Question

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

Answer 1

`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=√(29) \vee x-2=-√(29)\\ x=2+√(29) \vee x=2-√(29)\\ 2-√(29)\\ot>4\\ \Downarrow\\ \boxed{x=2+√(49)}`

Answer 2

`\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 = (4 \pm√(16+100))/(2) \\\\ x = (4\pm2√(29))/(2) \\\\ x = 2 \pm √(29) \\\\ \text{But } x > 0 \implies x = 2 + √(29)`