To solve \(\int_{1}^{k} \frac{1}{2x-1} \, dx = 1\), we first find the indefinite integral:

\(\int \frac{1}{2x-1} \, dx = \frac{1}{2} \ln|2x-1| + C\).

Applying the limits from 1 to \(k\), we have:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln|2 \times 1 - 1| = 1\).

This simplifies to:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln 1 = 1\).

Since \(\ln 1 = 0\), we have:

\(\frac{1}{2} \ln|2k-1| = 1\).

Solving for \(k\), we get:

\(\ln|2k-1| = 2\).

\(|2k-1| = e^2\).

Assuming \(2k-1 > 0\), we have:

\(2k-1 = e^2\).

\(2k = e^2 + 1\).

\(k = \frac{1}{2}(e^2 + 1)\).