$13 \cdot 46$ If $x$ is a positive number, and $\lg x \geqslant \lg 2+\frac{1}{2} \lg x$, then
(A) $x$ has no maximum or minimum value.
(B) The maximum value of $x$ is 1.
(C) The minimum value of $x$ is 1.
(D) The maximum value of $x$ is 4.
(E) The minimum value of $x$ is 4.
(9th American High School Mathematics Examination, 1958)
[Solution] From the given, we have $\lg x - \frac{1}{2} \lg x \geqslant \lg 2 \, (x > 0)$, which means $\lg \frac{x}{\sqrt{x}} \geqslant \lg 2$, or $\lg \sqrt{x} \geqslant \lg 2$.

Then $\sqrt{x} \geqslant 2$, yielding $\quad x \geqslant 4$. Therefore, the answer is $(E)$.