Answer :

naǫ
[tex]\log_{(x-1)} 9=2[/tex]

the domain:
[tex]x-1 >0 \ \land \ x-1 \not=1 \\ x>1 \ \land \ x \not= 2 \\ x \in (1; 2) \cup (2;+\infty)[/tex]

the equation:
[tex]\log_{(x-1)}9=2 \\ (x-1)^2=9 \\ \sqrt{(x-1)^2}=\sqrt{9} \\ |x-1|=3 \\ x-1=3 \ \lor \ x-1=-3 \\ x=4 \ \lor \ x=-2[/tex]

4 is in the domain
-2 is not in the domain

The answer:
[tex]x=4[/tex]

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Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
[tex]x=\sqrt{5x+24} \\ -3=\sqrt{5 \times (-3)+24} \\ -3=\sqrt{-15+24} \\ -3=\sqrt{9} \\ -3=3[/tex]
It's not true so -3 isn't a solution to this equation.
konrad509
[tex]D:x-1>0 \wedge x\n-1\not=1\\ D:x>1 \wedge x\not=2\\ D:x\in(1,2)\cup(2,\infty)\\\\ \log_{x-1}9=2\\ (x-1)^2=9\\ x-1=3 \vee x-1=-3\\ x=4 \vee x=-2\\-2\not \in D\\ \boxed{ x=4}[/tex]

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