Find the exact value solution of the equation: $$\log5x+\log(x-1)=2$$

$$\log5x+\log(x-1)=2\\ D:5x>0 \wedge x-1>0\\ D: x>0 \wedge x>1\\ D: x>1\\ \log5x(x-1)=2\\ 10^2=5x^2-5x\\ 5x^2-5x-100=0\\ x^2-x-20=0\\ x^2-5x+4x-20=0\\ x(x-5)+4(x-5)=0\\ (x+4)(x-5)=0\\ x=-4 \vee x=5\\ -4\not \in D\Rightarrow x=5$$

$$log_pa=b\ \ \ \Leftrightarrow\ \ \ p^b=a\\\\loga+logb=log(a\cdot b)\ \ \ \Rightarrow\ \ \ D:\ a>0\ \ \ and\ \ \ b>0\\ - \\log(5x)+log(x-1)=2\ \ \ \Rightarrow\ \ \ D:\ 5x>0\ \ \ and\ \ \ x-1>0\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0\ \ \ and\ \ \ \ \ \ \ \ x>1\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(1;+\infty)\\\\log[5x\cdot(x-1)]=2\ \ \ \Leftrightarrow\ \ \ 5x(x-1)=10^2$$

$$5x^2-5x=100\ /:5\\\\x^2-x-20=0\\\\x^2-5x+4x-20=0\\\\x(x-5)+4(x-5)=0\\\\(x-5)(x+4)=0\ \ \ \Leftrightarrow\ \ \ x-5=0\ \ \ or\ \ \ x+4=0\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5\ \ \ or\ \ \ \ \ \ \ \ x=-4\\\\5\in D;\ \ \ -4\notin D\\\\Ans. \ x=5$$

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