\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)

Recall, \( 1 = \log_{2}2\), so

\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)

= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)

\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)

\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)