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ser-zykov [4K]
3 years ago
9

Find solution to logarithm. 

Mathematics
2 answers:
sveta [45]3 years ago
8 0
\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

MAXImum [283]3 years ago
7 0
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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