Question

Solve the following equation:
$\frac{3\log x+19}{3\log x-1}=2\log x+1$

Answer 1

$\frac{3\log x+19}{3\log x-1}=2\log x+1\Rightarrow 3\log x+19=5(\log x{)}^2+\log x-1\Rightarrow {6\left(\log x\right)}^2-2\log x-20=0\Rightarrow 3(\log x{)}^2-\log x-10=0\Rightarrow \log x=\frac{1\pm \sqrt{121}}{6}\Rightarrow \log x=2,\frac{-5}{3}\Rightarrow x=100,\frac{-5}{3}$