Solve x in logarithm equationI am trying to solve x for 2 l o g...

Use the properties of logarithms
$k{\log }_{a}x={\log }_a(x^k)$
${\log }_{a}x-{\log }_{a}y={\log }_a\frac{x}{y}$
${\log }_{a}x=0⟺x=1$

Start with original problem:
$2{\log }_{10}(x-4)-{\log }_{10}4(x-1)=0$
A rule of logarithm: $2{\log }_{10}(x-4)={\log }_{10}(x-4{)}^2$
${\log }_{10}(x-4{)}^2-{\log }_{10}(4x-4)=0$
Another rule of logarithm: Express as the left side of the equation as a single product:
${\log }_{10}\frac{(x-4{)}^2}{4x-4}=0$
Convert to exponential form:
${10}^{{\log }_{10}\frac{(x-4{)}^2}{4x-4}}={10}^0$
or
$\frac{(x-4{)}^2}{4x-4}=1$
and when algebraically rearranging
$x^2-8x+16=4x-4$
or
$x^2-12x+20=0$
which is now a quadratic equation that you can solve.