{x}\right)=0
$$

A product is equal to zero only if one of its factors is zero, so either

$$
\lg x=0
$$

from which

$$
x_{1}=1
$$

or

$$
\frac{x}{2}-\sqrt{x}=0
$$

from which, dividing by $\sqrt{x} \neq 0$ (which we can do since we have excluded the value $x=0$),

$$
\frac{\sqrt{x}}{2}=1
$$

and thus

$$
x_{2}=4
$$

By substitution, we can verify that the two roots we have found satisfy our equation.