Solve the equation: $\;$ $\log 5 + \log \left(x + 10\right) - 1 = \log \left(21x - 20\right) - \log \left(2x - 1\right)$
Given equation: $\;\;$ $\log 5 + \log \left(x + 10\right) - 1 = \log \left(21x - 20\right) - \log \left(2x - 1\right)$
i.e. $\;$ $\log_{10} 5 + \log_{10} \left(x + 10\right) - \log_{10} \left(21x - 20\right) + \log_{10} \left(2x - 1\right) = 1$
i.e. $\;$ $\log_{10} \left[\dfrac{5 \left(x + 10\right) \left(2x - 1\right)}{21x - 20}\right] = \log_{10} 10$
i.e. $\;$ $\dfrac{5 \left(2x^2 + 19x - 10\right)}{21x - 20} = 10$
i.e. $\;$ $2x^2 + 19x - 10 = 2 \left(21x - 20\right)$
i.e. $\;$ $2x^2 + 19x - 10 = 42x - 40$
i.e. $\;$ $2x^2 - 23x + 30 = 0$
i.e. $\;$ $\left(x - 10\right) \left(2x - 3\right) = 0$
i.e. $\;$ $x = 10$ $\;$ or $\;$ $x = \dfrac{3}{2}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{3}{2}, \; 10 \right\}$