}

According to the definition of the root and the logarithm, the left side of the given inequality exists if and only if \(x-9>0\) and \(2 x-1>0\). This is true precisely for \(x>9\).

The given inequality is then equivalent to the following inequalities

\[
\begin{aligned}
\frac{1}{2} \lg (2 x-1)+\frac{1}{2} \lg (x-9) & >1 \\
\lg (2 x-1)(x-9) & >2 \\
\lg \left(2 x^{2}-19 x+9\right) & >\lg 100
\end{aligned}
\]

This inequality holds exactly when

\[
\left.2 x^{2}-19 x+9\right)>100 \quad \text { or } \quad x 2-\frac{19}{2} x-\frac{91}{2}>0
\]

Since \((x-13)\left(x+\frac{7}{2}\right)=x^{2}-\frac{19}{2} x-\frac{91}{2}\), the last inequality is equivalent to

\[
(x-13)\left(x+\frac{7}{2}\right)>0
\]

A product of two real numbers is positive if and only if either both factors are positive or both factors are negative. Therefore, (1) is satisfied if and only if either (2) \(x-13>0\) and \(x+\frac{7}{2}>0\) or (3) \(x-13<0\) and \(x+\frac{7}{2}<0\). Since \(x>9\), (2) implies \(x>13\) and vice versa; from (3) follows \(x<-\frac{7}{2}\), but since \(x>9\), only \(x>13\) is a solution to the given inequality; thus, the inequality is satisfied for all real numbers \(x>13\) and only for these.

Solutions to problems 1-5 of the third round in 1966 taken from [5]
\section*{Problem 5 - 061035}

Determine all real numbers \(x\) that satisfy the following inequality

\[
\frac{1}{2} \cdot \lg (2 x-1)+\lg \sqrt{x-9}>1
\]