Square both side of equation and simplify the equation.
[tex]\begin{gathered} (\sqrt[]{\log x-3})^2=(\log x-3)^2 \\ \log x-3=(\log x)^2-6\log x+9 \\ (\log x)^2-6\log x-\log x+9+3=0 \\ (\log x)^2-7\log x+12=0 \end{gathered}[/tex]
Assume log x = y. So equation is,
[tex]y^2-7y+12=0[/tex]
Simplify the equation to obtain the value of y.
[tex]\begin{gathered} y^2-7y+12=0 \\ y^2-4y-3y+12=0 \\ y(y-4)-3(y-4)=0 \\ (y-3)(y-4)=0 \\ y=3,4 \end{gathered}[/tex]
So the value of y is 3 or 4,
[tex]\begin{gathered} \log x=3 \\ x=e^3 \end{gathered}[/tex]
Or
[tex]\begin{gathered} \log x=4 \\ x=e^4 \end{gathered}[/tex]
