The general term in the expansion of \((x + 2k)^7\) is given by:

\(T_r = \binom{7}{r} (x)^{7-r} (2k)^r\).

For the term in \(x^5\), set \(7-r = 5\), so \(r = 2\):

\(T_5 = \binom{7}{2} x^5 (2k)^2 = 21 \cdot 4k^2 x^5 = 84k^2 x^5\).

For the term in \(x^4\), set \(7-r = 4\), so \(r = 3\):

\(T_4 = \binom{7}{3} x^4 (2k)^3 = 35 \cdot 8k^3 x^4 = 280k^3 x^4\).

Since the coefficients of \(x^4\) and \(x^5\) are equal, we equate them:

\(84k^2 = 280k^3\).

Divide both sides by \(k^2\) (since \(k\) is non-zero):

\(84 = 280k\).

Solve for \(k\):

\(k = \frac{84}{280} = \frac{3}{10} = 0.3\).