Question

Solve for x
logx+log(x-4)=2log5

Answer 1

$\log{x} + \log{(x-4)} = 2 \log{5}\\\\ \log{\big(x(x-4)\big)} = \log{5^2}\\\\ \log{(x^2-4x)} = \log{25} \\\\ x^2-4x = 25\\\\ x^2-4x-25=0 \\\\ x = \frac{4 \pm\sqrt{16+100}}{2} \\\\ x = \frac{4\pm2\sqrt{29}}{2} \\\\ x = 2 \pm \sqrt{29} \\\\ \text{But } x > 0 \implies x = 2 + \sqrt{29}$

Answer 2

$D:x>0\wedge x>4\\D:x>4\\\\ \log x+\log(x-4)=2\log5\\ \log x(x-4)=\log25\\ x(x-4)=25\\ x^2-4x-25=0\\ x^2-4x+4-29=0\\ (x-2)^2=29\\ x-2=\sqrt{29} \vee x-2=-\sqrt{29}\\ x=2+\sqrt{29} \vee x=2-\sqrt{29}\\ 2-\sqrt{29}\not>4\\ \Downarrow\\ \boxed{x=2+\sqrt{49}}$