Răspuns :
Aplici conditiile de existenta pt a determina domeniul de definitie al lui x
logₓ₋₂ 3
CE x-2 #1 ⇒ x# 3
x-2 >0 ⇒ x > 2
x apartine (2, infinit) /{3}
logₓ₋₁ (5-x)
CE x-1 # 1 ⇒ x # 2
x-1 > 0 ⇒ x > 1
5-x > 0 ⇒ -x > -5 ⇒ x < 5
x apartine (1,5) \{2}
[tex]\it log_{x-1} 5-x \Rightarrow \begin{cases} \it x-1\ne 1\ \d{s}i\ x-1>0 \Rightarrow x\in (1,\ \infty)\setminus\{2\}\\ \\ \it 5-x>0 \Rightarrow x<5\end{cases} \Rightarrow \\ \\ \\ \Rightarrow x\in(1,\ 5)\setminus\{2\}[/tex]
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[tex]\it log_{x-2} 3 \Rightarrow x-2\ne1\ \ \d{s}i\ x-2>0 \Rightarrow x\in(2,\ \ \infty)\setminus\{3\}[/tex]