[tex]\log_{10} (3y+2) - 1 = \log_{10}(y-4)\\\\\implies \log_{10} (3y+2) - \log_{10} 10 = \log_{10} (y-4)\\\\\implies \log_{10} \left( \dfrac{3y+2}{10} \right) = \log_{10} (y-4)\\\\\implies \dfrac{3y+2}{10} = y-4\\\\\implies 3y+2 = 10y-40\\\\\implies 10y-3y = 40+2\\\\\implies 7y = 42\\\\\implies y = \dfrac{42}7\\\\\implies y = 6[/tex]
