Solve the following equation:

$$
\left(\frac{x}{2}\right)^{\lg x}=40000
$$
I. solution: Two positive numbers are equal if and only if their logarithms are equal. We take the base-10 logarithm of both sides:

$$
\begin{aligned}
& \lg x(\lg x-\lg 2)=4+\lg 4 \\
& (\lg x)^{2}-\lg 2 \cdot \lg x-(4+2 \lg 2)=0
\end{aligned}
$$

This is a quadratic equation in $\lg x$, from which

$$
\begin{aligned}
& \lg x=\frac{\lg 2 \pm \sqrt{(\lg 2)^{2}+8 \lg 2+16}}{2}=\frac{\lg 2 \pm(\lg 2+4)}{2} \\
& \lg x_{1}=\frac{2 \lg 2+4}{2}=\lg 2+2 \\
& \lg x_{2}=\frac{-4}{2}=-2
\end{aligned}
$$

Thus,

$$
\begin{aligned}
& x_{1}=10^{2+\lg 2}=10^{2} \cdot 10^{\lg 2}=100 \cdot 2=200 \\
& x_{2}=10^{-2}=\frac{1}{100}
\end{aligned}
$$

II. solution: We transform the quadratic equation in $\lg x$:

$$
\begin{aligned}
& (\lg x)^{2}-4-\lg 2 \cdot \lg x-2 \lg 2=0 \\
& (\lg x+2)(\lg x-2)-\lg 2(\lg x+2)=0 \\
& \text { Factoring out } (\lg x+2) :
\end{aligned}
$$

$$
(\lg x+2)(\lg x-2-\lg 2)=0
$$

A product is zero only if one of its factors is zero. Thus,

$$
\lg x_{1}=2+\lg 2, \quad \text { from which } x_{1}=200
$$

or

$$
\lg x_{2}=-2, \quad \text { from which } x_{2}=\frac{1}{100}
$$

III. solution: Let's substitute $10^{y}$ for $x$.

$$
\begin{aligned}
\left(\frac{10^{y}}{2}\right)^{y} & =4 \cdot 10^{4} \\
\frac{10^{y^{2}}}{2^{y}} & =2^{2} \cdot 10^{4} \\
\frac{10^{y^{2}}}{10^{4}} & =2^{y} \cdot 2^{2}
\end{aligned}
$$

Thus,

$$
\begin{aligned}
10^{y^{2}-4} & =2^{y+2} \\
\left(10^{y-2}\right)^{y+2} & =2^{y+2}
\end{aligned}
$$

Two positive numbers raised to the $(y+2)$-th power are equal:

1. if the bases are equal, or

2. if the exponent is 0.

From this,

$$
10^{y-2}=2, \quad \frac{10^{y}}{10^{2}}=2, \quad x_{1}=10^{y}=200
$$

and

$$
y+2=0, \quad y_{2}=-2, \quad x_{2}=10^{-2}=\frac{1}{100}
$$