Rozwiązanie:
a)
[tex]log_{\sqrt{6}}(x+1)+log_{\sqrt{6} }x=2\\D: x+1>0 \wedge x>0\\x>0\\log_{\sqrt{6}}((x+1)x )=2\\log_{\sqrt{6} }(x^{2}+x)=log_{\sqrt{6} }6\\x^{2}+x=6\\x^{2}+x-6=0\\\Delta=1-4*1*(-6)=25\\x_{1}=\frac{-1+5}{2} =2\\x_{2}=\frac{-1-5}{2}=-3 \notin D[/tex]
b)
[tex]log_{5}(x^{2}-6)=log_{5}(x-2)+1\\D: x^{2}-6>0 \wedge x-2>0\\(x-\sqrt{6} )(x+\sqrt{6} )>0 \wedge x>2\\x \in (-\infty,-\sqrt{6}) \cup (\sqrt{6},\infty) \wedge x>2\\x \in (\sqrt{6},\infty)\\log_{5}(x^{2}-6)-log_{5}(x-2)=1\\log_{5}(\frac{x^{2}-6}{x-2} ) =log_{5}5\\\frac{x^{2}-6}{x-2}=5\\x^{2}-6=5x-10\\x^{2}-5x+4=0\\\Delta=25-4*1*4=9\\x_{1}=\frac{5-3}{2} =1 \notin D\\x_{2}=\frac{5+3}{2}=4[/tex]
