\(\log_{(x-1)} 9=2\)
the domain:
\(x-1 >0 \ \land \ x-1
ot=1 \\
x>1 \ \land \ x
ot= 2 \\
x \in (1; 2) \cup (2;+\infty)\)
the equation:
\(\log_{(x-1)}9=2 \\
(x-1)^2=9 \\
\sqrt{(x-1)^2}=\sqrt{9} \\
|x-1|=3 \\
x-1=3 \ \lor \ x-1=-3 \\
x=4 \ \lor \ x=-2\)
4 is in the domain
-2 is not in the domain
The answer:
\(x=4\)
***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
\(x=\sqrt{5x+24} \\
-3=\sqrt{5 \times (-3)+24} \\
-3=\sqrt{-15+24} \\
-3=\sqrt{9} \\
-3=3\)
It's not true so -3 isn't a solution to this equation.
