Answer :
[tex]\log{x} + \log{(x-4)} = 2 \log{5}\\\\
\log{\big(x(x-4)\big)} = \log{5^2}\\\\
\log{(x^2-4x)} = \log{25} \\\\
x^2-4x = 25\\\\
x^2-4x-25=0 \\\\
x = \frac{4 \pm\sqrt{16+100}}{2} \\\\
x = \frac{4\pm2\sqrt{29}}{2} \\\\
x = 2 \pm \sqrt{29} \\\\
\text{But } x > 0 \implies x = 2 + \sqrt{29}[/tex]
[tex]D:x>0\wedge x>4\\D:x>4\\\\
\log x+\log(x-4)=2\log5\\
\log x(x-4)=\log25\\
x(x-4)=25\\
x^2-4x-25=0\\
x^2-4x+4-29=0\\
(x-2)^2=29\\
x-2=\sqrt{29} \vee x-2=-\sqrt{29}\\
x=2+\sqrt{29} \vee x=2-\sqrt{29}\\
2-\sqrt{29}\not>4\\
\Downarrow\\
\boxed{x=2+\sqrt{49}}[/tex]
